### Core Logic Identify the central metal ion, its oxidation state, and outer electronic configuration: 1) [Cr(H_2O)_6]^3+$[Cr(H_2O)_6]^{3+}$ contains Cr^3+ : [Ar] 3d^3$Cr^{3+} : [Ar] 3d^3$. In an octahedral field with weak field ligands (H_2O$H_2O$), the configuration is t_2g^3 e_g^0$t_{2g}^3 e_g^0$. 2) [Fe(H_2O)_6]^3+$[Fe(H_2O)_6]^{3+}$ contains Fe^3+ : [Ar] 3d^5$Fe^{3+} : [Ar] 3d^5$. With weak field ligand (H_2O$H_2O$), it forms a high spin complex: t_2g^3 e_g^2$t_{2g}^3 e_g^2$. 3) [Ni(H_2O)_6]^2+$[Ni(H_2O)_6]^{2+}$ contains Ni^2+ : [Ar] 3d^8$Ni^{2+} : [Ar] 3d^8$. In an octahedral field, the configuration is t_2g^6 e_g^2$t_{2g}^6 e_g^2$. 4) [V(H_2O)_6]^3+$[V(H_2O)_6]^{3+}$ contains V^3+ : [Ar] 3d^2$V^{3+} : [Ar] 3d^2$. In an octahedral field, the configuration is t_2g^2 e_g^0$t_{2g}^2 e_g^0$. ### Step 1: Final Mapping A rightarrow$\rightarrow$ II (t_2g^3e_g^0$t_{2g}^3e_g^0$) B rightarrow$\rightarrow$ III (t_2g^3e_g^2$t_{2g}^3e_g^2$) C rightarrow$\rightarrow$ IV (t_2g^6e_g^2$t_{2g}^6e_g^2$) D rightarrow$\rightarrow$ I (t_2g^2e_g^0$t_{2g}^2e_g^0$) This matches option (4). ### Pattern Recognition Shortcut: Vanadium (V) in +3$+3$ state has 2$2$ electrons, so D rightarrow I$D \rightarrow I$. Only options (1) and (4) have D rightarrow I$D \rightarrow I$. Nickel (Ni) in +2$+2$ state has 8$8$ electrons, so C rightarrow IV$C \rightarrow IV$. Only option (4) has C rightarrow IV$C \rightarrow IV$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic For [Ni(CO)_4]$[Ni(CO)_4]$: Nickel is in 0$0$ oxidation state: Ni^0 = [Ar] 3d^8 4s^2$Ni^0 = [Ar] 3d^8 4s^2$. CO$CO$ is a strong field ligand, forcing pairing of electrons. The 4s$4s$ electrons move to the 3d$3d$ orbital, making the configuration 3d^10$3d^{10}$. With no unpaired electrons, the hybridization is sp^3$sp^3$ and it is diamagnetic. For [NiCl_4]^2-$[NiCl_4]^{2-}$: Nickel is in +2$+2$ oxidation state: Ni^2+ = [Ar] 3d^8$Ni^{2+} = [Ar] 3d^8$. Cl^-$Cl^-$ is a weak field ligand, so pairing does not occur against Hund's rule. The configuration remains t_2g^6 e_g^2$t_{2g}^6 e_g^2$ (or simply two unpaired electrons in the tetrahedral d$d$ splitting). The hybridization is sp^3$sp^3$ and with 2 unpaired electrons, it is paramagnetic. ### Step 1: Final Conclusion [Ni(CO)_4]$[Ni(CO)_4]$ is diamagnetic, while [NiCl_4]^2-$[NiCl_4]^{2-}$ is paramagnetic. This correctly corresponds to option (4). ### Pattern Recognition Strong field ligands (like CO$CO$, CN^-$CN^-$) usually lead to diamagnetism in d^8$d^8$ metal ions by forming square planar (dsp^2$dsp^2$) or forcing d^10$d^{10}$ configurations (for Ni^0$Ni^0$). Weak field halogens lead to paramagnetic sp^3$sp^3$ complexes for Ni(II)$Ni(II)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Step 1: Analyzing Statement A Crystal field theory considers ligands as point charges. Therefore, anionic ligands should exert the greatest splitting effect, but practically they are at the lower end of the spectrochemical series. CFT cannot explain this properly. Hence, Statement A is incorrect. ### Step 2: Analyzing Statement B Valence bond theory (VBT) is qualitative and does not give a quantitative interpretation of either thermodynamic or kinetic stability. Hence, Statement B is correct. ### Step 3: Analyzing Statement C In [Ni(CN)_4]^2-$[Ni(CN)_4]^{2-}$, Ni$Ni$ is in +2$+2$ state (3d^8$3d^8$). Since CN^-$CN^-$ is a strong field ligand, pairing of electrons takes place leaving one inner 3d orbital empty. Thus, the hybridization is dsp^2$dsp^2$ (square planar). Hence, Statement C is correct. ### Step 4: Analyzing Statement D cis-[PtCl_2(en)_2]^2+$cis-[PtCl_2(en)_2]^{2+}$ is an octahedral complex of type M(AA)_2a_2$M(AA)_2a_2$. The cis-isomer lacks a plane of symmetry and is optically active, thus it exists as a pair of enantiomers (d and l). Therefore, the number of possible isomers is 2, not 1. Statement D is incorrect. ### Final Conclusion Statements B and C are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Related Formula mu = sqrtn(n + 2) text BM$\mu = \sqrt{n(n + 2)} \text{ BM}$ ### Step 1: Electronic Configuration Atomic number of Ni = 28$Ni = 28$. Ni^2+ = [Ar] 3d^8$Ni^{2+} = [Ar] 3d^8$ Because the geometry is octahedral (coordination number 6) and NH_3$NH_3$ acts as a weak field ligand with Ni^2+$Ni^{2+}$ (or because d^8$d^8$ configuration always has 2 unpaired electrons in an octahedral field regardless of ligand strength): The configuration in the t_2g$t_{2g}$ and e_g$e_g$ levels is t_2g^6 e_g^2$t_{2g}^6 e_g^2$. ### Step 2: Calculating Magnetic Moment Number of unpaired electrons, n = 2$n = 2$. mu = sqrt2(2 + 2) = sqrt8 approx 2.828 text BM$\mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.828 \text{ BM}$ Representing in 10^-1$10^{-1}$ scale: mu approx 28.28 times 10^-1 text BM$\mu \approx 28.28 \times 10^{-1} \text{ BM}$ Rounding to the nearest integer as typically expected gives 28. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds