Spreadsheet: 0af0f2ba-c94c-4c9e-a310-6753b4ad7ae3
Sheet: Sheet1

Unnamed: 0 | Unnamed: 1 | Unnamed: 2 | 1st name | short name | surname | e-mail | Unnamed: 7 | Unnamed: 8 | Unnamed: 9 | Unnamed: 10 | Unnamed: 11 | Unnamed: 12 | Unnamed: 13 | Unnamed: 14 | Unnamed: 15 | Unnamed: 16 | Unnamed: 17 | Unnamed: 18 | Unnamed: 19 | Unnamed: 20 | Unnamed: 21 | Mark | Feedback 1 | Feedback 2 | Feedback 3 | Feedback 4 | Feedback 5 | Feedback 6 | Feedback 7 | Feedback 8 | Feedback 9 | Feedback 10 | Feedback 11 | Feedback 12 | Feedback 13 | Feedback 14 | Feedback 15 | Unnamed: 38 | Unnamed: 39 | pathname | Overall feedback
1352FD | A | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 5 | 5 | 1 | 4 | 5 | 1 | 1 | 3 | 4 | 1 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | E - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA1.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135257 | A | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 2 | 5 | 1 | 2 | 1 | 4 | 4 | 3 | 4 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | E - incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA2.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
1352C1 | A | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 5 | 5 | 4 | 1 | 4 | 2 | 1 | 1 | 4 | 4 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | E - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper [NAME] | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
0F8981 | A | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 5 | 4 | 3 | 1 | 2 | 1 | 5 | 4 | 3 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | E - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | E - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA4.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
135258 | A | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 4 | 2 | 3 | 1 | 4 | 2 | 1 | 1 | 4 | 4 | 4 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsA5.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13536F | A | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 5 | 2 | 1 | 1 | 4 | 2 | 6 | 1 | 3 | 5 | 6 | 6 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | No answer given: B is correct as this is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | No answer given: D is correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | No answer given: E is correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsA6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1352C4 | A | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 3 | 4 | 5 | 1 | 3 | 3 | 1 | 5 | 4 | 1 | 4 | 4 | 4 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | D - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsA7.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
0F88D6 | A | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 1 | 5 | 1 | 2 | 2 | 4 | 4 | 4 | 5 | 12 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | E - incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsA8.doc | Well done on scoring such a high mark. Keep up the good work!
0F897B | B | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 4 | 3 | 1 | 2 | 2 | 1 | 4 | 3 | 5 | 13 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB1.doc | Well done on scoring such a high mark. Keep up the good work!
0F88DE | B | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 5 | 2 | 1 | 2 | 2 | 1 | 4 | 4 | 3 | 13 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | B - incorrect. You have to choose the smallest cell possible within the highest symmetry. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | C - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsB2.doc | Well done on scoring such a high mark. Keep up the good work!
1352F3 | B | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsB3.doc
0F891E | B | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 3 | 2 | 1 | 3 | 1 | 5 | 1 | 2 | 4 | 2 | 4 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB4.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13536D | B | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 1 | 1 | 1 | 4 | 3 | 2 | 5 | 3 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 13 | A - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsB5.doc | Well done on scoring such a high mark. Keep up the good work!
0F896C | B | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 3 | 4 | 1 | 2 | 3 | 5 | 4 | 5 | 3 | 5 | 4 | 4 | 5 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | E - incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | E - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
0F8967 | B | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 4 | 1 | 4 | 4 | 5 | 2 | 2 | 1 | 4 | 5 | 5 | 4 | 5 | 1 | 2 | 3 | D - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | E - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | B - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsB7.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
1352A0 | B | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 3 | 1 | 5 | 4 | 3 | 2 | 2 | 3 | 5 | 5 | 2 | 2 | 3 | 3 | 1 | 7 | C - you got the charge right, but remember that Cl counts as 1- charge and O counts as 2- charge, hence Vanadium must have the oxidation state (IV). | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | E - incorrect as hexagonal cells have a = b ≠ c, α = β = 90° and γ ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsB8.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1353AD | C | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsC1.doc
135254 | C | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 1 | 4 | 1 | 3 | 1 | 1 | 5 | 1 | 1 | 2 | 4 | 1 | 5 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC2.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
0F8887 | C | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 5 | 1 | 1 | 2 | 2 | 1 | 3 | 1 | 4 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsC3.doc | Well done on scoring such a high mark. Keep up the good work!
