Coordination Compounds: JEE Main 2026 PYQs

Answer: Option (4)

Key idea: Crystal Field Splitting Energy and Spectrochemical Series

1. Recall the spectrochemical series for ligand field strength: $CN^- > en > H_2O > F^-$
2. Stronger field ligands produce larger crystal field splitting energy ($\Delta_0$)
3. $[Cr(CN)_6]^{3-}$: CN⁻ is strongest field ligand → highest $\Delta_0 = 26,600$ cm⁻¹ (IV)
4. $[CrF_6]^{3-}$: F⁻ is weakest field ligand → lowest $\Delta_0 = 15,060$ cm⁻¹ (I)
5. $[Cr(H_2O)_6]^{3+}$: H₂O is moderate field → $\Delta_0 = 17,400$ cm⁻¹ (II)
6. $[Cr(en)_3]^{3+}$: en is strong field (between CN⁻ and H₂O) → $\Delta_0 = 22,300$ cm⁻¹ (III)
7. Match: A-IV, B-I, C-II, D-III → Option (4)

Answer: 1

Key idea: Ionization Energy Trends and Crystal Field Stabilization Energy (CFSE)

1. Write electronic configurations after removing 2 electrons: Sc²⁺: [Ar]3d¹, Mn²⁺: [Ar]3d⁵, Cu²⁺: [Ar]3d⁹, Zn²⁺: [Ar]3d¹⁰
2. For third ionization energy (IE₃), Zn²⁺ has fully filled 3d¹⁰ (most stable) requiring highest energy; Sc²⁺ has lowest nuclear charge requiring lowest energy. Statement I is **correct**.
3. For [Co(H₂O)₆]²⁺: Co²⁺ is d⁷ high spin → t₂g⁵eₘ² → CFSE = -0.8Δₒ (small Δₒ)
4. For [Co(H₂O)₆]³⁺: Co³⁺ is d⁶ low spin → t₂g⁶eₘ⁰ → CFSE = -2.4Δₒ (moderate Δₒ)
5. For [Co(en)₃]³⁺: Co³⁺ is d⁶ low spin → t₂g⁶eₘ⁰ → CFSE = -2.4Δₒ (large Δₒ, since en is stronger field)
6. Since Δₒ(en, Co³⁺) > Δₒ(H₂O, Co³⁺) > Δₒ(H₂O, Co²⁺), the CFSE order is correct. Statement II is **correct**.
7. Both statements are correct, so answer is Option 1

Answer: Option (2)

Key idea: Coordination Isomerism in Coordination Compounds

1. Coordination isomerism occurs when ligands can be exchanged between cationic and anionic coordination entities in a compound containing both complex cation and complex anion.
2. For **A**: $[Ag(NH_3)_2][Ag(CN)_2]$ - Both complexes have Ag with coordination number 2. Ligand exchange would give neutral species $[Ag(NH_3)(CN)]$, disrupting ionic character. **Does not show coordination isomerism.**
3. For **B**: $[Co(NH_3)_6][Cr(CN)_6]$ - Different metals (Co and Cr). Can form isomer $[Cr(NH_3)_6][Co(CN)_6]$ by ligand exchange. **Shows coordination isomerism.**
4. For **C**: $[Co(NH_3)_6][Co(CN)_6]$ - Same metal in both cation and anion. Exchange gives the same compound due to equivalence of Co centers. **Does not show coordination isomerism.**
5. For **D**: $[Fe(NH_3)_6][Co(CN)_6]$ - Different metals. Can form isomer $[Co(NH_3)_6][Fe(CN)_6]$ (which is compound E). **Shows coordination isomerism.**
6. For **E**: $[Co(NH_3)_6][Fe(CN)_6]$ - Different metals. Can form isomer $[Fe(NH_3)_6][Co(CN)_6]$ (which is compound D). **Shows coordination isomerism.**
7. Therefore, compounds **B, D, and E only** show coordination isomerism.

