datavorous/entrance-exam-dataset huggingface dataset:
The oxidation number of phosphorus in ATP (adenosine triphosphate) is

Options:
A 4
B 3
C 5
D 2

The correct answer is:
5

Solution:
ATP has phosphate group, PO43-
So x+4(-2)=-3
or x=+5

Topic: Chemistry, Biomolecules, NEET
The oxidation number of sulphur in S8,S2F2,H2S respectively, are

Options:
A 0, +1 and – 2
B +2, +1 and – 2
C 0, +1 and + 2
D – 2, +1 and – 2

The correct answer is:
0, +1 and – 2

Solution:
Formula (i) oxidation state of element in its free state is zero.
(ii) sum of oxidation states of all atoms in compound is zero.
O.N. of S in S8=0;
O.N. of S in S2F2=+1;
O.N. of S in H2S=-2

Topic: Chemistry, Redox Reactions, NEET
The oxidation number of the underlined atom in the following species

(Identify the incorrect statement)

Options:
A Cu2O¯ is -1
B Cl¯O3- is +5
C KrCr2¯O7 is +6
D HAu¯Cl4 is +3

The correct answer is:
Cu2O¯ is -1

Solution:
Ox. state of "O" in Cu2O=-2
Ox. state of "Cl" in ClO3-=+5
Ox. state of "Cr" in K2Cr2O7=+6
Ox. state of "Au" in HAuCl4=+3
∴ (1) is incorrect

Topic: Chemistry, Redox Reactions, NEET, NEET 2020 (Phase 2)
The oxidation of benzene by V2O5 in the presence of air produces:

Options:
A Benzoic anhydride
B Maleic anhydride
C Benzoic acid
D Benzaldehyde

The correct answer is:
Maleic anhydride

Solution:

Topic: Chemistry, Hydrocarbons, NEET, NEET 2015 (Phase 2)
The oxidation of one NADH2 yields

Options:
A 18 ATP
B 6 ATP
C 3 ATP
D 2 ATP

The correct answer is:
3 ATP

Solution:
The oxidation of one NADH2 yields 3 ATP and one FADH2 yield 2 ATP molecules through oxidative phosphorylation method.

Topic: Biology, Respiration in Plants, NEET
The oxidation of SO2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if:

Options:
A Temperature is increased and pressure is kept constant
B Temperature is reduced and pressure is increased
C Both temperature and pressure are increased
D Both temperature and pressure are reduced

The correct answer is:
Temperature is reduced and pressure is increased

Solution:
SO 2 + 1 2 O 2 ⇌ SO 3 ; Δ H = - ve ;
Apply Le Chatelier principle.

Topic: Chemistry, Chemical Equilibrium, NEET
The oxidation potential of a hydrogen electrode at pH = 10 and PH2=1 atm

Options:
A 0.059 V
B 0.59 V
C 0.00 V
D 0.51 V

The correct answer is:
0.59 V

Solution:
EOP=EOPo-0.0591log[H+]PH2
So [H+]=10-10
PH2=1 atm, EOP=0.59 V

Topic: Chemistry, Electrochemistry, NEET
The oxidation reaction that happens in Daniel Cell is

Options:
A Zn →Zn+2+ 2e-
B Zn→Zn+2+ 4e-
C Cu+2+ 2e-→Cu
D Cu+2+ 4e-→Cu

The correct answer is:
Zn →Zn+2+ 2e-

Solution:
Correct Option is : 1
Zn →Zn+2+ 2e-

Topic: Chemistry, Redox Reactions, NEET
The oxidation state of $\mathrm{Cr}$ in $\mathrm{CrO}_5$ is

Options:
A -6
B +12
C +6
D +4

The correct answer is:
+6

Solution:
$\text {The structure of } \mathrm{CrO}_5 \text { is }$
Oxidation state of $\mathrm{Cr}$ is +6 due to the presence of two peroxide linkages, which can be calculated as. In $\mathrm{CrO}\left(\mathrm{O}_2\right)_2$, let the oxidation state of $\mathrm{Cr}$ be $\mathrm{x}$
$\begin{aligned}
\mathrm{x}+(-1) 4+(-2) & =0 \\
\mathrm{x}-6 & =0 \\
\mathrm{x} & =+6
\end{aligned}$

Topic: Chemistry, Redox Reactions, NEET, NEET 2019 (Odisha)
The oxidation state of molybdenum in its oxo complex species [Mo2O4(C2H4)2(NH3)]2- is

Options:
A 1
B 3
C 4
D 2

The correct answer is:
3

Solution:
[Mo2O4(C2H4)2(NH3)]2-
So 2x-8+0+0=-2
x=+3 .

