JEE Mains 2024

Jan + Apr

Coordination compounds
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 JEE Main 2024 : 9th April S2
Match List I with List II
1 List I List II
A. K2[Ni(CN)4] I. sp3
B. [Ni(CO)4] II. sp3d2
C. [Co(NH3)6Cl3 III. dsp2
D. Na3 [CoF6] IV. d2sp3
Choose the correct answer from the options given below:

A A-III, B-II, C-IV, D-I

B A-III, B-I, C-II, D-IV

C A-I, B-II, C-II, D-IV

D A-III, B-I, C-IV, D-II

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 JEE Main 2024 : 9th April S2
Match List I with List II
1 List I List II
A. K2[Ni(CN)4] I. sp3
B. [Ni(CO)4] II. sp3d2
C. [Co(NH3)6Cl3 III. dsp2
D. Na3 [CoF6] IV. d2sp3
Choose the correct answer from the options given below:

A A-III, B-II, C-IV, D-I

B A-III, B-I, C-II, D-IV

C A-I, B-II, C-II, D-IV

D A-III, B-I, C-IV, D-II

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 Solution:
+2
(A) K2[Ni(CN)4]
Ni2+ : [Ar]3d8 4so, (CN- is S.F.L)
Pre hybridization state of Ni+2

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 Solution:
(B) [Ni(CO)4]
Ni : [Ar]3d8 4s2
CO is S.F.L, so pairing occur
Pre hybridization state of Ni

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 Solution:
(C) [Co(NH3)6]Cl3
Co+3 : [Ar]3d6 4s0
with Co3+, NH3 act as S.F.L

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 Solution:
(D) Na3[CoF6]
Co3+ : [Ar]3d6 (F⊖ : W.F.L)

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 JEE Main 2024 : 9th April S2
2
The coordination environment of Ca2+ ion in its complex
with EDTA4- is :

A trigonal prismatic

B octahedral

C square planar

D tetrahedral

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 JEE Main 2024 : 9th April S2
2
The coordination environment of Ca2+ ion in its complex
with EDTA4- is :

A trigonal prismatic

B octahedral

C square planar

D tetrahedral

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 Solution:
The coordination environment of the Ca2+ ion when it
forms a complex with EDTA4-
(ethylenediaminetetraacetic acid) is octahedral.
Ethylenediaminetetraacetic acid (EDTA) is a
hexadentate ligand, which means it has six donor
atoms that can bind to a central metal ion. In the case
of EDTA, these donor atoms are four oxygen atoms
from its four carboxyl groups and two nitrogen atoms
from its two amine groups.

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 Solution:
When Ca2+ forms a complex with EDTA4-, all six donor
atoms from EDTA coordinate with the calcium ion. As a
result, the coordination number of the calcium ion is 6,
leading to an octahedral geometry. This is because the
octahedral geometry is the most common and
energetically favorable arrangement for a coordination
number of 6. In such a geometry, the six ligands are
placed at equal distances from the central ion and at
90° angles relative to adjacent ligands, maximizing the
distance between all ligands to minimize repulsion.
The correct answer is Option B, octahedral.

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 JEE Main 2024 : 9th April S1
3 Assertion (A): The total number of geometrical isomers
shown by [Co(en)2Cl2]+ complex ion is three.
Reason (R): [Co(en)2Cl2]+ complex ion has an
octahedral geometry.

A (A) is correct but (R) is not correct

B (A) is not correct but (R) is correct

Both (A) and (R) are correct but (R) is not the
C
correct explanation of (A)
Both (A) and (R) are correct and (R) is the
D
correct explanation of (A)

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 JEE Main 2024 : 9th April S1
3 Assertion (A): The total number of geometrical isomers
shown by [Co(en)2Cl2]+ complex ion is three.
Reason (R): [Co(en)2Cl2]+ complex ion has an
octahedral geometry.

A (A) is correct but (R) is not correct

B (A) is not correct but (R) is correct

Both (A) and (R) are correct but (R) is not the
C
correct explanation of (A)
Both (A) and (R) are correct and (R) is the
D
correct explanation of (A)

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 Solution:
[Co(en)2Cl2]- has octahedral geometry with two
geometrical isomers.

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 JEE Main 2024 : 8th April S2
Match List I with List II
4
List I List II
(Complex ion) (spin only magnetic moment in B.M.)
A. [Cr(NH3)6]3+ I. 4.90
B. [NiCl4]2- II. 3.87
C. [CoF6]3- III. 0.0
D. [Ni(CN)4]2- IV. 2.83
Choose the correct answer from the options given below:

A (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

B (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

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D (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
 JEE Main 2024 : 8th April S2
Match List I with List II
4
List I List II
(Complex ion) (spin only magnetic moment in B.M.)
A. [Cr(NH3)6]3+ I. 4.90
B. [NiCl4]2- II. 3.87
C. [CoF6]3- III. 0.0
D. [Ni(CN)4]2- IV. 2.83
Choose the correct answer from the options given below:

A (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

B (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

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D (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
 Solution:

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 JEE Main 2024 : 8th April S1
5 Number of complexes with even number of electrons in
t2g orbitals is-
[Fe(H2O)6]2+, [Co(H2O)6]2+., [Co(H2O)6]3+, [Cu(H2O)6]2+,
[Cr(H2O)6]2+

A 3

B 2

C 1

D 5

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 JEE Main 2024 : 8th April S1
5 Number of complexes with even number of electrons in
t2g orbitals is-
[Fe(H2O)6]2+, [Co(H2O)6]2+., [Co(H2O)6]3+, [Cu(H2O)6]2+,
[Cr(H2O)6]2+

A 3

B 2

C 1

D 5

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 Solution:

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 Solution:

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 JEE Main 2024 : 8th April S1
6 An octahedral complex with the formula CoCl3 . nNH3
upon reaction with excess of AgNO3 solution gives 2
moles of AgCl. Consider the oxidation state of Co in the
complex is 'x'. The value of "x + n" is __________.

A 6

B 5

C 3

D 8

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 JEE Main 2024 : 8th April S1
6 An octahedral complex with the formula CoCl3 . nNH3
upon reaction with excess of AgNO3 solution gives 2
moles of AgCl. Consider the oxidation state of Co in the
complex is 'x'. The value of "x + n" is __________.

