Given below are two statements: Statement I: Among [Cu(NH3)4]2+, [Ni(en)3]2+, [Ni(NH3)6]2+ and [Mn(H2O)6]2+, [Mn(H2O)6]2+ has the maximum number of unpaired electrons. Statement II: The number of pairs among [NiCl4]2-, [Ni(CO)4], [NiCl4]2-, [Ni(CN)4]2- and [Ni(CO)4], [Ni(CN)4]2- that contain only diamagnetic species is two. In the light of the above statements, choose the correct answer from the options given below:

Evaluating Statement I: - [Cu(NH3)4]2+: Cu2+ is 3d9, 1 unpaired electron. - [Ni(en)3]2+: Ni2+ is 3d8, in octahedral field, 2 unpaired electrons. - [Ni(NH3)6]2+: Ni2+ is 3d8, 2 unpaired electrons. - [Mn(H2O)6]2+: Mn2+ is 3d5, weak field ligand H2O leads to high spin, 5 unpaired electrons. So [Mn(H2O)6]2+ has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: - [Ni(CO)4]: Ni(0) is 3d8 4s2, strong field CO pairs electrons to 3d10, diamagnetic (0 unpaired). - [Ni(CN)4]2-: Ni2+ is 3d8, strong field CN^- forces pairing arrow dsp2 square planar, diamagnetic (0 unpaired). - [NiCl4]2-: Ni2+ is 3d8, weak field Cl^- does not pair arrow sp3 tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: - [NiCl4]2-, [Ni(CO)4] arrow 1 para, 1 dia (No) - [NiCl4]2-, [Ni(CN)4]2- arrow 1 para, 1 dia (No) - [Ni(CO)4], [Ni(CN)4]2- arrow Both dia (Yes) The number of such pairs is exactly ONE. Statement II says two, so it is false. Statement I is true, Statement II is false. Class 12 Chemistry: Coordination Compounds

Given below are two statements: Statement I: Among [Cu(NH3)4]2+, [Ni(en)3]2+, [Ni(NH3)6]2+ and [Mn(H2O)6]2+, [Mn(H2O)6]2+ has the maximum number of unpaired electrons. Statement II: The number of pairs among [NiCl4]2-, [Ni(CO)4], [NiCl4]2-, [Ni(CN)4]2- and [Ni(CO)4], [Ni(CN)4]2- that contain only diamagnetic species is two. In the light of the above statements, choose the correct answer from the options given below:

Evaluating Statement I: - [Cu(NH3)4]2+: Cu2+ is 3d9, 1 unpaired electron. - [Ni(en)3]2+: Ni2+ is 3d8, in octahedral field, 2 unpaired electrons. - [Ni(NH3)6]2+: Ni2+ is 3d8, 2 unpaired electrons. - [Mn(H2O)6]2+: Mn2+ is 3d5, weak field ligand H2O leads to high spin, 5 unpaired electrons. So [Mn(H2O)6]2+ has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: - [Ni(CO)4]: Ni(0) is 3d8 4s2, strong field CO pairs electrons to 3d10, diamagnetic (0 unpaired). - [Ni(CN)4]2-: Ni2+ is 3d8, strong field CN^- forces pairing arrow dsp2 square planar, diamagnetic (0 unpaired). - [NiCl4]2-: Ni2+ is 3d8, weak field Cl^- does not pair arrow sp3 tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: - [NiCl4]2-, [Ni(CO)4] arrow 1 para, 1 dia (No) - [NiCl4]2-, [Ni(CN)4]2- arrow 1 para, 1 dia (No) - [Ni(CO)4], [Ni(CN)4]2- arrow Both dia (Yes) The number of such pairs is exactly ONE. Statement II says two, so it is false. Statement I is true, Statement II is false. Class 12 Chemistry: Coordination Compounds