Topic: Chemistry, Coordination Compounds, NEET
The oxidation state of oxide formed by Scandium is:

Options:
A -2
B +2
C 0
D +3

The correct answer is:
+3

Solution:
Correct Option is : 4
+3

Topic: Chemistry, d and f Block Elements, NEET
The oxidation state of platinum in Na[PtBrCl(NO2)(NH3)] is

Options:
A +2
B +4
C +6
D 0

The correct answer is:
+2

Solution:
Na[PtBrCl(NO2)(NH3)]
Let x = oxidation state of Pt.
So 1(+1)+x+1(-1)+1(-1)+1(-1)+1(0)=0
Or x = +2.

Topic: Chemistry, Redox Reactions, NEET
The oxidation state of Vanadium in V2O5is:

Options:
A +5
B +4
C -2
D -4

The correct answer is:
+5

Solution:
Correct Option is : 1
+5

Topic: Chemistry, d and f Block Elements, NEET
The oxidation states not shown by Mn in given reaction is :
$3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
A. +6
B. +2
C. +4
D. +7
E. +3
Choose the most appropriate answer from the options given below :

Options:
A D and E only
B B and D only
C A and B only
D B and E only

The correct answer is:
B and E only

Solution:
In the following reaction
$3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
So +2 and +3 oxidation state is not shown by Mn .

Topic: Chemistry, d and f Block Elements, NEET, NEET 2024 (Re-NEET)
The oxidation states of Cr in [Cr(H2O)6]Cl3,[Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are

Options:
A +3, 0 and +4
B +3, +4 and +6
C +3, +2 and +4
D +3, 0 and +6

The correct answer is:
+3, 0 and +6

Solution:
Compound
Oxidation state of Cr
[Cr(H2O)6]Cl3 +3
[Cr(C6H6)2] 0
K2[Cr(CN)2(O)2(O2)(NH3)] +6
[Potassium amminedicyanodioxoperoxochromate (VI)]

Topic: Chemistry, Redox Reactions, NEET
The oxidation states of P in the following compounds
H3PO3, PCl3, Ca3P2, Na3PO4, âPOF3 are respectively be

Options:
A + 3, + 3, - 3, + 5, + 5
B + 5, + 5, - 3, + 3, + 3
C + 4, + 3, - 3, + 4, + 5
D + 3, + 3, - 2, + 4, + 5

The correct answer is:
+ 3, + 3, - 3, + 5, + 5

Solution:
Let the oxidation state of phosphorus be x.
(i) H3PO3 (3 × + 1) + x + (3 × - 2) = 0
x = + 3
(ii) PCl3 x + (3 × - 1)
x = + 3
(iii) Ca3P2 (3 × + 2) + 2x
x = - 3
(iv) Na3PO4 (3 × 1) + x + (4 × - 2)
x = + 5
(v) POF3 x + (- 2) + (3 × - 1)
x = + 5

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The oxidation states of S-atoms in Caro's and Marshell's acids are

Options:
A +6,+6
B +6,+4
C +6,-6
D +4,+6

The correct answer is:
+6,+6

Solution:
Caro's acid =H2SO5
2+x+2(-1)+3(-2)=0
x=+6
Marshell's acid =H2S2O8
2+2x+2(-1)+6(-2)=0
x=+6
In both these acids there is peroxy linkage (O-O) so sulphur atom is in +6 oxidation states.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The oxidation states of sulphur in the anions $\mathrm{SO}_3^2, \mathrm{~S}_2 \mathrm{O}_4^{2-}$ and $\mathrm{S}_2 \mathrm{O}_6^{2-}$ follow of order

