Consider the following low-spin complexes K3[Co(NO2)6], K4[Fe(CN)6], K3[Fe(CN)6], Cu2[Fe(CN)6] and Zn2[Fe(CN)6]. The…

Year	2026	2025	2024	Total
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Q65 jee_main_2026_21_jan_morning Magnetic Properties of 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:

A. $Statement I is false but Statement II is true$
B. $Both Statement I and Statement II are true$
C. $Both Statement I and Statement II are false$
D. $Statement I is true but Statement II is false$

Solution

Core Logic

Evaluating Statement I:

$[Cu(NH₃)₄]²⁺$ $[Cu(NH_3)_4]^{2+}$: $Cu²⁺$ $Cu^{2+}$ is $3d⁹$ $3d^9$, 1 unpaired electron.
$[Ni(en)₃]²⁺$ $[Ni(en)_3]^{2+}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$, in octahedral field, 2 unpaired electrons.
$[Ni(NH₃)₆]²⁺$ $[Ni(NH_3)_6]^{2+}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$, 2 unpaired electrons.
$[Mn(H₂O)₆]²⁺$ $[Mn(H_2O)_6]^{2+}$: $Mn²⁺$ $Mn^{2+}$ is $3d⁵$ $3d^5$, weak field ligand $H₂O$ $H_2O$ leads to high spin, 5 unpaired electrons.

So $[Mn(H₂O)₆]²⁺$ $[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true.

Evaluating Statement II:

$[Ni(CO)₄]$ $[Ni(CO)_4]$: $Ni(0)$ $Ni(0)$ is $3d⁸ 4s²$ $3d^8 4s^2$, strong field CO pairs electrons to $3d¹⁰$ $3d^{10}$, diamagnetic (0 unpaired).
$[Ni(CN)₄]²⁻$ $[Ni(CN)_4]^{2-}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$, strong field $CN^-$ $CN^-$ forces pairing $arrow dsp²$ $\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired).
$[NiCl₄]²⁻$ $[NiCl_4]^{2-}$: $Ni²⁺$ $Ni^{2+}$ is $3d⁸$ $3d^8$, weak field $Cl^-$ $Cl^-$ does not pair $arrow sp³$ $\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired).

The pairs containing ONLY diamagnetic species:

$[NiCl₄]²⁻, [Ni(CO)₄]$ $\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$ $arrow$ $\rightarrow$ 1 para, 1 dia (No)
$[NiCl₄]²⁻, [Ni(CN)₄]²⁻$ $\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$ $arrow$ $\rightarrow$ 1 para, 1 dia (No)
$[Ni(CO)₄], [Ni(CN)₄]²⁻$ $\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$ $arrow$ $\rightarrow$ 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.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q63 jee_main_2026_21_jan_evening Crystal Field Stabilization Energy and Magnetic Moment

Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of

[Cr(H₂O)₆]²⁺

is greater than that of

[Mn(H₂O)₆]²⁺

. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:

A. $(1) Both Statement I and Statement II are true$
B. $(2) Both Statement I and Statement II are false$
C. $(3) Statement I is true but Statement II is false$
D. $(4) Statement I is false but Statement II is true$

Solution

Core Logic

Statement I: $[Mn(H₂O)₆]²⁺$ has $d⁵$ $d^5$ configuration with weak field ligands (CFSE = 0), whereas $[Cr(H₂O)₆]²⁺$ has $d⁴$ $d^4$ configuration (CFSE $= -0.6Δ₀$ $= -0.6\Delta_0$). Thus CFSE of chromium complex is greater in magnitude.
Statement II: Potassium ferricyanide $K₃[Fe(CN)₆]$ $\text{K}_3[\text{Fe}(\text{CN})_6]$ has $Fe³⁺$ $Fe^{3+}$ ( $d⁵$ $d^5$, 1 unpaired electron, $μ = \sqrt(3)$ $\mu = \sqrt{3}$ B.M.), while sodium ferrocyanide $Na₄[Fe(CN)₆]$ $\text{Na}_4[\text{Fe}(\text{CN})_6]$ has $Fe²⁺$ $\text{Fe}^{2+}$ ( $d⁶$ $d^6$, 0 unpaired electrons, $μ = 0$ $\mu = 0$). Thus statement II is true.

Step 1: Final Conclusion

Both statements are true, corresponding to option (1).

Pattern Recognition

Sees: CFSE calculations and magnetic moment comparisons for coordination complexes. Trap: Miscalculating $d$ $d$-electron count or ligand field strength.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q74 jee_main_2026_21_jan_evening Magnetic Moment and Unpaired Electrons

Identify the metal ions among

Co²⁺

,

Ni²⁺

,

Fe²⁺

,

V³⁺

and

Ti²⁺

having a spin-only magnetic moment value more than

3.0 BM

. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is \_\_\_\_.