A 6

B 5

C 3

D 8

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 Solution:
+3
[Co(NH3)5Cl]Cl2 + excess AgNO3 ➝ 2AgCl
(2 moles)
x+0-1-2=0
x = +3
n=5
∴x+n=8

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 JEE Main 2024 : 8th April S1
7 Statement I : N(CH3)3 and P(CH3)3 can act as ligands to
form transition metal complexes.
Statement II : As N and P are from same group, the
nature of bonding of N(CH3)3 and P(CH3)3 is always
same with transition metals.

A Both Statement I and Statement II are incorrect

B Both Statement I and Statement II are correct

Statement I is incorrect but Statement II is

C
correct
Statement I is correct but Statement II is
D
incorrect
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 JEE Main 2024 : 8th April S1
7 Statement I : N(CH3)3 and P(CH3)3 can act as ligands to
form transition metal complexes.
Statement II : As N and P are from same group, the
nature of bonding of N(CH3)3 and P(CH3)3 is always
same with transition metals.

A Both Statement I and Statement II are incorrect

B Both Statement I and Statement II are correct

Statement I is incorrect but Statement II is

C
correct
Statement I is correct but Statement II is
D
incorrect
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 Solution:
Statement I: N(CH3)3 (trimethylamine) and P(CH3)3
(trimethylphosphine) can indeed act as ligands to form
transition metal complexes. Ligands are molecules or
ions that can donate a pair of electrons to a metal atom
to form a coordinate bond. Both nitrogen and
phosphorus have lone pairs of electrons that can be
donated to transition metals, making N(CH3)3 and
P(CH3)3 suitable ligands. Therefore, Statement I is
correct.

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 Solution:
Statement II: While N (Nitrogen) and P (Phosphorus) are
both from group 15 of the periodic table and share some
similar properties, the nature of their bonding with
transition metals is not always the same. Nitrogen
compounds typically form stronger bonds with metals
compared to phosphorus compounds. This difference is
due to several factors, including the difference in atomic
sizes, electronegativity, and the extent of orbital overlap.
Nitrogen, being smaller and more electronegative, can
form more effective overlap and stronger bonds with
metals compared to phosphorus. Therefore, Statement II
is incorrect.
Given this analysis, the most appropriate answer is Option
Join HP Gurukul :- Click Here D: Statement I is correct but Statement II is incorrect.
 JEE Main 2024 : 8th April S1
8 Match List I with List II
List I (Compound) List II (Colour)
A. Fe4[Fe(CN)6]3 . xH2O I. Violet
B. [Fe(CN)5NOS]4- II. Blood Red
C. [Fe(SCN)]2+ III. Prussian Blue
D. (NH4)3PO4 . 12MoO3 IV. Yellow
Choose the correct answer from the options given below:

A A-IV, B-I, C-II, D-III

B A-I, B-II, C-III, D-IV

C A-II, B-III, C-IV, D-I

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D A-III, B-I, C-II, D-IV
 JEE Main 2024 : 8th April S1
8 Match List I with List II
List I (Compound) List II (Colour)
A. Fe4[Fe(CN)6]3 . xH2O I. Violet
B. [Fe(CN)5NOS]4- II. Blood Red
C. [Fe(SCN)]2+ III. Prussian Blue
D. (NH4)3PO4 . 12MoO3 IV. Yellow
Choose the correct answer from the options given below:

A A-IV, B-I, C-II, D-III

B A-I, B-II, C-III, D-IV

C A-II, B-III, C-IV, D-I

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D A-III, B-I, C-II, D-IV
 Solution:
Fe4[Ffe(CN)6]3 . xH2O ➝ Prussian Blue
[Fe(CN)5NOS]4- ➝ Violet
[Fe(SCN)]2+ ➝ Blood Red
(NH4)3PO4 . 12MoO3 ➝ Yellow

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 JEE Main 2024 : 6th April S2
9 Statement I : PF5 and BrF5 both exhibit sp3d
hybridisation
Statement II : Both SF6 and [Co(NH3)6]3+ exhibit sp3d2
hybridisation.

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Statement I is false but Statement II is true

D Both Statement I and Statement II are true

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 JEE Main 2024 : 6th April S2
9 Statement I : PF5 and BrF5 both exhibit sp3d
hybridisation
Statement II : Both SF6 and [Co(NH3)6]3+ exhibit sp3d2
hybridisation.

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Statement I is false but Statement II is true

D Both Statement I and Statement II are true

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 Solution:
Hybridisation of P in PF5 is sp3d but the hybridisation of
Br in BrF5 is sp3d2. So, statement-I is false.
Hybridisation of S in SF6 is sp3d2 but the hybridisation
of Co3+ in [Co(NH3)6]3+ is d2sp3 as NH3 is a strong field
ligand forcing the unpaired electrons to pair up. So,
Statement-II is false.

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 JEE Main 2024 : 6th April S2
10 Match List I with List II
List I (Tetrahedral complex) List II (Electronic configuration)
A. TiCl4 I. e2, t20
B. [FeO4]2- II. e4, t23
C. [FeCl4]- III. e0, t20
D. [CoCl4]2- IV. e2, t23
Choose the correct answer from the options given below:

A A-IV, B-III, C-I, D-II

B A-III, B-I, C-IV, D-II

C A-I, B-III, C-IV, D-II

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D A-III, B-IV, C-II, D-I
 JEE Main 2024 : 6th April S2
10 Match List I with List II
List I (Tetrahedral complex) List II (Electronic configuration)
A. TiCl4 I. e2, t20
B. [FeO4]2- II. e4, t23
C. [FeCl4]- III. e0, t20
D. [CoCl4]2- IV. e2, t23
Choose the correct answer from the options given below:

A A-IV, B-III, C-I, D-II

B A-III, B-I, C-IV, D-II

C A-I, B-III, C-IV, D-II

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D A-III, B-IV, C-II, D-I
 Solution:
(A) TiCl4: Ti4+ : 3d04s0 or e0t20
(B) [FeO4]2- : Fe6+ : 3d24s0 or e2t20
(C) [FeCl4] : Fe3+ : 3d54s0 or e2t32
(D) [CoCl4]2- : Co2+ : 3d74s0 or e4 t23
∴ Correct matching of tetrahedral complexes given in
List-l with their electronic configuration given in List-ll is
(A)-(III); (B)-(I); (C)-(IV); (D)-(II)

