datavorous/entrance-exam-dataset huggingface dataset:
Among the following complexes (K - P), which of the following are diamagnetic?

K3[FeCN)6K,[CoNH3)6Cl3L,

Na3[Cooxalate)3 M,[NiH2O)6Cl2N,

K2[PtCN)4 O and [ZnH2O)6 (NO3)2P

Options:
A K, M, O, P
B K, L, M, N
C L, M, N, O
D L, M, O, P

The correct answer is:
L, M, O, P

Solution:
Following compounds are diamagnetic as in them metal ion has no unpaired electron.
L: [CoNH3)6Cl3
M: Na3[CoOx)3
O: K2[PtCN)4
P: [ZnH2O)6 (NO3)2
In L, M, O inner d-electrons get paired while in P Zn has already 3d10 configuration.
In K and N there are 1 and 2 unpaired electrons respectively so they are paramagnetic.

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following compound which can be dehydrated very easily is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Among the following compound which can be dehydrated very easily is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2004
Among the following compounds
2,3 - dimethylhexane is

Options:
A $i$
B ii
C iii
D iv

The correct answer is:
ii

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Among the following compounds, geometrical isomerism is exhibited by :

Options:
A
B
C
D

The correct answer is:

Solution:
Geometrical isomerism arises due to
(1) The presence of a restricted rotation (double bond or a ring structure).
(2) Two different groups should be attached to any two carbon atoms of restricted rotation.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Among the following compounds most basic amino acid is:

Options:
A Lysine
B Asparagine
C Serine
D Histidine

The correct answer is:
Lysine

Solution:
Lysine is the most basic amino acid among the following. Their pI values are as follows:
Lysine: 9.8
Histidine: 7.6
Serine: 5.7
Asparagine: 5.4

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Among the following compounds, the compound which shows the highest dipole-moment is

Options:
A II
B IV
C III
D I

The correct answer is:
I

Solution:
The compound (I) will show the highest dipole moment. It will have maximum charge separation. The dipole moment is the product of the magnitude of charge and distance between two charges.

The quasi aromatic ring is aromatic due to charge delocalization.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2017 (09 Apr Online)
Among the following compounds, the increasing order of their basic strength is:
(iii)
(IV)

Options:
A \( (\mathrm{II}) < (\mathrm{I}) < (\mathrm{III}) < (\mathrm{IV}) \)
B \( (\mathrm{II}) < (\mathrm{I}) < (\mathrm{IV}) < (\mathrm{III}) \)
C \( (\mathrm{I}) < (\mathrm{II}) < (\mathrm{IV}) < (\mathrm{III}) \)
D \( (\mathrm{I}) < (\mathrm{II}) < (\mathrm{III}) < (\mathrm{IV}) \)

The correct answer is:
\( (\mathrm{II}) < (\mathrm{I}) < (\mathrm{III}) < (\mathrm{IV}) \)

Solution:

Topic: Chemistry, Amines, JEE Main
Among the following compounds, the increasing order of their basic strength is:

Options:
A II<I<III<IV
B II<I<IV<III
C I<II<IV<III
D I<II<III<IV

The correct answer is:
II<I<III<IV

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2017 (09 Apr Online)
Among the following compounds the most acidic is

Options:
A P- hydroxybenzoic acid
B P- nitrophenol
C P- toluic acid
D O- hydroxybenzoic acid

The correct answer is:
O- hydroxybenzoic acid

Solution:
Acidic nature ∝ electron withdrawing group -I, -R , ortho effect
Acidic nature ∝1electron releasing group +I,+R
A carboxylic acid is more acidic than phenol due to more stability of conjugate base.
Due to ortho effect o- hydroxyl benzoic acid is strongest acid here and correct order of decreasing of acidic nature is

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the following compounds, which one has the shortest C–Cl bond?

Options:
A
B
C H3C-Cl
D

The correct answer is:

Solution:
Due to conjugation of lone pair of chlorine with π bond of C–C, partial double bond character decrease bond length that’s why CH2=CH–Cl have C–Clshortest bond length.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Among the following compounds, which one is found in RNA?