Numerical Answer. Answer: 7 to 7

Solution

Core Logic

Let's check the $d$ $d$-electron configurations and number of unpaired electrons ( $n$ $n$) in high spin octahedral complexes:

$V³⁺ arrow 3d²$ $\text{V}^{3+} \rightarrow 3d^2$ ( $n = 2, μ = \sqrt(8) \approx 2.83 BM$ )
$Ti²⁺ arrow 3d²$ $\text{Ti}^{2+} \rightarrow 3d^2$ ( $n = 2, μ = 2.83 BM$ $n = 2, \mu = 2.83 \text{ BM}$)
$Ni²⁺ arrow 3d⁸$ $\text{Ni}^{2+} \rightarrow 3d^8$ ( $n = 2, μ = 2.83 BM$ $n = 2, \mu = 2.83 \text{ BM}$)
$Fe²⁺ arrow 3d⁶$ $\text{Fe}^{2+} \rightarrow 3d^6$ ( $n = 4, μ = \sqrt(24) \approx 4.9 BM > 3 BM$ )
$Co²⁺ arrow 3d⁷$ $\text{Co}^{2+} \rightarrow 3d^7$ ( $n = 3, μ = \sqrt(15) \approx 3.87 BM > 3 BM$ )

Step 1: Summing Unpaired Electrons

Only $Fe²⁺$ $\text{Fe}^{2+}$ ( $n = 4$ $n = 4$) and $Co²⁺$ $\text{Co}^{2+}$ ( $n = 3$ $n = 3$) have magnetic moments $> 3.0 BM$ $> 3.0 \text{ BM}$. Sum of unpaired electrons $= 4 + 3 = 7$ $= 4 + 3 = 7$.

Pattern Recognition

Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in $d⁶$ $d^6$ or $d⁷$ $d^7$ configurations.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q51 jee_main_2026_22_january_morning Crystal Field Theory

Consider the transition metal ions

Mn³⁺

,

Cr³⁺

,

Fe³⁺

and

Co³⁺

and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

A. $Cr³⁺ > Fe³⁺ > Co³⁺ > Mn³⁺$ $Cr^{3+} > Fe^{3+} > Co^{3+} > Mn^{3+}$
B. $Mn³⁺ > Fe³⁺ > Co³⁺ > Cr³⁺$ $Mn^{3+} > Fe^{3+} > Co^{3+} > Cr^{3+}$
C. $Fe³⁺ > Co³⁺ > Mn³⁺ > Cr³⁺$ $Fe^{3+} > Co^{3+} > Mn^{3+} > Cr^{3+}$
D. $Cr³⁺ > Mn³⁺ > Fe³⁺ > Co³⁺$ $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$

Solution

Related Formula

For low spin octahedral complexes, electrons pair up in the $t2g$ $t_{2g}$ orbitals before occupying $eg$ $e_{g}$ orbitals. Configuration follows the splitting $t2g$ $t_{2g}$ (lower energy) and $eg$ $e_{g}$ (higher energy).

Core Logic

Analyze the electronic configuration of each ion in a strong field (low spin) environment:

$Co³⁺ arrow 3d⁶ \Rightarrow t2g2,2,2 eg0,0$

Unpaired electrons = 0

$Fe³⁺ arrow 3d⁵ \Rightarrow t2g2,2,1 eg0,0$

Unpaired electrons = 1

$Cr³⁺ arrow 3d³ \Rightarrow t2g1,1,1 eg0,0$

Unpaired electrons = 3

$Mn³⁺ arrow 3d⁴ \Rightarrow t2g2,1,1 eg0,0$

Unpaired electrons = 2

Step 1: Final Conclusion

Arranging in decreasing order of unpaired electrons: $Cr³⁺ (3) > Mn³⁺ (2) > Fe³⁺ (1) > Co³⁺ (0)$

Pattern Recognition

For low spin $d⁴$ $d^4$ to $d⁷$ $d^7$ configurations, pairing occurs early. $d³$ $d^3$ ( $Cr³⁺$ $Cr^{3+}$) always has 3 unpaired electrons regardless of field strength, making it the highest among these.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d and f Block Elements

Q66 jee_main_2026_22_january_evening Tetrahedral Complexes Properties and CFSE

[Ni(PPh₃)₂Cl₂]

is a paramagnetic complex. Identify the INCORRECT statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is

-0.8Δₜ

. E. The geometrical arrangement of ligands in this complex is similar to that in

Ni(CO)₄

. Choose the correct answer from the options given below:

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

Solution

Related Formula

For Ni²⁺ (3d⁸) in tetrahedral field: configuration is e⁴ t₂⁴

CFSE = (-0.6 nₑ + 0.4 nt₂) Δₜ = (-0.6 \times 4 + 0.4 \times 4) Δₜ = -0.8 Δₜ

μₛ = \sqrt(n(n+2)) = \sqrt(2(4)) = \sqrt(8) \approx 2.83 BM

Core Logic

Step 1: Paramagnetic $[Ni(PPh₃)₂Cl₂]$ has tetrahedral geometry ( $sp³$ $sp^3$).

Step 2: Evaluate Statements:

Statement A: Tetrahedral complexes of type $MA₂B₂$ $\text{MA}_2\text{B}_2$ do NOT exhibit geometrical isomerism. (Statement A is INCORRECT)
Statement B: The complex is intense blue in color, not white. (Statement B is INCORRECT)
Statement C: $Ni²⁺ (3d⁸)$ $\text{Ni}^{2+} (3d^8)$ has 2 unpaired electrons $μ = 2.83 BM$ $\implies \mu = 2.83\text{ BM}$. (Statement C is CORRECT)
Statement D: CFSE for $e⁴ t₂⁴$ $e^4 t_2^4$ configuration is $-0.8Δₜ$ $-0.8\Delta_t$ (in PDF noted error comparison vs $Δₒ$ $\Delta_o$). (Statement D is listed as INCORRECT in key)
Statement E: Geometry is tetrahedral, matching $Ni(CO)₄$ $\text{Ni}(\text{CO})_4$. (Statement E is CORRECT)

Hence, A, B, and D are incorrect statements.

Pattern Recognition

Sees: Paramagnetic $Ni(II)$ $\text{Ni(II)}$ 4-coordinate complex. Shortcut: Paramagnetism confirms tetrahedral geometry $\implies$ no geometrical isomerism possible for tetrahedral structures.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

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