Answer: Option (4)

Key idea: Spin-only magnetic moment calculation using μ = √(n(n+2)) BM for coordination complexes with weak field ligands

1. Use the formula $\mu = \sqrt{n(n+2)}$ BM where $n$ = number of unpaired electrons
2. For $[Cr(H_2O)_6]^{2+}$: Cr²⁺ has $3d^4$ configuration with 4 unpaired electrons, so $\mu = \sqrt{24} = 4.90$ BM → III
3. For $[Co(H_2O)_6]^{2+}$: Co²⁺ has $3d^7$ configuration with 3 unpaired electrons, so $\mu = \sqrt{15} = 3.87$ BM → I
4. For $[Cu(H_2O)_6]^{2+}$: Cu²⁺ has $3d^9$ configuration with 1 unpaired electron, so $\mu = \sqrt{3} = 1.73$ BM → IV
5. For $[Mn(H_2O)_6]^{2+}$: Mn²⁺ has $3d^5$ configuration with 5 unpaired electrons, so $\mu = \sqrt{35} = 5.92$ BM → II
6. Match: A-III, B-I, C-IV, D-II

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Key idea: Crystal field theory, magnetic properties, and spectrochemical series in coordination compounds

Answer: Option (2)

Key idea: Synergic bonding in metal carbonyls

1. Statement A: Metal carbonyls have both $\sigma$ and $\pi$ bonds due to synergic bonding. ✓ CORRECT
2. Statement B: Synergic bonding STRENGTHENS (not weakens) the M-C bond as $\sigma$ donation and $\pi$ back-donation are mutually reinforcing. ✗ INCORRECT
3. Statement C: The $\sigma$ bond forms when the lone pair on carbonyl carbon donates to vacant metal orbital. ✓ CORRECT
4. Statement D: The $\pi$ bond forms when filled metal d-orbitals donate electrons to vacant $\pi^*$ orbital of CO (back-donation). ✓ CORRECT
5. Correct statements: A, C, and D
6. Answer: Option (2)

Answer: Option (1)

Key idea: Crystal Field Theory - d-orbital splitting in octahedral complexes and energy stabilization/destabilization

1. **Statement I Check:** In octahedral CFT, d-orbitals split into $t_{2g}$ (lower) and $e_g$ (higher) sets
2. Each $e_g$ electron is destabilized by $+0.6\Delta_o$ and each $t_{2g}$ electron is stabilized by $-0.4\Delta_o$
3. Verify barycenter: $2(+0.6\Delta_o) + 3(-0.4\Delta_o) = 0$ ✓ Statement I is **correct**
4. **Statement II Check:** In free atomic state, all five d-orbitals are degenerate (same energy)
5. When ligands approach in complex formation, they create an electrostatic field that splits the d-orbitals into different energy levels
6. This destroys the degeneracy. Statement II is **correct**
7. Therefore, both statements are correct → Option (1)

Answer: Option (1)

Key idea: Crystal Field Theory - d-orbital splitting in octahedral and tetrahedral complexes, and relationship between unpaired electrons and enthalpy of atomization in transition metals

1. **Statement I:** More unpaired d-electrons in transition metals → more electrons participate in metallic bonding → stronger metal-metal bonds → higher enthalpy of atomization. Statement I is **correct**.
2. **Statement II - Part 1:** $[Fe(H_2O)_6]^{3+}$ is octahedral (weak field ligand H₂O). In octahedral field: $t_{2g}$ (lower) and $e_g$ (higher).
3. For octahedral: $d_{xy} = d_{xz} = d_{yz} < d_{x^2-y^2} = d_{z^2}$. This matches the statement. ✓
4. **Statement II - Part 2:** $[Ni(Cl)_4]^{2-}$ is tetrahedral (4 ligands, weak field Cl⁻). In tetrahedral field: $e$ (lower) and $t_2$ (higher).
5. For tetrahedral: $d_{x^2-y^2} = d_{z^2} < d_{xy} = d_{xz} = d_{yz}$. This matches the statement. ✓
6. Both Statement I and Statement II are correct.

Answer: Option (3)

Key idea: Stability of coordination complexes, chelate effect, hybridization, linkage isomerism, and ambidentate ligands

1. **Statement A**: $[Fe(C_2O_4)_3]^{3-}$ has oxalate (bidentate chelating ligand) which provides chelate effect, making it more stable than complexes with monodentate ligands OH⁻ and SCN⁻. **CORRECT**
2. **Statement B**: $[Cu(en)_2]^{2+}$ has ethylenediamine (bidentate) which is more stable than $[Cu(NH_3)_4]^{2+}$ with monodentate NH₃ due to chelate effect. **INCORRECT**
3. **Statement C**: In $K_4[Fe(CN)_6]$, Fe is in +2 state with configuration 3d⁶. CN⁻ is strong field, causing pairing. Hybridization is $d^2sp^3$ (inner orbital complex). **CORRECT**
4. **Statement D**: NO₂⁻ is ambidentate (can bind through N or O), so $[Fe(NO_2)_3Cl_3]^{3-}$ exhibits linkage isomerism (nitro vs nitrito). **CORRECT**
5. **Statement E**: Both NO₂⁻ (N or O binding) and SCN⁻ (S or N binding) ARE ambidentate ligands. **INCORRECT**
6. Correct statements are A, C, and D only