Options:
A

B

C

D

The correct answer is:

Solution:
RNA contains adenine (A), guanine (g), cytosine (c) and uracil (u).

(1) is cytosine, found in RNA and DNA, both.

(2) is uracil, found in RNA only.

(3) is thymine, found in DNA only.

(4) is not a pyrimidine, i.e., nat a constituents of RNA and DNA.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Among the following compounds, which one will react with 2,4-dinitrophenyl hydrazine reagent?

Options:
A CH3CONH2
B NH2CONH2
C CH3COOC2H5
D CCl3CH(OH)2

The correct answer is:
CCl3CH(OH)2

Solution:
CCl3CH(OH)2→CCl3CHO

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Among the following four aromatic compounds, which one will have the lowest melting point?

Options:
A

B

C

D

The correct answer is:

Solution:
Melting point ∝ force of attraction. In and

hydrogen bonding takes place. Hence, the maximum force of attraction is present among molecules. The compound

is polar in nature due to the carbonyl group and thus, will have moderately strong dipole-dipole interaction. This will give them a high melting point. In

only van der Waal's force of attraction (weak force) is present. Hence, it will have the lowest melting point.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Among the following four compounds
A. Phenol
B. Methyl phenol
C. meta-nitrophenol
D. para nitrophenol
the acidity order is

Options:
A D $>$ C $>$ A $>$ B
B C $>$ D $>$ A $>$ B
C A $>$ D $>$ C $>$ B
D B $>$ A $>$ C $>$ D

The correct answer is:
D $>$ C $>$ A $>$ B

Solution:
An electron withdrawing group (-I showing group like $\left.-\mathrm{NO}_2,-\mathrm{CN}\right)$ stabilises the phenoxide ion, thus when present, increases the acidity of phenol. On the other hand electron releasing groups ($+I$ showing group like $-\mathrm{CH}_3,-\mathrm{C}_2 \mathrm{H}_5$), when present, decrease the acidity of phenol by destabilising phenoxide ion. Hence, the correct order of acidity of given compounds is
\(\underset{(D)}{\text{p-nitrophenol}} > \underset{(C)}{\text{m-nitrophenol}} > \underset{(A)}{\text{phenol}} > \underset{(B)}{\text{methyl phenol}}\)
(due to intermolecular H -bonding)

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the following four compounds
A. Phenol
B. Methyl phenol
C. meta-nitrophenol
D. para nitrophenol
the acidity order is

Options:
A D $>$ C $>$ A $>$ B
B C $>$ D $>$ A $>$ B
C A $>$ D $>$ C $>$ B
D B $>$ A $>$ C $>$ D

The correct answer is:
D $>$ C $>$ A $>$ B

Solution:
An electron withdrawing group (-I showing group like $\left.-\mathrm{NO}_2,-\mathrm{CN}\right)$ stabilises the phenoxide ion, thus when present, increases the acidity of phenol. On the other hand electron releasing groups ($+I$ showing group like $-\mathrm{CH}_3,-\mathrm{C}_2 \mathrm{H}_5$), when present, decrease the acidity of phenol by destabilising phenoxide ion. Hence, the correct order of acidity of given compounds is
\(\underset{(D)}{\text{p-nitrophenol}} > \underset{(C)}{\text{m-nitrophenol}} > \underset{(A)}{\text{phenol}} > \underset{(B)}{\text{methyl phenol}}\)
(due to intermolecular H -bonding)

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2010 (Mains)
Among the following four structures I to IV.
It is true that

Options:
A
only I and II are chiral compounds

B
only III i a chiral compound

C
only II and IV are chiral compounds

D
all four are chiral compounds

The correct answer is:
only I and II are chiral compounds

Solution:
A chiral object or structure has four different groups attached to the carbocation.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2003
Among the following graphs, variation of rate constant \( (\mathrm{K}) \) with temperature \( (\mathrm{T}) \) according to Arrhenius theory is shown by:
(A= pre-exponential factor)

Options:
A
B
C
D

The correct answer is:

Solution:
K=A⋅e-EaRT
as T→∞
K→A

Topic: Chemistry, Chemical Kinetics, JEE Main
Among the following halogens
\(\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2\) and \(\mathrm{I}_2\)
Which can undergo disproportionation reactions?