Options:
A $\mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-} < \mathrm{S}_2 \mathrm{O}_6{ }^{2-}$
B $\mathrm{SO}_3^{2-} < \mathrm{S}_2 \mathrm{O}_3^{2-} < \mathrm{S}_2 \mathrm{O}_6^{2-}$
C $\mathrm{S}_2 \mathrm{O}_4{ }^{2-} < \mathrm{S}_2 \mathrm{O}_6{ }^{2-} < \mathrm{SO}_3{ }^{2-}$
D $\mathrm{S}_2 \mathrm{O}_6{ }^{2-} < \mathrm{S}_2 \mathrm{O}_4{ }^{2-} < \mathrm{SO}_3{ }^{2-}$

The correct answer is:
$\mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-} < \mathrm{S}_2 \mathrm{O}_6{ }^{2-}$

Solution:
Oxidation state of sulphur is $\mathrm{SO}_3^{2-}$ i 4 Oxidation state of sulphur in $\mathrm{S}_2 \mathrm{O}_4^{2-}$ is 3 Oxidation state of sulphur in $\mathrm{S}_2 \mathrm{O}_6^{2-}$ is 5 The correct order of oxidation state is as follows:
$$
\mathrm{S}_2 \mathrm{O}_4{ }^{2-} < \mathrm{SO}_3{ }^{2-} < \mathrm{S}_2 \mathrm{O}_6{ }^{2-}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2003
The oxidative decarboxylation occurs in the step

Options:
A Oxalosuccinate â Î±-ketoglutarate
B Acetyl Co-A â Citrate
C Î± -ketoglutarate â Succinyl Co-A
D Oxaloacetate â Citrate

The correct answer is:
α -ketoglutarate → Succinyl Co-A

Solution:
Î±-ketoglutarate undergoes simultaneous dehydrogenation and decarboxylation in the presence of enzyme Î±-ketoglutarate dehydrogenase complex. This enzyme complex contains TPP, Lipoic Acid, Mg2+, and trans-succinylase. NAD+ and CoA are required. The products formed are 4-carbon compound succinyl CoA, NADH2, and CO2.

Topic: Biology, Respiration in Plants, NEET
The oxygen dissociation curve of haemoglobin is:

Options:
A Similar to the growth curve under restricted environmental resources
B Sigmoid shape
C Hyperbola
D Both A and B

The correct answer is:
Both A and B

Solution:
When percentage saturation of haemoglobin with O2 is plotted against the pO2, a sigmoid curve is obtained which is termed as oxygen dissociation curve.
A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity. A plot of N in relation to time (t) results in a sigmoid curve.

Topic: Biology, Breathing and Exchange of Gases, NEET
The oxygen dissociation curve of hemoglobin is

Options:
A S shape
B Sigmoid shape
C Hyperbola
D Both a and (b)

The correct answer is:
Both a and (b)

Solution:
The relationship between the pO2 and %saturation of haemoglobin when represented on a graph is termed as the oxygenâhaemoglobin dissociation curve and is sigmoid (s-shaped) in shape.

Topic: Biology, Breathing and Exchange of Gases, NEET
The oxygen evolved during photosynthesis comes from water molecules. Which one of the following pairs of elements is involved in this reaction?

Options:
A Manganese and potassium
B Magnesium and molybdenum
C Magnesium and chlorine
D Manganese and chlorine

The correct answer is:
Manganese and chlorine

Solution:
During non-cyclic photophosphorylation, the PS-II helps in the photolysis of water molecules which evolves oxygen as a by-product. For the photolysis of water, some cations (like Mg+2 and Mn+2) and anions (like Cl-) are involved. The proton generated in this process is used to make the proton gradient for ATP synthesis.

Manganese, chlorine, and calcium help in the photolysis of water during the light reaction of photosynthesis.

Topic: Biology, Mineral Nutrition, NEET, NEET 2015 (Phase 2)
The oxygen – haemoglobin dissociation curve will show a right shift in case of

Options:
A High pCO₂
B High pO₂
C Low pCO₂
D Less H+ concentration

The correct answer is:
High pCO₂

Solution:
Oxygen – haemoglobin dissociation curve will show a right shift in case of High pCO₂ as formation of carboxyhaemoglobin occurs.