0F8962 | C | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 4 | 3 | 3 | 3 | 4 | 2 | 1 | 1 | 4 | 4 | 1 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | C - incorrect as trans isomer does not have optical isomers | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC4.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13524A | C | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 2 | 5 | 3 | 1 | 5 | 1 | 1 | 2 | 4 | 4 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsC5.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
135249 | C | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 3 | 4 | 1 | 3 | 1 | 1 | 1 | 1 | 1 | 1 | 2 | 4 | 4 | 6 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | C - incorrect as trans isomer does not have optical isomers | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | A - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1352F8 | C | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC7.doc
13FBD2 | C | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 2 | 2 | 1 | 3 | 3 | 2 | 3 | 1 | 4 | 4 | 4 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | B - incorrect as there's no c3 axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsC8.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
1352FE | D | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 2 | 4 | 2 | 1 | 1 | 1 | 1 | 5 | 2 | 1 | 4 | 4 | 4 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | B - incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F’s – this cannot give rise to 2 triplets. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | B - incorrect as there's no c3 axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD1.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135361 | D | 2 | [NAME] | Jo | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD2.doc
0F8989 | D | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 3 | 1 | 3 | 1 | 2 | 1 | 4 | 5 | 3 | 5 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | C - incorrect as trans isomer does not have optical isomers | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsD3.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1353BD | D | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 3 | 1 | 2 | 1 | 3 | 3 | 5 | 3 | 3 | 4 | 3 | 1 | 6 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | C - incorrect as Tc is a 4d element. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | C - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD4.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1353BE | D | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsD5.doc
0F88D3 | D | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 2 | 5 | 1 | 3 | 1 | 5 | 1 | 4 | 5 | 2 | 1 | 4 | 3 | 5 | 8 | B - correct, well done! | B - incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsD6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1352C8 | D | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsD7.doc
13529E | D | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 1 | 3 | 3 | 2 | 4 | 1 | 1 | 4 | 4 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | D - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | A - incorrect. See calculation at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsD8.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
1352C5 | E | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 3 | 3 | 1 | 2 | 3 | 1 | 2 | 1 | 5 | 4 | 2 | 1 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | C - incorrect as Tc is a 4d element. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | E - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsE1.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
0F8845 | E | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 3 | 4 | 3 | 2 | 1 | 1 | 4 | 3 | 4 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | D - incorrect. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsE2.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13524D | E | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 3 | 2 | 2 | 1 | 3 | 2 | 3 | 3 | 4 | 4 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | C - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsE3.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13FBDC | E | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 2 | 5 | 3 | 1 | 2 | 3 | 1 | 5 | 4 | 4 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsE4.doc | Well done on scoring such a high mark. Keep up the good work!
0F88BC | E | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 5 | 2 | 1 | 3 | 1 | 1 | 1 | 1 | 5 | 4 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | A - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsE5.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13537E | E | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 3 | 5 | 1 | 3 | 5 | 1 | 1 | 2 | 1 | 1 | 6 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | C - incorrect as trans isomer does not have optical isomers | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsE6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135324 | E | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 2 | 5 | 4 | 2 | 1 | 1 | 4 | 4 | 4 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | E - incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsE7.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13524B | E | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 1 | 3 | 1 | 1 | 3 | 2 | 1 | 1 | 4 | 4 | 4 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | C - incorrect as trans isomer does not have optical isomers | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsE8.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13532A | F | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 2 | 3 | 2 | 3 | 2 | 4 | 2 | 4 | 1 | 5 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | B - incorrect. You have to choose the smallest cell possible within the highest symmetry. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | D - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF1.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135296 | F | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 2 | 3 | 3 | 1 | 3 | 2 | 1 | 4 | 4 | 5 | 12 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | C - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF2.doc | Well done on scoring such a high mark. Keep up the good work!