Solution & Explanation

### Core Logic Evaluating Statement I: -

[Cu(NH_3)_4]^2+

Cu^2+

3d^9

, 1 unpaired electron. -

[Ni(en)_3]^2+

Ni^2+

3d^8

, in octahedral field, 2 unpaired electrons. -

[Ni(NH_3)_6]^2+

Ni^2+

3d^8

, 2 unpaired electrons. -

[Mn(H_2O)_6]^2+

Mn^2+

3d^5

, weak field ligand

H_2O

leads to high spin, 5 unpaired electrons. So

[Mn(H_2O)_6]^2+

has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: -

[Ni(CO)_4]

Ni(0)

3d^8 4s^2

, strong field CO pairs electrons to

3d^10

, diamagnetic (0 unpaired). -

[Ni(CN)_4]^2-

Ni^2+

3d^8

, strong field

CN^-

forces pairing

rightarrow dsp^2

square planar, diamagnetic (0 unpaired). -

[NiCl_4]^2-

Ni^2+

3d^8

, weak field

Cl^-

does not pair

rightarrow sp^3

tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: -

\[NiCl_4]^2-, [Ni(CO)_4]\

rightarrow

1 para, 1 dia (No) -

\[NiCl_4]^2-, [Ni(CN)_4]^2-\

rightarrow

1 para, 1 dia (No) -

\[Ni(CO)_4], [Ni(CN)_4]^2-\

rightarrow

Both dia (Yes) The number of such pairs is exactly ONE. Statement II says two, so it is false. ### Step 1: Final Conclusion Statement I is true, Statement II is false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

More Coordination Compounds Previous-Year Questions — Page 3

Q41 jee_main_2025_08_april_evening Valence Bond Theory

Determine the total number of chemical species from the list below that are specifically involved in an

sp^3d^2

hybridization state:

[textCo(NH_3)_6]^3+, text SF_6, \, [textCrF_6]^3-, \, [textCoF_6]^3-, \, [textMn(CN)_6]^3-, text and [textMnCl_6]^3-

A. 5
B. 6
C. 4
D. 3

Solution

### Core Logic Let us systematically determine the hybridization configuration of each species: 1. **

[textCo(NH_3)_6]^3+

**:

textCo^3+

has a

3d^6

configuration. Ammonia (

textNH_3

) acts as a Strong Field Ligand (SFL), forcing the pairing of

3d

electrons. This leaves two internal

3d

orbitals vacant, leading to an **inner orbital**

d^2sp^3

hybridization. 2. **

textSF_6

**: Central sulfur has 6 valence electrons, forming 6 single bonds. Steric number

= 6

, resulting in a regular outer octahedral **

sp^3d^2

** hybridization. 3. **

[textCrF_6]^3-

**:

textCr^3+

has a

3d^3

configuration. The

t_2g

subshell holds 3 unpaired electrons, leaving the two

e_g

orbitals empty regardless of ligand field strength. This results in a

d^2sp^3

hybridization. 4. **

[textCoF_6]^3-

**:

textCo^3+

has a

3d^6

configuration. Fluoride (

F^-

) is a Weak Field Ligand (WFL) and cannot induce spin pairing. Thus, the complex utilizes outer shell

4d

orbitals, yielding an **outer orbital** **

sp^3d^2

** configuration. 5. **

[textMn(CN)_6]^3-

**:

textMn^3+

has a

3d^4

configuration. Cyanide (

textCN^-

) is a Strong Field Ligand (SFL), inducing pairing to leave two

3d

slots vacant, giving a

d^2sp^3

hybridization. 6. **

[textMnCl_6]^3-

**:

textMn^3+

has a

3d^4

configuration. Chloride (

textCl^-

) is a Weak Field Ligand (WFL) and cannot cause pairing. It utilizes the outer

4d

shell, resulting in an **outer orbital** **

sp^3d^2

** hybridization. Counting the outer-orbital

sp^3d^2

species:

textSF_6

[textCoF_6]^3-

, and

[textMnCl_6]^3-

. Total count

= 3

. ### Pattern Recognition Outer orbital complexes (

sp^3d^2

) require weak field ligands (like

F^-, Cl^-

) paired with metal configurations where internal

d

-orbitals cannot be cleared by pairing (

d^4, d^5, d^6

SF_6

is a primary group molecule that always uses outer-shell d-orbitals. This brings our total to 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q44 jee_main_2025_08_april_evening Valence Bond Theory