Topic: Biology, Breathing and Exchange of Gases, NEET
The oxygenation activity of RuBisCO enzyme in photorespiration leads to the formation of

Options:
A 1 molecule of 3-C compound and 1 molecule of a 2-C compound
B 1 molecule of 6-C compound
C 1 molecule of 4-C compound and 1 molecule of 2-C compound
D 2 molecules of 3-C compound

The correct answer is:
1 molecule of 3-C compound and 1 molecule of a 2-C compound

Solution:
C2 cycle or glycolate pathway operates during the daytime in C3 plants & RuBisCO acts as oxygenase at higher concentration of O2 and low CO2 concentration in the C3-green cells. The light-dependent uptake of O2 & release of CO2 in C3 photosynthetic cells is called photorespiration.

RuBP + O2→Low CO2 concentration, High O2 concentration, High temperatureRuBisCO 1 PGA + 1 Phosphoglycolate(5C) (3C) (2C)

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2020 (Phase 1)
The pH value of decinormal solution of NH4OH, which is 20% ionized, is -

Options:
A 13.30
B 14.70
C 12.30
D 12.95

The correct answer is:
12.30

Solution:
OH â =0.1Ã 20 100
OHâ=0.1 Ã 0.2=2 Ã 10â2
pOH= -log[OH-] = 2- log2
pOH=1.7 , pH=14 â pOH = 14 - 1.7 = 12.3

Topic: Chemistry, Ionic Equilibrium, NEET
The pKa1 and pKa2 of an amino acid are 2.3 and 9.7 respectively. The isoelectric point of the amino acid is

Options:
A 7.4
B 3.5
C 12.0
D 6.0

The correct answer is:
6.0

Solution:
Isoelectric point =pKa1+pKa22=2.3+9.72=6

Topic: Chemistry, Biomolecules, NEET
The pKa of acetic acid and pKb of NH4OH are 4.76 and 4.75, respectively. Calculate the pH of ammonium acetate solution.

Options:
A 7.5
B 7.005
C 7.05
D 7.55

The correct answer is:
7.005

Solution:
Acetic acid and NH4OH are weak acid (WA) and weak base (WB) respectively. When they combine a salt of WA and WB i.e., ammonium acetate is formed. So by using hydrolysis equation for salts of WA and WB ,
pH=12(pKw+pKa-pKb)
= 1 2 14+4.76−4.75
=7+0.005=7.005 .

Topic: Chemistry, Ionic Equilibrium, NEET
The pKa values for the three ionisable groups X, Y and Z of glutamic acid are 4.3 9.7 and 2.2 respectively.

The isoelectric point for the amino acid is

Options:
A 7.00
B 3.25
C 4.95
D 5.95

The correct answer is:
3.25

Solution:
In acidic medium
Isoelectric point =4.3+2.22
=3.25

Topic: Chemistry, Biomolecules, NEET
The pKb value of ammonium hydroxide is 4.75. An aqueous solution of ammonium hydroxide is titrated with HCl. The pH of the solution at the point where half of the ammonium hydroxide has been neutralized will be

Options:
A 9.25
B 8.25
C 7.50
D 4.75

The correct answer is:
9.25

Solution:
pOH=pKb+log[Salt][Base]
when [Salt] = [Base]
So, pOH =pKb=4.75
So, pH = 14 – 4.75 = 9.25.

Topic: Chemistry, Ionic Equilibrium, NEET
The P-V diagram for an ideal gas in a piston cylinder assembly undergoing a thermodynamic process is shown in the figure. The process is

Options:
A adiabatic
B isochoric
C isobaric
D isothermal

The correct answer is:
isobaric

Solution:
In thermodynamics, an isobaric process is a type of thermodynamic process in which the pressure of the system stays constant: ΔP=0. The heat transferred to the system does work but also changes the internal energy of the system.
P=constant ⇒Isobaric process.