135299 | F | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 5 | 2 | 2 | 4 | 3 | 1 | 2 | 5 | 1 | 4 | 4 | 4 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | E - incorrect: Fe is a 3d element so OK there, but [NAME] is in Gp 8, so Fe(IV) is d4. | B - incorrect as there's no c3 axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | E - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsF3.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
0F895D | F | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 3 | 4 | 1 | 1 | 4 | 1 | 1 | 4 | 4 | 3 | 3 | 5 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | D - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | A - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | C - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF4.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
0F8932 | F | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 2 | 4 | 4 | 2 | 5 | 3 | 1 | 4 | 5 | 3 | 4 | 4 | 2 | 1 | 3 | B - correct, well done! | B - incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | B - incorrect as there's no c3 axis. | E - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsF5.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
0F8837 | F | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 4 | 3 | 3 | 3 | 4 | 1 | 1 | 1 | 4 | 4 | 1 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | C - incorrect as trans isomer does not have optical isomers | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | A - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsF6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135349 | F | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 4 | 2 | 3 | 3 | 4 | 2 | 1 | 3 | 4 | 1 | 2 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | C - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | B - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsF7.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
[NAME] | F | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 1 | 1 | 4 | 1 | 4 | 1 | 2 | 3 | 4 | 2 | 1 | 1 | 4 | 4 | 5 | 7 | A - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsF8.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135297 | G | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 1 | 1 | 3 | 1 | 2 | 2 | 1 | 4 | 2 | 5 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG1.doc | Well done on scoring such a high mark. Keep up the good work!
135371 | G | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 5 | 1 | 4 | 1 | 3 | 1 | 2 | 1 | 4 | 5 | 1 | 1 | 4 | 4 | 4 | 5 | E - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsG2.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
13FBD4 | G | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG3.doc
135250 | G | 4 | [NAME] | Jo | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG4.doc
135377 | G | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 4 | 1 | 3 | 3 | 4 | 5 | 1 | 2 | 2 | 3 | 3 | 4 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | C - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsG5.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
13FBD5 | G | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 3 | 2 | 4 | 3 | 1 | 5 | 1 | 1 | 4 | 3 | 5 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG6.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
135253 | G | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 1 | 1 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 13 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsG7.doc | Well done on scoring such a high mark. Keep up the good work!
1352F7 | G | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 1 | 1 | 2 | 1 | 5 | 5 | 1 | 3 | 4 | 2 | 1 | 1 | 5 | 1 | 5 | 5 | A - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | B - incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F’s – this cannot give rise to 2 triplets. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | E - incorrect - you've spotted way too many isomers. Make sure sure you understand why B is the correct answer. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsG8.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
135344 | H | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 3 | 2 | 1 | 1 | 4 | 2 | 2 | 1 | 2 | 3 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH1.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13FBE0 | H | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 3 | 5 | 4 | 4 | 2 | 1 | 3 | 1 | 2 | 3 | 4 | 4 | 4 | 4 | 9 | B - correct, well done! | C - incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH2.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13FBE5 | H | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 2 | 1 | 3 | 5 | 2 | 2 | 1 | 4 | 4 | 3 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | E - incorrect as hexagonal cells have a = b ≠ c, α = β = 90° and γ ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | C - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH3.doc | Well done on scoring such a high mark. Keep up the good work!
1353BF | H | 4 | [NAME] | Jo | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsH4.doc
0F8871 | H | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 5 | 1 | 3 | 4 | 1 | 2 | 1 | 1 | 4 | 4 | 5 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | E - incorrect as although there is a c2 axis, a tetrahedron cannot have a horizontal mirror plane. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | D - incorrect. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsH5.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
0F888F | H | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 4 | 2 | 2 | 3 | 1 | 3 | 4 | 4 | 4 | 4 | 1 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | C - incorrect as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | D - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsH6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13524C | H | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 2 | 5 | 3 | 4 | 2 | 4 | 1 | 2 | 4 | 5 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | D - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is half of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsH7.doc | Well done on scoring such a high mark. Keep up the good work!
0F8833 | H | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 2 | 3 | 5 | 1 | 1 | 2 | 1 | 1 | 4 | 4 | 5 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | B - incorrect as there's no c3 axis. | C - incorrect as trans isomer does not have optical isomers | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsH8.doc | Well done on scoring such a high mark. Keep up the good work!
[NAME] | I | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 4 | 3 | 2 | 3 | 3 | 2 | 1 | 3 | 4 | 4 | 1 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | C - incorrect as trans isomer does not have optical isomers | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | C - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsI1.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13534E | I | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 2 | 4 | 3 | 2 | 3 | 1 | 4 | 2 | 2 | 1 | 4 | 4 | 5 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | B - incorrect since although there are 2 F environments, there is only 1 F in one environment (trans NMe3) and 2 mutually trans F’s – this cannot give rise to 2 triplets. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI2.doc | Well done on scoring such a high mark. Keep up the good work!
0F89AC | I | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 3 | 1 | 1 | 3 | 1 | 2 | 1 | 1 | 4 | 4 | 5 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsI3.doc | Well done on scoring such a high mark. Keep up the good work!
