The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is

Core Logic

Analyze the oxidation state, electronic configuration, and ligand field strength for each Nickel complex:\n$\bullet$ [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$. CN^-$CN^-$ is a strong field ligand. It causes pairing of electrons, resulting in dsp²$dsp^2$ hybridization and 0 unpaired electrons (Diamagnetic).\n$\bullet$ Ni(CO)₄$Ni(CO)_4$: Ni⁰$Ni^0$ is 3d⁸ 4s²$3d^8 4s^2$. CO$CO$ is a strong field ligand, forcing the 4s$4s$ electrons into the 3d$3d$ subshell, resulting in a 3d¹⁰$3d^{10}$ configuration. It undergoes sp³$sp^3$ hybridization with 0 unpaired electrons (Diamagnetic).\n$\bullet$ [NiCl₄]²⁻$[NiCl_4]^{2-}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$. Cl^-$Cl^-$ is a weak field ligand. No pairing occurs, resulting in sp³$sp^3$ hybridization with 2 unpaired electrons (Paramagnetic).

Final Conclusion

Ni(CO)₄$Ni(CO)_4$ and [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$ are diamagnetic, while [NiCl₄]²⁻$[NiCl_4]^{2-}$ is paramagnetic.

Pattern Recognition

Ni with strong ligands (CN^-, CO$CN^-, CO$) collapses into paired diamagnetic states. Ni with weak halogens (Cl^-$Cl^-$) stays paramagnetic and tetrahedral.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

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

Related Formula

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

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

Related Formula

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$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.



• [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$\mathrm{t}_{2\mathrm{g}}^{4}\mathrm{e}_{\mathrm{g}}^{2}$.
• [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$\mathrm{t}_{2\mathrm{g}}^{3}\mathrm{e}_{\mathrm{g}}^{2}$.
• [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$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{4}$.

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

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$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$.

Pattern Recognition

For octahedral complexes of d⁶$\mathrm{d^6}$ metals, a low-spin configuration (t2g⁶eg⁰$\mathrm{t}_{2\mathrm{g}}^{6}\mathrm{e}_{\mathrm{g}}^{0}$) 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