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 JEE Main 2024 : 6th April S2
11

The correct IUPAC name of [PtBr2(PMe3)2] is :

A bis(trimethylphosphine)dibromoplatinum(II)

B didbromodi(trimethylphosphine)platinum(II)

C dibromobis(trimethylphosphine)platinum(II)

D bis[bromo(trimethylphosphine)]platinum(II)

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 JEE Main 2024 : 6th April S2
11

The correct IUPAC name of [PtBr2(PMe3)2] is :

A bis(trimethylphosphine)dibromoplatinum(II)

B didbromodi(trimethylphosphine)platinum(II)

C dibromobis(trimethylphosphine)platinum(II)

D bis[bromo(trimethylphosphine)]platinum(II)

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 Solution:
The correct IUPAC name of [PtBr2(PMe3)2]is
dibromobis(trimethylphosphine)platinum(II).

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 JEE Main 2024 : 6th April S1
12 Consider the following complexes
(A) [CoCl(NH3)5]2+, (B) [Co(CN)6]3-, (C) [Co(NH3)5
(H2O)]3+, (D) [Cu(H2O)4]2+
The correct order of A, B, C and D in terms of
wavenumber of light absorbed is:

A D<A<C<B

B B<C<A<D

C C<D<A<B

D A<C<B<D

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 JEE Main 2024 : 6th April S1
12 Consider the following complexes
(A) [CoCl(NH3)5]2+, (B) [Co(CN)6]3-, (C) [Co(NH3)5
(H2O)]3+, (D) [Cu(H2O)4]2+
The correct order of A, B, C and D in terms of
wavenumber of light absorbed is:

A D<A<C<B

B B<C<A<D

C C<D<A<B

D A<C<B<D

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 Solution:
Wavenumber = 1/λ ∝ Frequency ∝ Δ0
Wavenumber order : D < A < C < B

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 JEE Main 2024 : 6th April S1
13 Match List I with List II
List I (Hybridization) List II (Orientation in Shape)
A. sp3 I. Trigonal bipyramidal
B. dsp2 II. Octahedral
C. sp3d III. Tetrahedral
D. sp3d2 IV. Square planar
Choose the correct answer from the options given below:

A A-III, B-I, C-IV, D-II

B A-III, B-IV, C-I, D-II

C A-IV, B-III, C-I, D-II

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D A-II, B-I, C-IV, D-III
 JEE Main 2024 : 6th April S1
13 Match List I with List II
List I (Hybridization) List II (Orientation in Shape)
A. sp3 I. Trigonal bipyramidal
B. dsp2 II. Octahedral
C. sp3d III. Tetrahedral
D. sp3d2 IV. Square planar
Choose the correct answer from the options given below:

A A-III, B-I, C-IV, D-II

B A-III, B-IV, C-I, D-II

C A-IV, B-III, C-I, D-II

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D A-II, B-I, C-IV, D-III
 Solution:
A ➝ Tetrahedral (III)
B ➝ Square planar (IV)
C ➝ Trigonal bipyramidal (I)
D ➝ Octahedral (II)

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 JEE Main 2024 : 5th April S2
14
The number of complexes from the following with no
electrons in the t2 orbital is_____.
TiCl4, [MnO4]-, [FeO4]2-, [FeCl4]-, [CoCl4]2-

A 4

B 2

C 3

D 1

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 JEE Main 2024 : 5th April S2
14
The number of complexes from the following with no
electrons in the t2 orbital is_____.
TiCl4, [MnO4]-, [FeO4]2-, [FeCl4]-, [CoCl4]2-

A 4

B 2

C 3

D 1

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 Solution:
TiCl4 ➝ Ti+4 ➝ [Ar]4s03d0 ➝ e0t20
[MnO4]⊖ ➝ Mn+7 ➝ [Ar]4s03d0 ➝ e0t20
[FeO4]2- ➝ Fe+6 ➝ [Ar]4s03d0 ➝ e0t20
[FeCl4]⊖ ➝ Fe3+ ➝ [Ar]4s0 3d5 ➝ e2t23
[CoCl4]2- ➝ Co+2 ➝ [Ar]4s03d7 ➝ e4t23
Three complexes have no electrons in t2 orbital

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 JEE Main 2024 : 5th April S2
15 The metal atom present in the complex MABXL (where
A, B, X and L are unidentate ligands and M is metal)
involves sp3 hybridization. The number of geometrical
isomers exhibited by the complex is :

A 0

B 2

C 3

D 4

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 JEE Main 2024 : 5th April S2
15 The metal atom present in the complex MABXL (where
A, B, X and L are unidentate ligands and M is metal)
involves sp3 hybridization. The number of geometrical
isomers exhibited by the complex is :

A 0

B 2

C 3

D 4

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 Solution:
To determine the number of geometrical isomers for the
given complex MABXL, where M is a metal and A, B, X,
and L are unidentate ligands, we need to consider the
geometry implied by the sp3 hybridization of the metal.
The sp3 hybridization suggests a tetrahedral geometry
for the metal complex.

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 Solution:
In a tetrahedral complex, geometrical isomerism does
not arise. This is because geometrical isomerism (also
known as cis-trans isomerism) typically occurs in
square planar or octahedral complexes where ligands
can be positioned differently around the central metal
atom to give distinct isomers. In a tetrahedral complex,
every position is equivalent due to the symmetric
arrangement of the ligands around the metal center,
making it impossible to distinguish between different
sides of the molecule as you would with a square
planar or octahedral complex.

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 Solution:
Given that the complex follows tetrahedral geometry
because of sp3 hybridization and that tetrahedral
complexes do not exhibit geometrical isomerism, the
number of geometrical isomers exhibited by the
complex is:
Option A: 0
This is because in a tetrahedral arrangement of ligands
around a metal center, no geometrical isomers can be
formed due to the lack of cis or trans arrangements
possible within such a symmetric geometry.