Topic: Physics, Thermodynamics, NEET, NEET 2020 (Phase 2)
The P-V diagram of a system undergoing thermodynamic transformation is shown in figure. The work done on the system in going from AâBâC is 50 J and 20 cal heat is given to the system. The change in internal energy between A and C is -

Options:
A 34 J
B 70Â J
C 84Â J
D 134Â J

The correct answer is:
34 J

Solution:
From first law of thermodynamics,
ÎU=ÎQ-ÎW
ÎU=20Ã4.2-54=34 J

Topic: Physics, Thermodynamics, NEET
The p50 of myoglobin is less than the p50 of adult hemoglobin, the affinity of adult hemoglobin for oxygen is

Options:
A is less than myoglobin
B is more than myoglobin
C is equal to myoglobin
D is nil

The correct answer is:
is less than myoglobin

Solution:
Myoglobin has higher affinity for oxygen and hence it retards the release of the oxygen. But, hemoglobin is a more efficient oxygen carrier since its affinity for oxygen is lower and because of this it correspondingly releases the oxygen easily to the correct areas of the body. The P50 represents the partial pressure at which hemoglobin is 50 percent saturated with oxygen. and normal P50 is 27 mm Hg.
P50 provides a quantifying relation of the hemoglobin's affinity (willingness to bond) with oxygen and this is reflected in the shifts of the dissociation curve.
Right shift - hemoglobin has decreased affinity, increased P50 - takes more oxygen to reach 50% (higher partial pressure to get 50% saturated).
Left shift - increased affinity, decreased P50 - less oxygen to reach 50% (less partial pressure to get 50% saturated).

Topic: Biology, Breathing and Exchange of Gases, NEET
The packaging of chromatin at a higher level requires an additional set of proteins that are collectively referred to as

Options:
A Small nuclear ribonucleoprotein (SnRNP)
B Nucleoplasmin protein
C Ribophorin protein
D Non histone chromosomal proteins

The correct answer is:
Non histone chromosomal proteins

Solution:
The nucleosome in chromatin are seen as 'beads on string' structure when viewed under electron microscope. It is further packaged to form Chromatin fibres that are further coiled and condensed at metaphase stage of cell division to form chromosome, this packaging at higher level requires an additional set of proteins collectively referred as NHC protein.

Topic: Biology, Molecular Basis of Inheritance, NEET
The packaging of chromatin at a higher level requires an additional set of proteins that are collectively referred to as

Options:
A Non-histone chromosomal proteins
B Small nuclear ribonucleoprotein (SnRNP)
C Nucleoplasmin protein
D Ribophorin protein

The correct answer is:
Non-histone chromosomal proteins

Solution:
The nucleosomes in chromatin are seen as 'beads on string' structure when viewed under electron microscope. It is further packaged to form chromatin fibres that are further coiled and condensed at metaphase stage of cell division to form chromosome. This packaging at higher level requires an additional set of proteins, collectively referred as NHC protein.

Topic: Biology, Molecular Basis of Inheritance, NEET
The packaging of chromatin at a higher level requires an additional set of proteins that collectively are referred to as

Options:
A Histones
B NHC proteins
C Tubulin
D Actin

The correct answer is:
NHC proteins

Solution:
The packaging of chromatin at a higher level requires an additional set of proteins that collectively are referred to as Non-histone Chromosomal (NHC) proteins.

Topic: Biology, Molecular Basis of Inheritance, NEET
The pair acids in which phosphorous atom has a formal oxidation state of +3 is

Options:
A Pyrophosphorous and hypophosphoric acids
B Orthophosphorous and hypophosphoric acids
C Pyrophosphorous and pyrophosphoric acids
D Orthophosphorous and pyrophosphorous acids

The correct answer is:
Orthophosphorous and pyrophosphorous acids

Solution:
Acid
Formula
Formal oxidation state of phosphorous

Pyrophosphorous acids
H4P2O5
+3

Pyrophosphoric acid
H4P2O7
+5

Orthophosphorous acid
H3PO3
+3

Hypophosphoric acid
H4P2O6
+4

Both pyrophosphorous and orhtophosphorous acids have +3 formal oxidation state.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The pair in which both species have same magnetic moment is -