Options:
A \(\mathrm{F}_2, \mathrm{Cl}_2\) and \(\mathrm{Br}_2\)
B \(\mathrm{F}_2\) and \(\mathrm{Cl}_2\)
C Only \(\mathrm{I}_2\)
D \(\mathrm{Cl}_2, \mathrm{Br}_2\) and \(\mathrm{I}_2\)

The correct answer is:
\(\mathrm{Cl}_2, \mathrm{Br}_2\) and \(\mathrm{I}_2\)

Solution:
$\mathrm{F}_2$ do not disproportionate because fluorine do not exist in positive oxidation state however $\mathrm{Cl}_2, \mathrm{Br}_2 \& \mathrm{I}_2$ undergoes disproportionation.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Among the following how many of the molecules are linear.
CO 2 , NO 2 − , SO 2 , N 3 − , C 2 H 2 , I 3 − , XeF 2 , CS 2

Options:

The correct answer is:
6

Solution:
CO 2 ,  N 3 − ,  C 2 H 2 ,  I 3 − ,  XeF 2 ,  CS 2
these are all linear in structure

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among the following,
I
II
III
IV
V
the antiaromatic compounds are

Options:
A I and IV
B III and V
C II and V
D I and III

The correct answer is:
III and V

Solution:
III.
All C, sp $^{2}$ hybrid anti aromatic
V.
All C, sp $^{2}$ hybrid anti aromatic

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the following, increase in body mass can be taken as a criterion for growth:

Options:
A in living organisms
B in non-living objects
C both 1 and 2
D none of the above

The correct answer is:
both 1 and 2

Solution:
Living creatures share some characters which are not found in the non-living.
Growth is one of such features or characteristic property which is present in the living. The numerical changes or structural changes leading to enhancement in the object, in terms of an increase in size, number, volume, mass, etc. are termed as Growth. There are two main types of physiological reactions, Anabolism and Catabolism. When the cell (or living being) makes up (builds) complex material from simple substances, it is called Anabolism. On the contrary, when the living being breaks complex forms into simpler forms, it is termed as Catabolism.
Growth occurs when Anabolism > Catabolism. This is a characteristic feature of living. Growth in living takes place from within (called Intrinsic growth).
However, non-living beings, such as Mountain, River, Desert, etc. also grows. This is not intrinsic. The growth seen in non-living beings is external or extrinsic. Additional material is deposited outside and there is an increase in area, volume, mass, or size.

Topic: Biology, The Living World, NEET
Among the following ions which one has the highest paramagnetism

Options:
A $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
B $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
C $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
D $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

The correct answer is:
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Solution:
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has four unpaired electrons, $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ and $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ have 3, 1,0 unpaired electrons respectively.

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following marked proton of which compound shows lowest pKa value?

Options:
A

B

C

D

The correct answer is:

Solution:
The smaller the value of pKa, the stronger the acid. The acidic nature of the compound depends on the stability of conjugate base.

The stability of conjugate base formed by above compound is more stable than other three compounds given.

So it has least pKa value.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Among the following metal oxides, which is most basic?

Options:
A ZnO
B Al2O3
C As2O3
D K2O

The correct answer is:
K2O

Solution:
The basic character of metal oxides decreases from left to right in a period due to decrease in electropositive character which in turn decreases the polarity of bond as well as the internuclear distance between the oxygen and metal atom. Therefore, alkali metal oxides are most basic and halogen oxide (oxygen halides) are most acidic ∴K2O is most basic metal oxide.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among the following metals the number of metals that are extracted by self-reduction method from their respective ores is
Hg, Zn, Cu, Al, Mg, Pb, Fe, Na.