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 JEE Main 2024 : 5th April S1
16
The correct order of ligands arranged in increasing field
strength.

A F- < Br- < I- < NH3

B H2O < -OH < CN- < NH3

C Br- < F- < H2O < NH3

D Cl- < -OH < Br- < CN-

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 JEE Main 2024 : 5th April S1
16
The correct order of ligands arranged in increasing field
strength.

A F- < Br- < I- < NH3

B H2O < -OH < CN- < NH3

C Br- < F- < H2O < NH3

D Cl- < -OH < Br- < CN-

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 Solution:
Experimental order Br- < F- < H2O < NH3

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 JEE Main 2024 : 5th April S1
17
Which one of the following complexes will exhibit the
least paramagnetic behaviour ? [Atomic number,
Cr = 24, Mn = 25, Fe = 26, Co = 27]

A [Fe(H2O)6]2+

B [Mn(H2O)6]2+

C [Co(H2O)6]2+

D [Cr(H2O)6]2+

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 JEE Main 2024 : 5th April S1
17
Which one of the following complexes will exhibit the
least paramagnetic behaviour ? [Atomic number,
Cr = 24, Mn = 25, Fe = 26, Co = 27]

A [Fe(H2O)6]2+

B [Mn(H2O)6]2+

C [Co(H2O)6]2+

D [Cr(H2O)6]2+

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 Solution:

So, the complex with minimum number of unpaired e⊖

is option (2) [Co(H2O)6]2+.

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 JEE Main 2024 : 4th April S2
18
If an iron (III) complex with the formula [Fe(NH3)x (CN)y]-
has no electron in its eg orbital, then the value of x + y is

A 6

B 4

C 3

D 5

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 JEE Main 2024 : 4th April S2
18
If an iron (III) complex with the formula [Fe(NH3)x (CN)y]-
has no electron in its eg orbital, then the value of x + y is

A 6

B 4

C 3

D 5

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 Solution:
Balancing charges,
3 - y = -1
⇒y=4
x=2
x+y=6

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 JEE Main 2024 : 4th April S2
19
The number of unpaired d-electrons in [Co(H2O)6]3+
is__

A 0

B 2

C 1

D 4

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 JEE Main 2024 : 4th April S2
19
The number of unpaired d-electrons in [Co(H2O)6]3+
is__

A 0

B 2

C 1

D 4

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 Solution:
[Co(H2O)6]3+ is an inner orbital complex.
So, electronic configuration is t26 g eg0.
All electrons are paired

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 JEE Main 2024 : 4th April S2
20
A first row transition metal in its +2 oxidation state has a
spin-only magnetic moment value of 3.86 BM. The
atomic number of the metal is

A 22

B 23

C 26

D 25

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 JEE Main 2024 : 4th April S2
20
A first row transition metal in its +2 oxidation state has a
spin-only magnetic moment value of 3.86 BM. The
atomic number of the metal is

A 22

B 23

C 26

D 25

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 Solution:
22
Ti+2 ⇒ [Ar]3d2

23
V+2 ⇒ [Ar]3d3

25
Mn+2 ⇒ [Ar]3d5
+2 6
26
Fe ⇒ [Ar]3d

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 JEE Main 2024 : 4th April S1
21 Number of complexes from the following with even
number of unpaired “d” electrons is ______.
[V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+, [Ni(H2O)6]3+,
[Cu(H2O)6]2+ [Given atomic numbers : V = 23, Cr = 24,
Fe = 26, Ni = 28, Cu = 29]

A 1

B 5

C 2

D 4
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 JEE Main 2024 : 4th April S1
21 Number of complexes from the following with even
number of unpaired “d” electrons is ______.
[V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+, [Ni(H2O)6]3+,
[Cu(H2O)6]2+ [Given atomic numbers : V = 23, Cr = 24,
Fe = 26, Ni = 28, Cu = 29]

A 1

B 5

C 2

D 4
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 Solution:
[V(H2O)6]3+ ➝ 2 unpaired electrons
[Cr(H2O)6]2+ ➝ 4 unpaired electrons
Above 2 complex have even number of unpaired
electrons.

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 JEE Main 2024 : 4th April S1
22
The correct sequence of ligands in the order of
decreasing field strength is :

A NCS- > EDTA4- > CN- > CO

B CO > H2O > F- > S2-

C S2- > -OH > EDTA4- > CO

D -
OH > F- > NH3 > CN-

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 JEE Main 2024 : 4th April S1
22
The correct sequence of ligands in the order of
decreasing field strength is :

A NCS- > EDTA4- > CN- > CO

B CO > H2O > F- > S2-

C S2- > -OH > EDTA4- > CO

D -
OH > F- > NH3 > CN-

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 Solution:
Field strength order : CO > H2O > F- > S2-

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 JEE Main 2024 : 1st Feb S2
23
[Co(NH3)6]3+ and [CoF6]3- are respectively known as:

A Inner orbital complex, spin paired complex

B Spin paired complex, spin free complex

C Spin free complex, spin paired complex

D Outer orbital complex, inner orbital complex

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 JEE Main 2024 : 1st Feb S2
23
[Co(NH3)6]3+ and [CoF6]3- are respectively known as:

A Inner orbital complex, spin paired complex

B Spin paired complex, spin free complex

C Spin free complex, spin paired complex

D Outer orbital complex, inner orbital complex

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 Solution:
[Co(NH3)6]3+
Co3+ (strong field ligand) ⇒ 3d6 (t26g, eg0),
Hybridisation : d2sp3
Inner orbital complex (spin paired complex)
Pairing will take place
[CoF6]3-
Co3+ (weak field ligand) ⇒ 3d6 (t24g, eg2)
Hybridisation : sp3d2
Outer orbital complex (spin free complex)
No pairing will take place

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 JEE Main 2024 : 1st Feb S2
24 Statement (I) : Dimethylglyoxime forms a six-membered
covalent chelate when treated with NiCl2 solution in presence
of NH4OH.
Statement (II) : Prussian blue precipitate contains iron both in
(+2) and (+3) oxidation states.
In the light of the above statements, choose the most
appropriate answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