0F899C | I | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 1 | 4 | 3 | 2 | 1 | 1 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 12 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsI4.doc | Well done on scoring such a high mark. Keep up the good work!
0F8847 | I | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 1 | 3 | 3 | 4 | 2 | 1 | 2 | 4 | 2 | 1 | 8 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsI5.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
0F8849 | I | 6 | [NAME] | Jo | Bloggs | [EMAIL] | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI6.doc
1353BB | I | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 3 | 1 | 1 | 1 | 2 | 2 | 1 | 4 | 4 | 4 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | C - incorrect as trans isomer does not have optical isomers | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsI7.doc | Well done on scoring such a high mark. Keep up the good work!
13FC0D | I | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 5 | 2 | 1 | 1 | 3 | 1 | 1 | 1 | 4 | 5 | 1 | 1 | 2 | 2 | 1 | 2 | E - this is incorrect since the complex is neutral overall. | B - incorrect as the maximum possible oxidation state of a TM ion is the Group number, and this corresponds to complete removal of the d electrons. | A - incorrect as there are 2 F environments, hence must be 2 resonances. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsI8.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
0F8994 | J | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 3 | 1 | 4 | 1 | 3 | 5 | 4 | 2 | 1 | 1 | 3 | 3 | 4 | 4 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | D - incorrect as there is no c3 rotation axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | E - incorrect. Choose the highest symmetry cell even if it is much larger than one of lower symmetry. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ1.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
13524F | J | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 4 | 3 | 1 | 5 | 2 | 4 | 4 | 4 | 3 | 11 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | D - incorrect as NEt3 (an amine) is a sigma donor ligand which favours medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | C - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsJ2.doc | Well done on scoring such a high mark. Keep up the good work!
13FBE6 | J | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 3 | 4 | 1 | 3 | 2 | 1 | 4 | 5 | 1 | 3 | 4 | 2 | 4 | 4 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | C - incorrect as trans isomer does not have optical isomers | B - incorrect as Cl- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | C - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ3.doc | Not too good! The pass mark was 6, so make sure you take notice of the feedback below, and ask for help if you need it. You should make sure you do some work over the Easter vacation to get up to scratch with this material.
13FBD9 | J | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 4 | 1 | 3 | 3 | 4 | 5 | 1 | 2 | 3 | 2 | 1 | 5 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | D - incorrect as there is no c3 rotation axis. | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | B - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | C - incorrect. See calculation at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ4.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
13FBDE | J | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 5 | 3 | 1 | 5 | 3 | 4 | 1 | 3 | 4 | 9 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | C - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is halfway along one edge. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | A - incorrect. See calculation at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ5.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13FC52 | J | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 3 | 4 | 1 | 2 | 3 | 3 | 1 | 2 | 1 | 1 | 5 | 4 | 5 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | A - incorrect as a tetrahedral ion cannot have a c4 rotation axis. | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ6.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
1353BC | J | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 5 | 3 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 15 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ7.doc | Excellent effort - you scored full marks. Keep up the good work!
0F883B | J | 8 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 4 | 4 | 3 | 1 | 3 | 1 | 3 | 5 | 1 | 1 | 4 | 2 | 1 | 6 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | B - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | A - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsJ8.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135330 | K | 1 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 1 | 3 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 14 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] [NAME] | Well done on scoring such a high mark. Keep up the good work!
0F8974 | K | 2 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 1 | 1 | 3 | 1 | 2 | 1 | 1 | 4 | 3 | 2 | 10 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | C - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  Here the labels are swapped around. | B - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsK2.doc | You've done well - this is a 2:1, but with a bit more effort to sort out the parts you were not so clear about you can do even better!
13FBD3 | K | 3 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 5 | 1 | 3 | 3 | 3 | 3 | 3 | 5 | 1 | 1 | 4 | 1 | 4 | 7 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | A - incorrect: Co is a 3d element, so OK there, but [NAME] is in Gp 9, so Co(IV) is d5 | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | C - incorrect as trans isomer does not have optical isomers | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | A - incorrect as the tetrahedral holes are always 3/8 of a unit cell above and below every atom.  [NAME] and ¾ are the heights of octahedral holes. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsK3.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
135369 | K | 4 | [NAME] | Jo | Bloggs | [EMAIL] | 2 | 1 | 5 | 4 | 3 | 2 | 5 | 3 | 1 | 2 | 2 | 1 | 4 | 4 | 5 | 15 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | E - correct as CO is a sigma donor/ pi acceptor – the latter part of the bonding removes electron density from the M, so stabilises species with high electron density (low ox states). | C - correct. First choose the highest symmetry then the smallest cell within it. | A - correct as monoclinic cells have a ≠ b ≠ c, α = γ = 90°and β ≠ 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | B - correct. This is a quarter of the way from one corner to the opposite corner across the cube. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK4.doc | Excellent effort - you scored full marks. Keep up the good work!
