The correct statement among the following is :

Q: The correct statement among the following is :

Analyze the oxidation state, electronic configuration, and ligand field strength for each Nickel complex: [Ni(CN)4]2-: Ni2+ is 3d8. CN^- is a strong field ligand. It causes pairing of electrons, resulting in dsp2 hybridization and 0 unpaired electrons (Diamagnetic). Ni(CO)4: Ni0 is 3d8 4s2. CO is a strong field ligand, forcing the 4s electrons into the 3d subshell, resulting in a 3d10 configuration. It undergoes sp3 hybridization with 0 unpaired electrons (Diamagnetic). [NiCl4]2-: Ni2+ is 3d8. C

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More Coordination Compounds Previous-Year Questions — Page 4

Q73 jee_main_2026_28_january_morning Isomerism in Coordination Compounds

X is the number of geometrical isomers exhibited by

[Pt(NH₃)(H₂O)BrCl]

. Y is the number of optically inactive isomer(s) exhibited by

[CrCl₂(ox)₂]³⁻

Z is the number of geometrical isomers exhibited by

[Co(NH₃)₃(NO₂)₃]

The value of

X + Y + Z

is _____.

Numerical Answer. Answer: 6 to 6

Solution

Step 1: Evaluate X

$[Pt(NH₃)(H₂O)BrCl]$ $[Pt(NH_3)(H_2O)BrCl]$ is a square planar complex of the type $[Mabcd]$ $[Mabcd]$. Such complexes exhibit exactly 3 geometrical isomers (by fixing one ligand and placing the other three opposite to it, creating two cis-like and one trans-like variations relative to the fixed ligand).\nThus, $X = 3$ $X = 3$.

Step 2: Evaluate Y

$[CrCl₂(ox)₂]³⁻$ $[CrCl_2(ox)_2]^{3-}$ is an octahedral complex of the type $[M(AA)₂b₂]$ $[M(AA)_2b_2]$. It has two geometrical isomers: cis and trans.\n$\bullet$ The cis-isomer lacks a plane of symmetry and is optically active (exists as a pair of enantiomers).\n$\bullet$ The trans-isomer has a plane of symmetry and is optically inactive (meso).\nThe question asks for the number of optically inactive isomers. Thus, $Y = 1$ $Y = 1$.

Step 3: Evaluate Z

$[Co(NH₃)₃(NO₂)₃]$ $[Co(NH_3)_3(NO_2)_3]$ is an octahedral complex of the type $[Ma₃b₃]$ $[Ma_3b_3]$. It exhibits exactly 2 geometrical isomers: Facial (fac) and Meridional (mer).\nThus, $Z = 2$ $Z = 2$.

Final Conclusion

$X + Y + Z = 3 + 1 + 2 = 6$ $X + Y + Z = 3 + 1 + 2 = 6$

Pattern Recognition

$[Mabcd]$ $[Mabcd]$ planar = 3 G.I. $[Ma₃b₃]$ $[Ma_3b_3]$ octahedral = 2 G.I. (fac/mer). $[M(AA)₂b₂]$ $[M(AA)_2b_2]$ octahedral trans isomer = ALWAYS optically inactive.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q69 jee_main_2026_28_january_evening Magnetic Moments Of Complexes

The correct increasing order of spin-only magnetic moment values of the complex ions

[MnBr₄]²⁻

(A),

[Cu(H₂O)₆]²⁺

(B),

[Ni(CN₄)]²⁻

[Ni(H₂O)₆]²⁺

(D) is:

A. $(1) A = B < C < D$ $(1)\text{ } A = B < C < D$
B. $(2) A = B < D < C$ $(2)\text{ } A = B < D < C$
C. $(3) C = D < B < A$ $(3)\text{ } C = D < B < A$
D. $(4) C < B < D < A$ $(4)\text{ } C < B < D < A$

Solution

Related Formula

μ = \sqrt(n(n+2)) B.M.

where $n$ $n$ is the number of unpaired electrons.

Core Logic

(A) $[MnBr₄]²⁻$ $[MnBr_{4}]^{2-}$: $Mn²⁺$ $Mn^{2+}$ has $3d⁵$ $3d^5$ configuration. $Br^-$ $Br^-$ is a weak field ligand (WFL). It forms a tetrahedral complex with 5 unpaired electrons ( $n=5$ $n=5$).