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 JEE Main 2024 : 1st Feb S2
24 Statement (I) : Dimethylglyoxime forms a six-membered
covalent chelate when treated with NiCl2 solution in presence
of NH4OH.
Statement (II) : Prussian blue precipitate contains iron both in
(+2) and (+3) oxidation states.
In the light of the above statements, choose the most
appropriate answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

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 Solution:
Let's analyze each statement separately:
Statement (I):

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 Solution:
So, Statement I is false.
Statement (II): Prussian blue, also known as ferric
ferrocyanide, is a famous pigment that contains iron in
both +2 and +3 oxidation states. Its chemical formula
can be represented as Fe4[Fe(CN)6]3. In this formula,
the iron within the square brackets, [Fe(CN)64-], is in the
+2 oxidation state (ferrocyanide ion), while the iron
outside the square brackets is in the +3 oxidation state
(ferric ion). The compound is a complex salt that arises
from the reaction of ferric and ferrous ions with cyanide
ions. Therefore, Statement II is also true.
With both statements being true, the correct answer is:
Join HP Gurukul :- Click Here Option A : Statement I is false but Statement II is true.
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 JEE Main 2024 : 1st Feb S2
25
Which of the following compounds show colour due to
d-d transition?

A K2Cr2O7

B CuSO4 . 5H2O

C KMnO4

D K2CrO4

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 JEE Main 2024 : 1st Feb S2
25
Which of the following compounds show colour due to
d-d transition?

A K2Cr2O7

B CuSO4 . 5H2O

C KMnO4

D K2CrO4

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 Solution:
CuSO4 . 5H2O
Cu2+ : 3d9 4s0
Unpaired electron present so it show colour due to d-d
transition.

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 JEE Main 2024 : 1st Feb S1
26
Which of the following complex is homoleptic?

A [Ni(NH3)2Cl2]

B [Co(NH3)4Cl2]+

C [Fe(NH3)4Cl2]+

D [Ni(CN)4]2-

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 JEE Main 2024 : 1st Feb S1
26
Which of the following complex is homoleptic?

A [Ni(NH3)2Cl2]

B [Co(NH3)4Cl2]+

C [Fe(NH3)4Cl2]+

D [Ni(CN)4]2-

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 Solution:
In Homoleptic complex all the ligand attached with the
central atom should be the same. Hence [Ni(CN)4]2- is
a homoleptic complex.

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 JEE Main 2024 : 1st Feb S1
27
Statement (I) : A solution of [Ni(H2O)6]2+ is green in
colour
Statement (II) : A solution of [Ni(CN)4]2- is colourless.

A Statement I incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Both Statement I and Statement II are incorrect

Statement I is correct but Statement II is

D
incorrect

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 JEE Main 2024 : 1st Feb S1
27
Statement (I) : A solution of [Ni(H2O)6]2+ is green in
colour
Statement (II) : A solution of [Ni(CN)4]2- is colourless.

A Statement I incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Both Statement I and Statement II are incorrect

Statement I is correct but Statement II is

D
incorrect

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 Solution:
In order to determine the correctness of the given
statements, we need to analyze the nature of the
mentioned complexes.
Statement (I) asserts that a solution of [N(H2O)6]2-is green
in color. This complex involves a nickel(II) cation
coordinated with six water molecules. Nickel(II) has the
electronic configuration [Ar]3d8. In an octahedral field
created by water ligands, the d-orbitals split into two sets,
t2g and eg, due to crystal field splitting. The presence of
unpaired electrons allows for d-d transitions when light is
absorbed, which is responsible for the characteristic color
of the complex. Since typical [Ni(H2O)6]2+ complexes are
indeed green in color due to such transitions, Statement (I)
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 Solution:
Statement (II) states that a solution of [Ni(CN)4]2- is
colorless. In this complex, nickel(II) is surrounded by four
cyanide ligands. Cyanide is a strong field ligand, which
leads to a large crystal field splitting that exceeds the
pairing energy of the electrons. Hence, all the electrons in
the nickel(II) ion are paired, and there are no unpaired
electrons available for d-d transitions. As there are no d-d
transitions to absorb light in the visible region, the complex
appears colorless. Therefore, Statement (II) is also correct.
Given that both statements are correct, the most
appropriate answer would be :
Option B : Both Statement I and Statement II are correct.

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 JEE Main 2024 : 31st Jan S2
28
Select the option with correct property -

A [Ni(CO)4] and [NiCl4]2- both paramagnetic

B [Ni(CO)4] and [NiCl4]2- both diamagnetic

C [NiCl4]2- diamagnetic, [Ni(CO)4] paramagnetic

D [Ni(CO)4] Diamagnetic [NiCl4]2- paramagnetic

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 JEE Main 2024 : 31st Jan S2
28
Select the option with correct property -

A [Ni(CO)4] and [NiCl4]2- both paramagnetic

B [Ni(CO)4] and [NiCl4]2- both diamagnetic

C [NiCl4]2- diamagnetic, [Ni(CO)4] paramagnetic

D [Ni(CO)4] Diamagnetic [NiCl4]2- paramagnetic

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 Solution:
[Ni(CO)4] ➝ diamagnetic, sp3 hybridisation, number of
unpaired electrons = 0
[NiCl4]2-, ➝ paramagnetic, sp3 hybridisation, number of
unpaired electrons = 2

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 JEE Main 2024 : 31st Jan S2
29 Match List I with List II
List I (Complex ion) List II (Electronic Configuration)
A. [Cr(H2O)6]3+ I. t2g2 eg0
B. [Fe(H2O)6]3+ II. t2g3 eg0
C. [Ni(H2O)6]2+ III. t2g3 eg2
D. [V(H2)6]3+ IV. t2g6 eg2
Choose the correct answer from the options given below:

A A-IV, B-I, C-II, D-III

B A-III, B-II, C-IV, D-I

C A-II, B-III, C-IV, D-I

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D A-IV, B-III, C-I, D-II
 JEE Main 2024 : 31st Jan S2
29 Match List I with List II
List I (Complex ion) List II (Electronic Configuration)
A. [Cr(H2O)6]3+ I. t2g2 eg0
B. [Fe(H2O)6]3+ II. t2g3 eg0
C. [Ni(H2O)6]2+ III. t2g3 eg2
D. [V(H2)6]3+ IV. t2g6 eg2
Choose the correct answer from the options given below:

A A-IV, B-I, C-II, D-III

B A-III, B-II, C-IV, D-I

C A-II, B-III, C-IV, D-I

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D A-IV, B-III, C-I, D-II
 Solution:
[Cr(H2O)6]3+ Contains Cr3+ : [Ar]3d3 : t2g3 : t2g3 eg0
[Fe(H2O)6]3+ Contains Fe3+ : [Ar]3d5 : t2g3 eg2
[Ni(H2O)6]2+ Contains Ni2+ : [Ar]3d8 : t2g6 eg2
[V(H2O)6]3+ Contains V3+ : [Ar]3d2 : t2g2 eg0

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 JEE Main 2024 : 31st Jan S1
30 The correct statements from following are:
A. The strength of anionic ligands can be explained by crystal
field theory.
B. Valence bond theory does not give a quantitative
interpretation of kinetic stability of coordination compounds.
C. The hybridization involved in formation of [Ni(CN)4]2- complex
is dsp2.
D. The number of possible isomer(s) of cis-[PtCl2(en)2]2+ is one
Choose the correct answer from the options given below:

A B, C only

B B, D only

C A, C only

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D A, D only
 JEE Main 2024 : 31st Jan S1
30 The correct statements from following are:
A. The strength of anionic ligands can be explained by crystal
field theory.
B. Valence bond theory does not give a quantitative
interpretation of kinetic stability of coordination compounds.
C. The hybridization involved in formation of [Ni(CN)4]2- complex
is dsp2.
D. The number of possible isomer(s) of cis-[PtCl2(en)2]2+ is one
Choose the correct answer from the options given below:

A B, C only

B B, D only

C A, C only

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D A, D only
 Solution:
B. VBT does not explain stability of complex
C. Hybridisation of [Ni(CN)4]-2 is dsp2.

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 JEE Main 2024 : 30th Jan S2
31
The coordination geometry around the manganese in
decacarbonyldimanganese (0) is

A Trigonal bipyramidal

B Square pyramidal

C Square planar

D Octahedral

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 JEE Main 2024 : 30th Jan S2
31
The coordination geometry around the manganese in
decacarbonyldimanganese (0) is

A Trigonal bipyramidal

B Square pyramidal

C Square planar

D Octahedral

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 Solution:
Mn2(CO)10

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 JEE Main 2024 : 30th Jan S1
32 Choose the correct statements from the following :
(A) Ethane-1, 2-diamine is a chelating ligand.
(B) Metallic aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite.
(C) Cyanide ion is used as ligand for leaching of silver.
(D) Phosphine act as a ligand in Wilkinson catalyst.
(E) The stability constants of Ca2+ and Mg2+ are similar with
EDTA complexes.
Choose the correct answer from the options given below :

A (B), (C), (E) only B (A), (D), (E) only

C (C), (D), (E) only B (A), (B), (C) only

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 JEE Main 2024 : 30th Jan S1
32 Choose the correct statements from the following :
(A) Ethane-1, 2-diamine is a chelating ligand.
(B) Metallic aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite.
(C) Cyanide ion is used as ligand for leaching of silver.
(D) Phosphine act as a ligand in Wilkinson catalyst.
(E) The stability constants of Ca2+ and Mg2+ are similar with
EDTA complexes.
Choose the correct answer from the options given below :

A (B), (C), (E) only B (A), (D), (E) only

C (C), (D), (E) only B (A), (B), (C) only

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 Solution:

Based on Hall-Heroults process

[Rh(PPh3)3Cl] Wilkinson’s catalyst
Air
Ag2S + NaCNn ⇌ Na[Ag(CN)2] + Na2 S

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 JEE Main 2024 : 29th Jan S2
33
A reagent which gives brilliant red precipitate with
Nickel ions in basic medium is

A Diimethyl glyoxime

B Sodium nitroprusside

C meta-dinitrobenzene

D Neutral FeCl3

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 JEE Main 2024 : 29th Jan S2
33
A reagent which gives brilliant red precipitate with
Nickel ions in basic medium is

A Diimethyl glyoxime

B Sodium nitroprusside

C meta-dinitrobenzene

D Neutral FeCl3

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 Solution:
Ni2+ + 2dmg- ➝ [Ni(dmg)2]
Rosy red/Bright Red precipitate

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 JEE Main 2024 : 29th Jan S1
34 Match List I with List II
List I (Substances) List II (Element Present)
(A) Ziegler catalyst (I) Rhodium
(B) Blood Pigment (II) Cobalt
(C) Wilkinson catalyst (III) Iron
(D) Vitamin B12 (IV) Titanium
Choose the correct answer from the options given below:

A A-IV, B-III, C-I, D-II

B A-II, B-IV, C-I, D-III

C A-III, B-II, C-IV, D-I

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D A-II, B-III, C-IV, D-I
 JEE Main 2024 : 29th Jan S1
34 Match List I with List II
List I (Substances) List II (Element Present)
(A) Ziegler catalyst (I) Rhodium
(B) Blood Pigment (II) Cobalt
(C) Wilkinson catalyst (III) Iron
(D) Vitamin B12 (IV) Titanium
Choose the correct answer from the options given below:

A A-IV, B-III, C-I, D-II

B A-II, B-IV, C-I, D-III

C A-III, B-II, C-IV, D-I

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D A-II, B-III, C-IV, D-I
 Solution:
Ziegler catalyst ➝ Titanium
Blood pigment ➝ Iron
Wilkinson catalyst ➝ Rhodium
Vitamin B12 ➝ Cobalt

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 JEE Main 2024 : 29th Jan S1
35
In which one of the following metal carbonyls, CO forms
a bridge between metal atoms?

A [Os3(CO)12]

B [Ru3(CO)12]

C [Mn2(CO)10]

D [Co2(CO)8]

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 JEE Main 2024 : 29th Jan S1
35
In which one of the following metal carbonyls, CO forms
a bridge between metal atoms?

