Total number of molecules/species from following which will be paramagnetic is O₂, O₂⁺, NO, NO₂, CO, K₂[NiCl₄], [Co(NH₃)₆]Cl₃, K₂[Ni(CN)₄]$$O_{2},\ O_{2}^{+},\ NO,\ NO_{2},\ CO,\ K_{2}[NiCl_{4}],\ [Co(NH_{3})_{6}]Cl_{3},\ K_{2}[Ni(CN)_{4}]$$

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

Core Logic

• Statement I: [Mn(H₂O)₆]²⁺$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ has d⁵$d^5$ configuration with weak field ligands (CFSE = 0), whereas [Cr(H₂O)₆]²⁺$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ 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 = √(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

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 = √(8) ≈ 2.83 BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$)
• 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 = √(24) ≈ 4.9 BM > 3 BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$)
• Co²⁺ arrow 3d⁷$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, μ = √(15) ≈ 3.87 BM > 3 BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$)

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

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 ⇒ t2g2,2,2 eg0,0$Co^{3+} \rightarrow 3d^{6} \Rightarrow t_{2g}^{2,2,2} e_{g}^{0,0}$

Unpaired electrons = 0

• Fe³⁺ arrow 3d⁵ ⇒ t2g2,2,1 eg0,0$Fe^{3+} \rightarrow 3d^{5} \Rightarrow t_{2g}^{2,2,1} e_{g}^{0,0}$

Unpaired electrons = 1

• Cr³⁺ arrow 3d³ ⇒ t2g1,1,1 eg0,0$Cr^{3+} \rightarrow 3d^{3} \Rightarrow t_{2g}^{1,1,1} e_{g}^{0,0}$

Unpaired electrons = 3

• Mn³⁺ arrow 3d⁴ ⇒ t2g2,1,1 eg0,0$Mn^{3+} \rightarrow 3d^{4} \Rightarrow t_{2g}^{2,1,1} e_{g}^{0,0}$

Unpaired electrons = 2

Step 1: Final Conclusion

Arranging in decreasing order of unpaired electrons: Cr³⁺ (3) > Mn³⁺ (2) > Fe³⁺ (1) > Co³⁺ (0)$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$

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

Related Formula

Core Logic

Step 1: Paramagnetic [Ni(PPh₃)₂Cl₂]$[\text{Ni}(\text{PPh}_3)_2\text{Cl}_2]$ 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