(B) $[Cu(H₂O)₆]²⁺$ $[Cu(H_{2}O)_{6}]^{2+}$: $Cu²⁺$ $Cu^{2+}$ has $3d⁹$ $3d^9$ configuration. It has exactly 1 unpaired electron ( $n=1$ $n=1$).

(C) $[Ni(CN₄)]²⁻$ $[Ni(CN_{4})]^{2-}$: $Ni²⁺$ $Ni^{2+}$ has $3d⁸$ $3d^8$ configuration. $CN^-$ $CN^-$ is a strong field ligand (SFL). It forms a square planar complex ( $dsp²$ $dsp^2$) forcing pairing. Unpaired electrons $n=0$ $n=0$.

(D) $[Ni(H₂O)₆]²⁺$ $[Ni(H_{2}O)_{6}]^{2+}$: $Ni²⁺$ $Ni^{2+}$ has $3d⁸$ $3d^8$ configuration. $H₂O$ $H_2O$ is a weak field ligand (WFL). In an octahedral field, it leaves 2 unpaired electrons ( $n=2$ $n=2$).

Step 1: Final Conclusion

Unpaired electrons (n): (C) = 0 (B) = 1 (D) = 2 (A) = 5 Increasing order of magnetic moment: C < B < D < A.

Pattern Recognition

Nickel + Strong Ligand ( $CN^-$ $CN^-$) = Square Planar, Diamagnetic ( $n=0$ $n=0$). Nickel + Weak Ligand ( $H₂O$ $H_2O$) = Octahedral, Paramagnetic ( $n=2$ $n=2$). Copper (II) is always $d⁹$ $d^9$ ( $n=1$ $n=1$). Manganese (II) with WFL is high spin $d⁵$ $d^5$ ( $n=5$ $n=5$).

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q75 jee_main_2026_28_january_evening Crystal Field Theory

The number of isoelectronic species among

Sc³⁺

Cr²⁺

Mn³⁺

Co³⁺

and

Fe³⁺

is 'n'. If 'n' moles of AgCl is formed during the reaction of complex with formula

CoCl₃(en)₂NH₃

with excess of

AgNO₃

solution, then the number of electrons present in the

t2g

orbital of the complex is ____.

Numerical Answer. Answer: 6 to 6

Solution

Core Logic

Step 1: Find 'n' (number of isoelectronic species) Electrons count: $Sc³⁺ = 21 - 3 = 18$ $Sc^{3+} = 21 - 3 = 18$ $Cr²⁺ = 24 - 2 = 22$ $Cr^{2+} = 24 - 2 = 22$ $Mn³⁺ = 25 - 3 = 22$ $Mn^{3+} = 25 - 3 = 22$ $Co³⁺ = 27 - 3 = 24$ $Co^{3+} = 27 - 3 = 24$ $Fe³⁺ = 26 - 3 = 23$ $Fe^{3+} = 26 - 3 = 23$ The isoelectronic species are $Cr²⁺$ $Cr^{2+}$ and $Mn³⁺$ $Mn^{3+}$ (both have 22 electrons). So, $n = 2$ $n = 2$.

Step 2: Complex formulation. Reaction produces 'n' moles of AgCl, so 2 moles of $Cl^-$ $Cl^-$ are ionizable outside the coordination sphere. Complex formula is $[Co(en)₂(NH₃)Cl]Cl₂$ $[Co(en)_2(NH_3)Cl]Cl_2$.

Step 3: Crystal field splitting. The central metal ion is $Co³⁺$ $Co^{3+}$ ( $3d⁶$ $3d^6$ configuration). The ligands are $en$ $en$ (strong field) and $NH₃$ $NH_3$ (strong field), which will cause complete pairing of electrons. $Cl^-$ $Cl^-$ inside the sphere is weak, but the overwhelming presence of $en$ $en$ and $NH₃$ $NH_3$ combined with $+3$ $+3$ oxidation state makes it a strong field (low spin) complex. $Co³⁺$ $Co^{3+}$ ( $3d⁶$ $3d^6$) low spin splitting: $t2g⁶ eg⁰$ $t_{2g}^{6} \; e_g^{0}$.

Step 1: Final Conclusion

The number of electrons present in the $t2g$ $t_{2g}$ orbital of the complex is 6.