A [Os3(CO)12]

B [Ru3(CO)12]

C [Mn2(CO)10]

D [Co2(CO)8]

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 Solution:

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 JEE Main 2024 : 27th Jan S2
36
Identity the incorrect pair from the following :

A Haber process - Iron

B Polythene preparation - TiCl4, Al(CH3)3

C Photography - AgBr

D Wacker process - PtCl2

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 JEE Main 2024 : 27th Jan S2
36
Identity the incorrect pair from the following :

A Haber process - Iron

B Polythene preparation - TiCl4, Al(CH3)3

C Photography - AgBr

D Wacker process - PtCl2

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 Solution:
The catalyst used in Wacker's process is PdCl2

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 JEE Main 2024 : 27th Jan S2
37
Identify from the following species in which d2sp3
hybridization is shown by central atom :

A [Co(NH3)6]3+

B SF6

C [Pt(Cl4)]2-

D BrF5

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 JEE Main 2024 : 27th Jan S2
37
Identify from the following species in which d2sp3
hybridization is shown by central atom :

A [Co(NH3)6]3+

B SF6

C [Pt(Cl4)]2-

D BrF5

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 Solution:
[Co(NH3)6]+3 - d2sp3 hybridization
BrF5 - sp3d2 hybridization
[PtCl4]-2 -dsp2 hybridization
SF6 - sp3d2 hybridization

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 JEE Main 2024 : 27th Jan S1
38 Consider the following complex ions
P = [FeF6]3-
Q = [V(H2O)6]2+
R = [Fe(H2O)6]2+
The correct order of the complex ions, according to their
spin only magnetic moment values (in B.M.) is :

A R<Q<P

B R<P<Q

C Q<R<P

D Q<P<R
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 JEE Main 2024 : 27th Jan S1
38 Consider the following complex ions
P = [FeF6]3-
Q = [V(H2O)6]2+
R = [Fe(H2O)6]2+
The correct order of the complex ions, according to their
spin only magnetic moment values (in B.M.) is :

A R<Q<P

B R<P<Q

C Q<R<P

D Q<P<R
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 Solution:
[FeF6]3- : Fe+3 : [Ar] 3d5
F : Weak field Ligand
No. of unpaired electrons = 5
μ = √5(5 + 2)
μ = √35 BM
[V(H2O)6]+2 : V+2 : 3d3

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 Solution:
No. of unpaired electrons = 3
μ = √3(3 + 2)
μ = √15 BM
[Fe(H2O)6]+2 : Fe+2 : 3d6
H2O : Weak field Ligand
No. of unpaired electrons = 4
μ = √4(4 + 2)
μ = √24 BM

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 JEE Main 2024 : 27th Jan S1
39
Yellow compound of lead chromate gets dissolved on
treatment with hot NaOH solution. The product of lead
formed is a :

Tetra Anionic complex with coordination number

A
six

B Neutral complex with coordination number four

C Dianionic complex with coordination number six

Dianionic complex with coordination number

D
four

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 JEE Main 2024 : 27th Jan S1
39
Yellow compound of lead chromate gets dissolved on
treatment with hot NaOH solution. The product of lead
formed is a :

Tetra Anionic complex with coordination number

A
six

B Neutral complex with coordination number four

C Dianionic complex with coordination number six

Dianionic complex with coordination number

D
four

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 Solution:
Lead chromate (PbCrO4) dissolves in hot NaOH solution to
form products due to a chemical reaction. The reaction
involves the formation of a compound where lead (Pb) is
coordinated by hydroxide ions (OH-).
The balanced chemical reaction is:
PbCrO4 + 4NaOH(hot, excess) ➝ [Pb(OH)4]2- + Na2CrO4
In this reaction, lead forms a dianionic complex [Pb(OH)4]2-
with hydroxide ions. This indicates that the lead is in the
center of a complex with a coordination number of four,
meaning it is directly bound to four hydroxide ions.
Therefore, the correct description of the product formed
with lead is a dianionic complex with a coordination
number of four.
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 JEE Main 2024 : 8th April S2
40 Consider the following test for a group-IV cation.
M2+ + H2S ➝ A (Black precipitate) + byproduct
A + aqua regia ➝ B + NOCl + S + H2O
B + KNO2 + CH3COOH ➝ C + byproduct
The spin-only magnetic moment value of the metal
complex C is _____ BM (Nearest Integer)

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 JEE Main 2024 : 8th April S2
40 Consider the following test for a group-IV cation.
M2+ + H2S ➝ A (Black precipitate) + byproduct
A + aqua regia ➝ B + NOCl + S + H2O
B + KNO2 + CH3COOH ➝ C + byproduct
The spin-only magnetic moment value of the metal
complex C is _____ BM (Nearest Integer)

Correct Answer : 0

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 Solution:

ln K3[Co(NO2)6], Co+3 : 3d6 4s0

Co3+ : d2sp3 hybridisation
Number of unpaired e- = 0
Magnetic moment = √n(n + 2) = 0 B.M
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 JEE Main 2024 : 9th April S1
41
Number of ambidentate ligands among the following is_
NO2-, SCN-, C2O42-, NH3, CN-, SO42-, H2O

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 JEE Main 2024 : 9th April S1
41
Number of ambidentate ligands among the following is_
NO2-, SCN-, C2O42-, NH3, CN-, SO42-, H2O

Correct Answer : 3

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 Solution:
Ambidentate ligands are ligands that can attach to a central
metal atom through two different atoms. Let's analyze the
given ligands one by one:
NO2- : This ligand can bind through either the nitrogen (N) or
the oxygen (O). Therefore, NO2- is an ambidentate ligand.
SCN- : This ligand can bind through either the sulfur (S) or the
nitrogen (N). Therefore, SCN- is an ambidentate ligand.
C2O42- (oxalate): This is a bidentate ligand, but it coordinates
through both oxygen atoms only. Therefore, it is not an
ambidentate ligand.
NH3 : This ligand can only bind through nitrogen (N).
Therefore, it is not an ambidentate ligand.
CN- : This ligand can bind through either carbon (C) or
nitrogen (N). Therefore, CN- is an ambidentate ligand.
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 Solution:
SO42- (sulfate): This ligand typically coordinates through
oxygen atoms. It is not considered an ambidentate ligand.
H2O: This ligand can only bind through oxygen (O). Therefore,
it is not an ambidentate ligand.
Based on this analysis, the ambidentate ligands are NO2-,
SCN-, and CN-.
Therefore, the number of ambidentate ligands among the
given options is 3.