Match the coordination complexes listed in LIST-I with their geometric shape and magnetic moment characteristics described in LIST-II:

LIST-I (Complex/Species)	LIST-II (Shape & magnetic moment)
A. $[textNi(CO)_4]$ $[\text{Ni(CO)}_4]$	I. Tetrahedral, 2.8 BM
B. $[textNi(CN)_4]^2-$ $[\text{Ni(CN)}_4]^{2-}$	II. Square planar, 0 BM
C. $[textNiCl_4]^2-$ $[\text{NiCl}_4]^{2-}$	III. Tetrahedral, 0 BM
D. $[textMnBr_4]^2-$ $[\text{MnBr}_4]^{2-}$	IV. Tetrahedral, 5.9 BM

Choose the correct answer from the options given below:

A. $textA-III, B-IV, C-II, D-I$ $\text{A-III, B-IV, C-II, D-I}$
B. $textA-I, B-II, C-III, D-IV$ $\text{A-I, B-II, C-III, D-IV}$
C. $textA-III, B-II, C-I, D-IV$ $\text{A-III, B-II, C-I, D-IV}$
D. $textA-IV, B-I, C-III, D-II$ $\text{A-IV, B-I, C-III, D-II}$

Solution

### Core Logic Let us apply Valence Bond Theory (VBT) and crystal field rules to evaluate each coordination complex: * **A.

[textNi(CO)_4]

**: Nickel is in the

0

oxidation state (

3d^8 4s^2

). Carbon monoxide (

textCO

) is a strong field ligand, forcing the

4s

electrons into the

3d

shell to produce a fully paired

3d^10

configuration. The vacant

4s

and three

4p

orbitals hybridize into an **

sp^3

tetrahedral** geometry. All spins are paired, so

mu = 0 text BM

. Thus,

textA rightarrow textIII

Valence orbital diagram for nickel tetracarbonyl sp3 system

* **B.

[textNi(CN)_4]^2-

**: Nickel is in the

+2

state (

3d^8

). Cyanide (

textCN^-

) is a strong field ligand, forcing the pairing of the two unpaired

3d

electrons. This leaves one internal

3d

orbital vacant, leading to **

dsp^2

square planar** hybridization with zero unpaired electrons (

mu = 0 text BM

). Thus,

textB rightarrow textII

* **C.

[textNiCl_4]^2-

**: Nickel is in the

+2

state (

3d^8

). Chloride (

textCl^-

) is a weak field ligand, leaving the two

3d

electrons unpaired (

n = 2

). The system adopts **

sp^3

tetrahedral** hybridization with a spin-only moment of

mu = sqrt2(2+2) = sqrt8 approx 2.8 text BM

. Thus,

textC rightarrow textI

* **D.

[textMnBr_4]^2-

**: Manganese is in the

+2

state (

3d^5

). Bromide (

textBr^-

) is a weak field ligand, preserving five unpaired parallel spins (

n = 5

). The geometry is **

sp^3

tetrahedral** with a maximum spin-only moment of

mu = sqrt5(5+2) = sqrt35 approx 5.9 text BM

. Thus,

textD rightarrow textIV

### Step 1: Alignment Summary Consolidating our results:

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

This matches Option (3). ### Pattern Recognition Nickel complexes provide classic benchmarks: Nickel zero tetracarbonyl is always tetrahedral diamagnetic. Nickel

+2

tetracyanide is square planar diamagnetic due to strong ligand field pairing. Spotting these properties cuts down the problem solving time significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q26 jee_main_2025_29_jan_evening Magnetic Properties of Coordination Compounds

The calculated spin-only magnetic moments of

K_3[Fe(OH)_6]

and

K_4[Fe(OH)_6]

respectively are: (1) 4.90 and 4.90 B.M. (2) 5.92 and 4.90 B.M. (3) 3.87 and 4.90 B.M. (4) 4.90 and 5.92 B.M.