1353AB | K | 5 | [NAME] | [NAME] | Bloggs | [EMAIL] | 1 | 1 | 5 | 3 | 3 | 2 | 3 | 1 | 4 | 2 | 1 | 1 | 5 | 4 | 5 | 8 | A - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | E - correct: 2 F environments – the unique F couples to the other 2 mutually trans F’s to give a triplet, while the 2 mutually trans F’s resonate together but couple to the one unique F to give a doublet. | C - incorrect as Tc is a 4d element. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | D - incorrect as orthorhombic cells have a ≠ b ≠ c and α = β = γ = 90°. | B - correct as there are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | E - incorrect as this is the packing density of a face centred cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | E - correct as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK5.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
13FC53 | K | 6 | [NAME] | Jo | Bloggs | [EMAIL] | 5 | 1 | 4 | 4 | 3 | 2 | 3 | 3 | 2 | 5 | 1 | 4 | 4 | 4 | 4 | 7 | E - this is incorrect since the complex is neutral overall. | A - this is correct since [NAME] is in Group 6 - well done. | D - incorrect: there will be 2 resonances, but these are independent (a doublet of triplets means the nucleus you are observing is coupling to 2 different types of nuclei – one in one environment and two in another environment – ie sequential couplings). | D - correct as Mn is a 3d element and Mn is in Gp 7, so Mn(IV) is d3. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | B - correct as there are trans and cis geometric isomers, and the cis form also has optical isomers by virtue of the chelating ligand backbone. | C - incorrect as H2O is a sigma donor ligand which stabilises medium (usually +2, +3) oxidation states. | C - correct. First choose the highest symmetry then the smallest cell within it. | B - incorrect as cubic cells have a = b = c and α = β = γ = 90°. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | D - incorrect as the ratio of spheres in the close packed array to octahedral holes is 1:1. | D - correct, as shown at the end of lecture 2. | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsK6.doc | You've passed, so well done, but make sure you think carefully about what you can do to improve.
1353C0 | K | 7 | [NAME] | [NAME] | Bloggs | [EMAIL] | 2 | 1 | 3 | 3 | 3 | 1 | 1 | 1 | 3 | 5 | 1 | 1 | 2 | 4 | 4 | 5 | B - correct, well done! | A - this is correct since [NAME] is in Group 6 - well done. | C - incorrect as there are 2 F environments, hence must be 2 resonances. | C - incorrect as Tc is a 4d element. | C - correct: 2 pairs of ligands on the tetrahedral metal ion means there is a c2 rotation axis, and there is also a mirror plane bisecting the M-L axes | A - incorrect as you have probably spotted either the cis/trans geometric isomers or the optical isomers for the cis form, but need to see both! | A - incorrect as O2- is a strong sigma donor/pi donor ligand which is best for stabilising high ox states (both components pushing electron density onto the metal ion). | A - incorrect. Symmetry is the important point, choose the highest symmetry possible then the smallest cell within it. | C - incorrect as triclinic cells have a ≠ b ≠ c and α ≠ β ≠ γ. | E - incorrect as you have just counted the number of spheres!  There are 4 large spheres (8×1/8 + 6×½) and 8 small spheres (all inside the cell). | A - incorrect. The answer is a quarter of the way from one corner to the opposite corner across the cube, this is a quarter of the diagonal across one face. | A - correct as the ratio of spheres in the close packed array to octahedral holes is 1:1. | B - incorrect. This is the packing density of a primitive cubic cell (see calculation at the end of lecture 2). | D - correct as the tetrahedral holes are always 3/8 of a unit cell above and below every atom. | D - incorrect as CsCl is formed from [NAME] occupying cubic holes in a primitive cubic lattice. | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\BloggsK7.doc | You were close to passing, but there's definitely work to be done! Ask for help if you need it and make sure you read the sections in the textbook on the questions you got wrong
135343 | K | 8 | [NAME] | Jo | Bloggs | [EMAIL] |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | 0 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss [NAME] feedback\Reports\ | .doc | C:\Southampton\Curriculum and teaching\0708\CHEM1006\1st in calss test\Zapper feedback\Reports\BloggsK8.doc