Options:

The correct answer is:
3

Solution:
Self-Reduction Process: The cations of the less electropositive metals like Pb, Hg and Cu may be reduced without the use of any additional reducing agent.
Self-reduction of Cu:
2Cu2O + Cu2Sâ 6Cu + SO2g
It is called self-reducing because copper oxide itself reduced the copper sulfide to give copper and sulfur dioxide.
Similar reactions occurs for Pb and Hg.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Among the following mixtures, dipole-dipole as the major interaction, is present in

Options:
A
benzene and ethanol

B
acetonitrile and acetone

C
KCl and water

D
benzene and carbon tetrachloride

The correct answer is:
acetonitrile and acetone

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2006
Among the following molecules/ ions,

C22-, N22-,O22-, O2

Which one is diamagnetic and has the shortest bond length?

Options:
A N22-
B O22-
C O2
D C22-

The correct answer is:
C22-

Solution:
O2(16)= σ1s2σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2=π2py2 π* 2px1=π2py1

Paramagnetic; Bond order = 2

O22-(18)= σ1s2σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2=π2py2 π* 2px2=π*2py2

Diamagnetic; Bond order = 1

N22-(16)= σ1s2σ*1s2 σ 2s2 σ*2s2 π2px2 π2py2 σ2pz2π*2px1 =π*2py1

Paramagnetic; Bond order = 2

C22-(14)= σ1s2σ*1s2 σ 2s2 σ*2s2 π2px2=π2py2 σ2pz2

Diamagnetic; Bond order =3

Bond order ∝1Bond length

Hence C22 is diamagnetic and is having shortest bond length.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Among the following nitrates, the one that decomposes without leaving any solid residue is

Options:
A Lead nitrate
B Ammonium nitrate
C Silver nitrate
D Sodium nitrate

The correct answer is:
Ammonium nitrate

Solution:
\(\begin{aligned}
& \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow{\Delta} 2 \mathrm{H}_2 \mathrm{O} \uparrow+\mathrm{N}_2 \mathrm{O} \uparrow \\
& \text { (a) } \\
& \mathrm{NaNO}_3 \xrightarrow{\Delta} \mathrm{NaNO}_2+\mathrm{O}_2 \uparrow \\
& \text { (a) } \qquad\qquad \text { (b) } \\
\end{aligned}\)
\(\begin{aligned}
& 2 \mathrm{AgNO}_3(\mathrm{s}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{NO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \\
& \text { Lunar caustic } \\
& 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow 2 \underset{(s)}{2 \mathrm{PbO}}+4 \mathrm{NO}_2 \uparrow+\mathrm{O}_2 \uparrow
\end{aligned}\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ___ . (Integer answer)
(A) Li
(B) Na
(C) Rb
(D) Cs

Options:

The correct answer is:
1

Solution:
Cs is used as electrodes in the photoelectric cell.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Among the following observations, the correct one that differentiates between $\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ is

Options:
A both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ dissolves in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not
B $\mathrm{SO}_{3}^{2-}$ forms precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{4}^{2-}$ does not
C $\mathrm{SO}_{4}^{2-}$ forms precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ does not
D both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{4}^{2-}$ dissolves in HCl but $S O_{3}^{2-}$ does not

The correct answer is:
both form precipitate with $\mathrm{BaCl}_{2}, \mathrm{SO}_{3}^{2-}$ dissolves in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not

Solution:
$\mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ when treated with $\mathrm{BaCl}_{2}$, give white ppt of $BaSO_{3}$ and $BaSO_{4}$ respectively.
$$
\begin{array}{c}
\mathrm{BaCl}_{2}+\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{BaSO}_{3}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphide} \\
\mathrm{BaCl}_{2}+\mathrm{SO}_{4}^{2-} \longrightarrow \begin{array}{c}
\text { } \\
\mathrm{BaSO}_{4}+2 \mathrm{Cl}^{-} \\
\text {Barium sulphate }
\end{array}
\end{array}
$$
Out of these $\mathrm{two}, \mathrm{SO}_{3}^{2-}$ is soluble in $\mathrm{HCl}$ but $\mathrm{SO}_{4}^{2-}$ does not.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores of iron is