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 JEE Main 2024 : 8th April S2
42
Total number of unpaired electrons in the complex ions
[Co(NH3)6]3+ and [NiCl4]2- is

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 JEE Main 2024 : 8th April S2
42
Total number of unpaired electrons in the complex ions
[Co(NH3)6]3+ and [NiCl4]2- is

Correct Answer : 2

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 Solution:
[Co(NH3)6]3+ : Co3+ : t2g6 eg0 : n = 0
[NiCl4]2- : Ni2+ : t2g6 eg2 : n = 2

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 JEE Main 2024 : 8th April S1
43 The 'spin only' magnetic moment value of MO42-
is____BM. (Where M is a metal having least metallic
radii. among Sc, Ti, V, Cr, Mn and Zn).
(Given atomic number: Sc = 21, Ti = 22, V = 23,
Cr = 24, Mn = 25 and Zn = 30)

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 JEE Main 2024 : 8th April S1
43 The 'spin only' magnetic moment value of MO42-
is____BM. (Where M is a metal having least metallic
radii. among Sc, Ti, V, Cr, Mn and Zn).
(Given atomic number: Sc = 21, Ti = 22, V = 23,
Cr = 24, Mn = 25 and Zn = 30)

Correct Answer : 0

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 Solution:
Metal having least metallic radii among Sc, Ti, V, Cr,
Mmn & Zn is Cr.
Spin only magnetic moment of CrO42-.
Here Cr+6 is in d0 configuration (diamagnetic).

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 JEE Main 2024 : 6th April S1
44 The difference in the 'spin-only' magnetic moment
values of KMnO4 and the manganese product formed
during titration of KMnO4 against oxalic acid in acidic
medium is______BM. (nearest integer)

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 JEE Main 2024 : 6th April S1
44 The difference in the 'spin-only' magnetic moment
values of KMnO4 and the manganese product formed
during titration of KMnO4 against oxalic acid in acidic
medium is______BM. (nearest integer)

Correct Answer : 6

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 Solution:
The 'spin-only' magnetic moment for Mn7+ is 0 BM.
During the titration of KMnO4 with oxalic acid in an
acidic medium, manganese is reduced to Mn2+, which
has a 'spin-only' magnetic moment of 5.91 BM.
Thus, the difference in magnetic moment values is
calculated as follows:
0 BM (for Mn7+) subtracted from 5.91 BM (for Mn2+),
which results in 5.91 BM.
Rounding this to the nearest integer, we get a
difference of approximately 6 BM.

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 JEE Main 2024 : 5th April S1
45 The spin-only magnetic moment value of the ion among
Ti2+, Ti2+, V2+, Co3+ and Cr2+, that acts 3+ BM (Near
integer). as strong oxidising agent in aqueous solution
is_____ BM (Near Integer).
(Given atomic numbers: Ti : 22, V : 23, Cr : 24, Co : 27)

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 JEE Main 2024 : 5th April S1
45 The spin-only magnetic moment value of the ion among
Ti2+, Ti2+, V2+, Co3+ and Cr2+, that acts 3+ BM (Near
integer). as strong oxidising agent in aqueous solution
is_____ BM (Near Integer).
(Given atomic numbers: Ti : 22, V : 23, Cr : 24, Co : 27)

Correct Answer : 5

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 Solution:
The ion which acts as strong oxidising agent in
aqueous solution is Cr2+ : [Ar] 4s°3d4
μ = √4(4 + 2) = 4.89 ⇒ 5

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 JEE Main 2024 : 31st Jan S1
46
The ‘Spin only’ Magnetic moment for [Ni(NH3)6]2+
is___× 10-1 BM. (given = Atomic number of Ni : 28)

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 JEE Main 2024 : 31st Jan S1
46
The ‘Spin only’ Magnetic moment for [Ni(NH3)6]2+
is___× 10-1 BM. (given = Atomic number of Ni : 28)

Correct Answer : 28

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 Solution:
NH3 act as WFL with Ni2+
Ni2+ = 3d8

No. of unpaired electron = 2

μ = √n(n + 2) = √8 = 2.82 BM
= 28.2 × 10-1 BM
x = 28

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 JEE Main 2024 : 30th Jan S2
47 Number of complexes which show optical isomerism
among the following is _______
cis- [Cr(ox)2Cl2]3-, [Co(en)3]3+, cis- [Pt(en)2Cl2]2+,
cis- [Co(en)2Cl2]+, trans- [Pt(en)2Cl2]2+,
trans- [Cr(ox)2Cl2]3-

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 JEE Main 2024 : 30th Jan S2
47 Number of complexes which show optical isomerism
among the following is _______
cis- [Cr(ox)2Cl2]3-, [Co(en)3]3+, cis- [Pt(en)2Cl2]2+,
cis- [Co(en)2Cl2]+, trans- [Pt(en)2Cl2]2+,
trans- [Cr(ox)2Cl2]3-

Correct Answer : 4

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 Solution:
cis-[Cr(ox)2Cl2]3- ➝ can show optical isomerism
(no POS & COS)
[Co(en)3]3+ ➝ can show (no POS & COS)
cis-[Pt(en)2Cl2]2+ → can show (no POS & COS)
cis-[Co(en)2Cl2]+ ➝ can show (no POS & COS)
trans-[Pt(en)2Cl2]2+ ➝ can't show
(contains POS & COS)
trans[Cr(ox)2Cl2]3- ➝ can't show (contains POS & COS)

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 JEE Main 2024 : 27th Jan S2
48 The spin only magnetic moment value of square planar
complex [Pt(NH3)2Cl(NH2CH3)]Cl is _____ B.M.
(nearest Integer) (Given atomic number for Pt = 78)

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 JEE Main 2024 : 27th Jan S2
48 The spin only magnetic moment value of square planar
complex [Pt(NH3)2Cl(NH2CH3)]Cl is _____ B.M.
(nearest Integer) (Given atomic number for Pt = 78)

Correct Answer : 0

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 Solution:
Pt2 + (d8)

Pt2+ ➝ dsp2 hybridization and have no unpaired e- s.

∴ Magnetic moment = 0

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