Options:
A [Cr(H2O)6]2+,[CoCl4]2-
B [Cr(H2O)6]2+,[Fe(H2O)6]2+
C [Mn(H2O)6]2+,[Cr(H2O)6]2+
D [CoCl4]2-,[Fe(H2O)6]2+

The correct answer is:
[Cr(H2O)6]2+,[Fe(H2O)6]2+

Solution:
[Cr(H2O)6]2+ have Cr2+
Cr2+=3d4 4sÂ° (four unpaired electrons)
Fe2+=3d6 4sÂ° (in presence of weak ligand also have four unpaired electrons)

Topic: Chemistry, d and f Block Elements, NEET
The pair of compounds in which both the compounds give positive test with Tollen’s reagent is

Options:
A glucose and sucrose
B fructose and sucrose
C acetophenone and hexanal
D glucose and fructose

The correct answer is:
glucose and fructose

Solution:
Both aldoses (glucose) and fructose can reduce Tollen’s reagent. So option (D) is correct.
In other options only 1 can do it for example in option (a) glucose, in option (b) fructose and option (c) only hexanal. Fructose can reduce Tollen's reagent due to its enolisation.

Topic: Chemistry, Biomolecules, NEET
The pair of compounds that can exist together is:

Options:
A FeCl3,SnCl2
B HgCl2, SnCl2
C FeCl2,SnCl2
D FeCl3 KI

The correct answer is:
FeCl2,SnCl2

Solution:
FeCl2 and SnCl2 (both are reducing agent and have lower oxidation no.)

Topic: Chemistry, Redox Reactions, NEET, NEET 2014
The pair of compounds that can exist together is

Options:
A FeCl3, Kl
B FeCl3,SnCl2
C FeCl2,SnCl2
D HgCl2,SnCl2

The correct answer is:
FeCl2,SnCl2

Solution:
As FeCl2,SnCl2 both are reducing agents, so reaction is not possible between them. Hence, they can exist together.
In rest of other pairs redox reaction occurs so they cannot exist together.

Topic: Chemistry, Redox Reactions, NEET
The pair of d orbitals involved in octahedral structure having d2sp3 hybridization, are

Options:
A dxy,dyz
B dx2,dxz
C dx2-y2,dz2
D dxz,dx2-y2

The correct answer is:
dx2-y2,dz2

Solution:
In the formation of d2sp3 hybrid orbitals two (n-1)d orbitals of e.g., set [i.e., (n-1)dz2 and (n-1)dx2-y2 orbitals] one ns and three np [ npx,npy and npz ] orbitals combine together and form six d2sp3 hybrid orbitals.

Topic: Chemistry, Coordination Compounds, NEET
The pair of electron in the given carbanion, CH3C≡C⊖ is present in which of the following orbitals?

Options:
A 2p
B sp3
C sp2
D sp

The correct answer is:
sp

Solution:
CH3-C≡C⊖ : In the carbanion, the carbon is having 1 sigma bond, 2π bonds and 1 lone pair therefore C is sp hybridized and lone pair is present in hybridised orbital.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2016 (Phase 1)
The pair of ions having same electronic configuration is

Options:
A Cr3+,Fe3+
B Fe3+,Mn2+
C Fe3+,Co3+
D Sc3+,Cr3+

The correct answer is:
Fe3+,Mn2+

Solution:
Cr(24):3d54s1,Cr3+:3d3
Fe(26):3d64s2,Fe3+:3d5
Mn(25):3d54s2,Mn2+:3d5
Co(27):3d74s2,Co3+:3d6
Sc(21):3d14s2,Sc3+:3d0
Thus, Fe3+ and Mn+ have same electronic configuration.