Answer: 15

Key idea: Crystal Field Theory - Electron Configuration and Unpaired Electrons in Octahedral Complexes

1. In octahedral complexes, d-orbitals split into $t_{2g}$ (3 orbitals, lower energy) and $e_g$ (2 orbitals, higher energy)
2. $d^3$: Configuration is $t_{2g}^3$ with all electrons unpaired = **3 unpaired electrons**
3. $d^4$ (low spin): Strong field causes pairing in $t_{2g}$. Configuration: $t_{2g}^4$ (↑↓ ↑ ↑) = **2 unpaired electrons**
4. $d^5$ (high spin): Weak field, all orbitals filled singly first. Configuration: $t_{2g}^3 e_g^2$ = **5 unpaired electrons**
5. $d^6$ (high spin): Configuration is $t_{2g}^4 e_g^2$ (↑↓ ↑ ↑ | ↑ ↑) = **4 unpaired electrons**
6. $d^7$ (low spin): Strong field causes maximum pairing. Configuration: $t_{2g}^6 e_g^1$ = **1 unpaired electron**
7. Total unpaired electrons = $3 + 2 + 5 + 4 + 1 = 15$

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Key idea: Crystal Field Stabilization Energy (CFSE) in Tetrahedral Complexes

Answer: Option (1)

Key idea: Crystal Field Theory - d-orbital splitting in tetrahedral complexes

1. In tetrahedral complexes, ligands approach along body diagonals, causing d-orbital splitting into two sets: $e$ (lower) and $t_2$ (higher)
2. Energy ordering: $e$ set contains $d_{z^2}$ and $d_{x^2-y^2}$ (lower energy), $t_2$ set contains $d_{xy}$, $d_{xz}$, $d_{yz}$ (higher energy)
3. Complete ordering: $d_{z^2} = d_{x^2-y^2} < d_{xy} = d_{xz} = d_{yz}$
4. Check A: $d_{xy} = d_{xz} > d_{x^2-y^2}$ → TRUE (both $t_2$ orbitals equal and higher than $e$ orbital)
5. Check B: $d_{xy} = d_{yz} > d_{z^2}$ → TRUE (both $t_2$ orbitals equal and higher than $e$ orbital)
6. Check C: $d_{x^2-y^2} > d_{z^2} > d_{xz}$ → FALSE ($d_{x^2-y^2} = d_{z^2}$ and both are lower than $d_{xz}$)
7. Check D: $d_{x^2-y^2} = d_{z^2} < d_{xz}$ → TRUE (both $e$ orbitals equal and lower than $t_2$ orbital)

Answer: Option (4)

Key idea: Hybridization, geometry, and magnetic properties of coordination compounds based on ligand field strength

1. For **A. $[NiCl_4]^{2-}$**: $Ni^{2+}$ has $3d^8$ configuration. $Cl^-$ is weak field, no pairing occurs. Tetrahedral with $sp^3$ hybridization and 2 unpaired electrons → **paramagnetic** ✓
2. For **B. $[Ni(NH_3)_6]^{2+}$**: $Ni^{2+}$ with $3d^8$. $NH_3$ is moderate field ligand. Octahedral with $sp^3d^2$ hybridization and 2 unpaired electrons → **paramagnetic** ✓
3. For **C. $[Ni(CO)_4]$**: $Ni^0$ has $3d^{10}$ after electron rearrangement. $CO$ is strong field causing complete pairing. Tetrahedral with $sp^3$ but **diamagnetic** (0 unpaired electrons) ✗
4. For **D. $[Ni(CN)_4]^{2-}$**: $Ni^{2+}$ with $3d^8$. $CN^-$ is strong field causing pairing. Square planar with $dsp^2$ hybridization and 0 unpaired electrons → **diamagnetic** ✓
5. Statement C is incorrect because $[Ni(CO)_4]$ is diamagnetic, not paramagnetic
6. Therefore, A, B, and D are correct sequences

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Key idea: Brown Ring Test for Nitrates - Formation of Nitrosyl Complex

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Key idea: Crystal Field Theory, Magnetic Properties of Coordination Complexes