Pattern Recognition

$Co³⁺$ $Co^{3+}$ in predominantly N-donor environment forms low spin octahedral complexes.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Structure of Atom

Q27 jee_main_2025_02_april_evening Crystal Field Stabilization Energy

The d-orbital electronic configuration of the complex among

[Co(en)₃]³⁺

[CoF₆]³⁻

[Mn(H₂O)₆]²⁺

and

[Zn(H₂O)₆]²⁺

that has the highest CFSE is:

A. $t2g⁶eg⁰$
B. $t2g⁶eg⁴$
C. $t2g³eg²$
D. $t2g⁴eg²$

Solution

Related Formula

CFSE = ( -0.4 n_t2g + 0.6 n_eg ) Δₒ + nₚ P

Core Logic

Crystal Field Stabilization Energy (CFSE) is maximized (becomes most negative) when electrons populate lower-energy $t2g$ $\mathrm{t_{2g}}$ orbitals and stay out of higher-energy $eg$ $\mathrm{e_g}$ orbitals. This is favored by strong-field ligands (SFL) that induce large $Δₒ$ $\Delta_o$ splitting, leading to low-spin configurations.

Step 1: Analyze Ligand Strength and Configuration

Let us check each of the given complexes:

$[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$: Here $Co³⁺$ $\mathrm{Co^{3+}}$ has a $3d⁶$ $3\mathrm{d^6}$ configuration. Since ethylenediamine ( $en$ $\mathrm{en}$) is a strong-field ligand, it causes pairing of all $6$ $6$ electrons in the $t2g$ $\mathrm{t_{2g}}$ subshell. The configuration is $t2g⁶eg⁰$ .

d-orbital splitting diagram for low-spin Co3+ complex with t2g6 configuration

$[CoF₆]³⁻$ $[\mathrm{CoF_6}]^{3-}$: $Co³⁺$ $\mathrm{Co^{3+}}$ is $3d⁶$ $3\mathrm{d^6}$. Since $F^-$ $\mathrm{F^-}$ is a weak-field ligand (WFL), no pairing occurs. The configuration is $t2g⁴eg²$ .
$[Mn(H₂O)₆]²⁺$ $[\mathrm{Mn(H_2O)_6}]^{2+}$: $Mn²⁺$ $\mathrm{Mn^{2+}}$ is $3d⁵$ $3\mathrm{d^5}$. Since $H₂O$ $\mathrm{H_2O}$ is a weak-field ligand, the configuration is high-spin: $t2g³eg²$ .
$[Zn(H₂O)₆]²⁺$ $[\mathrm{Zn(H_2O)_6}]^{2+}$: $Zn²⁺$ $\mathrm{Zn^{2+}}$ is $3 d¹⁰$ $3\mathrm{d^{10}}$. The d-subshell is fully filled, yielding $t2g⁶eg⁴$ .

Step 2: Compare CFSE Values

Calculating CFSE (neglecting pairing energy term for simplicity):

For $[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$: $CFSE = 6 \times (-0.4 Δₒ) = -2.4 Δₒ$
For $[CoF₆]³⁻$ $[\mathrm{CoF_6}]^{3-}$: $CFSE = [4(-0.4) + 2(0.6)] Δₒ = -0.4 Δₒ$
For $[Mn(H₂O)₆]²⁺$ $[\mathrm{Mn(H_2O)_6}]^{2+}$: $CFSE = [3(-0.4) + 2(0.6)] Δₒ = 0$
For $[Zn(H₂O)₆]²⁺$ $[\mathrm{Zn(H_2O)_6}]^{2+}$: $CFSE = [6(-0.4) + 4(0.6)] Δₒ = 0$

Hence, $[Co(en)₃]³⁺$ $[\mathrm{Co(en)_3}]^{3+}$ has the highest crystal field stabilization energy, corresponding to the d-orbital electronic configuration $t2g⁶eg⁰$ .

Pattern Recognition

For octahedral complexes of $d⁶$ $\mathrm{d^6}$ metals, a low-spin configuration ( $t2g⁶eg⁰$ ) achieves the theoretical maximum orbital stabilization since the $eg$ $\mathrm{e_g}$ levels are completely empty and $t2g$ $\mathrm{t_{2g}}$ is fully filled.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Solution & Explanation

Core Logic

Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Coordination Compounds Previous-Year Questions — Page 4

Solution

Step 1: Evaluate X

Step 2: Evaluate Y

Step 3: Evaluate Z

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyze Ligand Strength and Configuration

Step 2: Compare CFSE Values

Pattern Recognition

Chapter Mix

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