Options:

The correct answer is:
4

Solution:
Siderite: FeCO3

Malachite: CuCO3.CuOH2

Magnetite: Fe3O4

Haematite: Fe2O3

Cryolite: Na3AlF6

Cuprite: Cu2O

Limonite: FeOOH.nH2O

Kaolinite: Al2OH4Si2O5

Sphalerite: ZnS

Bauxite: AlOxOH3-2x

So, as we can see that among the given options 4 ores (Siderite, magnetite, Haematite, Limonite) contain iron.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Among the following organic acids, the acid present in rancid butter is :

Options:
A Pyruvic acid
B Acetic acid
C Butyric acid
D Lactic acid

The correct answer is:
Butyric acid

Solution:
Butyric acid is a fatty acid occuring in the form of esters in animal fats when butter goes rancid, butyric acid is liberated from the glyceride by hydrolysis, leading to unpleasant odor.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2014 (19 Apr Online)
Among the following oxide of p - block elements, Cl2O7,CO,PbO2, N2O,NO,Al2O3,SiO2, N2O5, SnO2

Options:

The correct answer is:
3

Solution:
Amphoteric oxides are the oxides that behave as both acidic and basic oxides. Amphoteric Oxides have features of acidic as well as basic oxides that neutralise both acids and bases.

Acidic oxide: Cl2O7,SiO2, N2O5
Neutral oxide: CO,NO,N2O
Amphoteric oxide: Al2O3,SnO2,PbO2

Hence the answer is 3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (01 Feb Shift 1)
Among the following oxoacids, the correct decreasing order of acid strength is :

Options:
A HOCl > HClO2 > HClO3 > HClO4
B HClO4 > HOCl > HClO2 > HClO3
C HClO4 > HClO3 > HClO2 > HOCl
D HClO2 > HClO4 > HClO3 > HOCl

The correct answer is:
HClO4 > HClO3 > HClO2 > HOCl

Solution:
In oxyacids of chlorine as oxygen content increases. The acidic strength increases.
HClO4 > HClO3 > HClO2 > HOCl
Conjugate base
Cl O 4 − − 4 R.S.
Cl O 3 − − 3 R.S.
Cl O 2 − − 2 R.S.
Cl O − − no R.S.
(conjugate base) anion stability increase due to resonance. So acidic character increase.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2014 (06 Apr)
Among the following pair of complexes in which case the central atoms are having some hybridisation and have same values of E.A.N. also.

Options:
A NiCO4 and NiCN42-
B FeF63- and FeH2O63+
C FeCN63- and FeCN62-
D CuCN42- and CuCN43-

The correct answer is:
FeF63- and FeH2O63+

Solution:
In case of FeF63- and FeH2O63+ central atom has sp3d2 hybridisation and the value of E.A.N. is 35 in both the complexes.

Topic: Chemistry, Coordination Compounds, JEE Main
Among the following reactions, a mixture of diastereomers is produced from

Options:
A
B
C
D

The correct answer is:

Solution:
Diastereomers are optical isomers, which are not mirror images of one each other.

The product obtained from the given reactions are as follows.

(a)

(b)

(c)

(d)

Thus the reaction given in option a is correct

Topic: Chemistry, Hydrocarbons, JEE Main
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is:

Options:
A H2+Cl2→2 HCl
B H2+I2→2 HI
C H2+F2→2 HF
D H2+Br2→2 HBr

The correct answer is:
H2+I2→2 HI

Solution:
I2 is the least reactive halogen. So, its reaction with H2 requires a catalyst.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (10 Jan Shift 2)
Among the following series of transition metal ions, how many metals ions have d 2 electronic configuration

Fe4+,V3+,Sn2+,Au3+Ti2+,As3+,Cr3+,Mn5+Co3+,Zr2+,Cd2+,Mo4+,La2+,Ta4+,Co6+,Ni4+

Options:

The correct answer is:
6

Solution:
Ti22=3d24s2 ; Ti2+=3d2
V23=3d34s2 ; V3+=3d2
Cr24=3d54s1 ; Cr4+=3d2
Mn25=3d54s2 ; Mn5+=3d2
Zr40= 4d25s2 ; Zr2+= 4d2
Mo42= 4d55s1;Mo4+= 4d2

Topic: Chemistry, d and f Block Elements, JEE Main
Among the following series of transition metal ions, the one where all metals ions have $3 d^2$ electronic configuration is

Options:
A $\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}$
B $\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}$
C $\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}$
D $\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}$

The correct answer is:
$\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}$

Solution:
$T i_{22}=3 d^2 4 s^2 ;
T i^{2+}=3 d^2$
$V_{23}=3 d^3 4 s^2 ; \quad V^{2+}=3 d^2$
$\mathrm{Cr}_{24}=3 d^4 4 s^2 ; \quad \mathrm{Cr}^{4+}=3 d^2$
$M n_{25}=3 d^5 4 s^2 ; M n^{5+}=3 d^2$

Topic: Chemistry, d and f Block Elements, JEE Main
Among the following species, the diamagnetic molecule is:

Options:
A CO
B B2
C O2
D NO

The correct answer is:
CO

Solution:
(1) B2(10)⇒σ1s2, σ*1s2, σ2s2,σ*2s2,π2px1=π2py1 , paramagnetic
(2) NO15⇒σ1s2σ*1s2σ*2s2σ2pz2π2px2=π2py2=π*2px1 , paramagnetic
(3) O216⇒ σ1s2σ*1s2σ2s2σ*2pz2π2px2=π2py2π*2px1=π*2py, paramagnetic
(4) CO14⇒ σ1s2σ*1s2,σ2s2,σ*2s2,πpx2=π2py2,σ2pz2, diamagnetic (No unpaired electron)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Among the following species the one which causes the highest CFSE, $\Delta_0$ as a ligand is:

Options:
A
$\mathrm{CN}^{-}$

B
$\mathrm{NH}_3$

C
$\mathrm{F}^{-}$

D
$\mathrm{CO}$

The correct answer is:
$\mathrm{CO}$

Solution:
Following is the order of increasing value of $\Delta_0$ for the ligands.
$$
\begin{aligned}
&\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-} < \mathrm{F}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{Py}=\mathrm{NH}_3 < \text { en } \\
& < \mathrm{NO}_2-< < \mathrm{CN}^{-} < \mathrm{CO}
\end{aligned}
$$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (12 Apr Online)
Among the following species which option contains all isoelectronic species?

Options:
A O-,Â F-,Â Â Na,Â Mg+
B O2-,Â F-,Â Na,Â Mg2+
C O-,Â F-,Â Na+,Â Mg2+
D O2-, F-, Na+, Mg2+

The correct answer is:
O2-, F-, Na+, Mg2+

Solution:
O-2 = 8+2 = 10
F- = 9+1 = 10
Na+ = 11-1 = 10
Mg+2 = 12-2 = 10
All have 10 electrons

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2017 (02 Apr)
Among the following species which two have trigonal pyramidal shape?
(I) $\mathrm{NI}_3$
(II) $\mathrm{I}_3^{-}$
(III) $\mathrm{SO}_3^{2-}$
(IV) $\mathrm{NO}_3^{-}$

Options:
A
I and III

B
III and IV

C
I and IV

D
II and III

The correct answer is:
I and III

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (26 May Online)
Among the following statements about the molecules X and Y, which is incorrect?

Options:
A X and Y are diastereomers
B X and Y are enantiomers
C X and Y are both aldohexoses
D X is a D-sugar and Y is an L-sugar

The correct answer is:
X and Y are enantiomers

Solution:
They are mirror images of each other hence, are enantiomers. They are aldohexoses too (aldehyde group + six carbon).

Topic: Chemistry, General Organic Chemistry, JEE Main