A. 4.90 and 4.90 B.M.
B. 5.92 and 4.90 B.M.
C. 3.87 and 4.90 B.M.
D. 4.90 and 5.92 B.M.

Solution

### Related Formula

mu = sqrtn(n+2)text B.M.

### Core Logic In

K_3[Fe(OH)_6]

, iron is in the

+3

oxidation state (

Fe^3+ = 3d^5

). Since

OH^-

is a weak field ligand, no pairing of electrons takes place. The number of unpaired electrons (

n

) is

5

mu = sqrt5(5+2) = sqrt35 approx 5.92text B.M.

K_4[Fe(OH)_6]

, iron is in the

+2

oxidation state (

Fe^2+ = 3d^6

). Since

OH^-

is a weak field ligand, no pairing occurs. The number of unpaired electrons (

n

) is

4

mu = sqrt4(4+2) = sqrt24 approx 4.90text B.M.

### Pattern Recognition Identify the ligand field strength first.

OH^-

is a weak field ligand in the spectrochemical series, so it does not cause pairing in either

Fe^2+

Fe^3+

configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q37 jee_main_2025_29_jan_evening Homoleptic Complexes and Electronic Configurations

Identify the homoleptic complexes with odd number of d electrons in the central metal. (A)

[FeO_4]^2-

(B)

[Fe(CN)_6]^3-

(C)

[Fe(CN)_5NO]^2-

(D)

[CoCl_4]^2-

(E)

[Co(H_2O)_3F_3]

Choose the correct answer from the options given below:

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

Solution

### Core Logic A complex is homoleptic if the metal is bound to only one kind of donor ligand group. * (A)

[FeO_4]^2-

is homoleptic, but

Fe^+6

corresponds to a

3d^2

(even) electronic configuration. * (B)

[Fe(CN)_6]^3-

is homoleptic.

Fe^+3

corresponds to a

3d^5

(odd) configuration. * (C)

[Fe(CN)_5NO]^2-

is heteroleptic (contains two types of ligands). * (D)

[CoCl_4]^2-

is homoleptic.

Co^+2

corresponds to a

3d^7

(odd) configuration. * (E)

[Co(H_2O)_3F_3]

is heteroleptic. ### Pattern Recognition Filter by 'homoleptic' first to instantly eliminate multi-ligand mixed structures like options (C) and (E). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Q46 jee_main_2025_29_jan_evening Crystal Field Theory and Colors of Complexes

Consider the following low-spin complexes

K_3[Co(NO_2)_6]

K_4[Fe(CN)_6]

K_3[Fe(CN)_6]

Cu_2[Fe(CN)_6]

and

Zn_2[Fe(CN)_6]

. The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer is nearest integer)

Numerical Answer. Answer: 0 to 0

Solution

### Core Logic From the given list, the complexes exhibiting a distinct yellow color are

K_3[Co(NO_2)_6]

and

K_4[Fe(CN)_6]

. Let's calculate the spin-only magnetic moments for these low-spin configurations: 1) For

K_3[Co(NO_2)_6]

, cobalt is in

+3

oxidation state (

Co^3+ = 3d^6

). In the presence of the strong ligand field (

NO_2^-

), all six electrons pair up completely in the

t_2g

orbitals:

t_2g^6 e_g^0 implies n = 0 text unpaired electrons implies mu = 0text BM

Crystal Field Theory and Colors of Complexes diagram for Q46 - JEE Main 2025 Evening

2) For

K_4[Fe(CN)_6]

, iron is in

+2

oxidation state (

Fe^2+ = 3d^6

). In the strong field of cyanide ligands (

CN^-

), pairing is complete:

t_2g^6 e_g^0 implies n = 0 text unpaired electrons implies mu = 0text BM

Therefore, the sum of their spin-only magnetic moments is

0 + 0 = 0

. ### Pattern Recognition Low-spin

d^6

octahedral complexes always yield a fully closed-shell

t_2g^6

arrangement with zero unpaired electrons, leading deterministically to a magnetic moment of 0 BM. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

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