Topic: Chemistry, Structure of Atom, NEET
The pair of lanthanoid ions which are diamagnetic is

Options:
A $\mathrm{Ce}^{3+}$ and $\mathrm{Eu}^{2+}$
B $\mathrm{Gd}^{3+}$ and $\mathrm{Eu}^{3+}$
C $\mathrm{Pm}^{3+}$ and $\mathrm{Sm}^{3+}$
D $\mathrm{Ce}^{4+}$ and $\mathrm{Yb}^{2+}$

The correct answer is:
$\mathrm{Ce}^{4+}$ and $\mathrm{Yb}^{2+}$

Solution:
Magnetic moment $\mu=\sqrt{n(n+2)}$
$\mathrm{n} \rightarrow$ number of unpaired electron
Hence $\mathrm{Ce}^{4+}$ and $\mathrm{Yb}^{2+}$ are only diamagnetic.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2024
The pair of quantities having same dimensions is:

Options:
A Impulse and Surface Tension
B Angular momentum and Work
C Work and Torque
D Young's modulus and Energy

The correct answer is:
Work and Torque

Solution:
(1)
$$
\begin{aligned}
\text { Torque }=\text { F.S. } & =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
correct option is (3)
(2) Angular moment $=$ M.I. $\times$ angular velocity
$$
\begin{aligned}
=\left[\mathrm{ML}^2\right]\left[\mathrm{T}^{-1}\right] & =\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\
\text { Work }=F . S . & =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
(3) Work $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$ from
Related Theory
(1)
$$
\begin{aligned}
& \text { Impluse }=\text { force } \times \text { time } \\
&=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\
& \text { Surface tension }=\frac{\text { Force }}{\text { length }}= \\
& \frac{\left[M L T^{-2}\right]}{[L]}=\left[M T^{-1}\right]
\end{aligned}
$$

Topic: Physics, Units and Dimensions, NEET, NEET 2013 (Karnataka)
The pair of species having identical shapes for molecules of both species is

Options:
A CF4,Â SF4
B XeF2, CO2
C BF3,Â PCl3
D PF5,Â IF5

The correct answer is:
XeF2, CO2

Solution:
In CO2, C-atom is sp-hybridised, thus it has linear structure. In XeF2, Xe is sp3d hybridised with three lone-pairs of electrons on equatorial positions. This mtextnimtextses repulsion, hence it has also linear structure.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The pair of species that has the same bond order in the following is:

Options:
A $\mathrm{CO}, \mathrm{NO}^{+}$
B $\mathrm{NO}^{-}, \mathrm{CN}^{-}$
C $\mathrm{O}_2, \mathrm{~N}_2$
D $\mathrm{O}_2, \mathrm{~B}_2$

The correct answer is:
$\mathrm{CO}, \mathrm{NO}^{+}$

Solution:
Both $\mathrm{CO}$ and $\mathrm{NO}^{+}$have 14 electrons. The electronic configuration is $\sigma 1 s^2$ $\sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p^2 z \pi 2 p_z^2=\pi 2 p_y^2$ Bond order $=\frac{10-4}{2}=3$

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2013 (Karnataka)
The pair of species with the same bond order is

Options:
A $\mathrm{O}_2^{2-}, \mathrm{B}_2$
B $\mathrm{O}_2^{+}, \mathrm{NO}^{+}$
C $\mathrm{NO}, \mathrm{CO}$
D $\mathrm{N}_2, \mathrm{O}_2$

The correct answer is:
$\mathrm{O}_2^{2-}, \mathrm{B}_2$

Solution:
Bond order of \(\mathrm{O}_2{ }^{2-}\) :
Electronic : \(\sigma 1 \mathrm{~s}^2 \sigma^{\star}\left(\mathrm{s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{P}_{\mathrm{z}}{ }^2\left(\pi 2 \mathrm{P}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{P}_{\mathrm{y}}{ }^2\right)\right.\)
\(\left(\pi 2 \mathrm{P}_{\mathrm{x}}^2=\pi 2^{\star} \mathrm{P}_{\mathrm{y}}^2\right) \sigma^{\star} 2 \mathrm{pz}{ }^{\circ}\)
Bond order \(=\frac{\text { Bonding }- \text { Antibonding electrons }}{2}=\frac{10-8}{2}=1\)
Bond order of \(B_2\) :
\(\sigma 1 s^2 \sigma_{1 s}{ }^2 2 \sigma_{2 s} 2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2\)
Bond order \(=\frac{6-4}{2}=1\)
\(\therefore \mathrm{O}_2{ }^{2-}\) and \(\mathrm{B}_2\) has same bond order 1.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2012 (Screening)