Match the coordination complexes listed in LIST-I with their geometric shape and magnetic moment characteristics described in LIST-II:

LIST-I (Complex/Species)	LIST-II (Shape & magnetic moment)
A. [textNi(CO)_4]$[\text{Ni(CO)}_4]$	I. Tetrahedral, 2.8 BM
B. [textNi(CN)_4]^2-$[\text{Ni(CN)}_4]^{2-}$	II. Square planar, 0 BM
C. [textNiCl_4]^2-$[\text{NiCl}_4]^{2-}$	III. Tetrahedral, 0 BM
D. [textMnBr_4]^2-$[\text{MnBr}_4]^{2-}$	IV. Tetrahedral, 5.9 BM

Choose the correct answer from the options given below:

### Core Logic Let us apply Valence Bond Theory (VBT) and crystal field rules to evaluate each coordination complex: * **A. [textNi(CO)_4]$[\text{Ni(CO)}_4]$**: Nickel is in the 0$0$ oxidation state (3d^8 4s^2$3d^8 4s^2$). Carbon monoxide (textCO$\text{CO}$) is a strong field ligand, forcing the 4s$4s$ electrons into the 3d$3d$ shell to produce a fully paired 3d^10$3d^{10}$ configuration. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridize into an **sp^3$sp^3$ tetrahedral** geometry. All spins are paired, so mu = 0 text BM$\mu = 0 \text{ BM}$. Thus, textA rightarrow textIII$\text{A} \rightarrow \text{III}$.

* **B. [textNi(CN)_4]^2-$[\text{Ni(CN)}_4]^{2-}$**: Nickel is in the +2$+2$ state (3d^8$3d^8$). Cyanide (textCN^-$\text{CN}^-$) is a strong field ligand, forcing the pairing of the two unpaired 3d$3d$ electrons. This leaves one internal 3d$3d$ orbital vacant, leading to **dsp^2$dsp^2$ square planar** hybridization with zero unpaired electrons (mu = 0 text BM$\mu = 0 \text{ BM}$). Thus, textB rightarrow textII$\text{B} \rightarrow \text{II}$.

* **C. [textNiCl_4]^2-$[\text{NiCl}_4]^{2-}$**: Nickel is in the +2$+2$ state (3d^8$3d^8$). Chloride (textCl^-$\text{Cl}^-$) is a weak field ligand, leaving the two 3d$3d$ electrons unpaired (n = 2$n = 2$). The system adopts **sp^3$sp^3$ tetrahedral** hybridization with a spin-only moment of mu = sqrt2(2+2) = sqrt8 approx 2.8 text BM$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 \text{ BM}$. Thus, textC rightarrow textI$\text{C} \rightarrow \text{I}$.

* **D. [textMnBr_4]^2-$[\text{MnBr}_4]^{2-}$**: Manganese is in the +2$+2$ state (3d^5$3d^5$). Bromide (textBr^-$\text{Br}^-$) is a weak field ligand, preserving five unpaired parallel spins (n = 5$n = 5$). The geometry is **sp^3$sp^3$ tetrahedral** with a maximum spin-only moment of mu = sqrt5(5+2) = sqrt35 approx 5.9 text BM$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM}$. Thus, textD rightarrow textIV$\text{D} \rightarrow \text{IV}$.

### Step 1: Alignment Summary Consolidating our results: textA-III, B-II, C-I, D-IV$$\text{A-III, B-II, C-I, D-IV}$$ This matches Option (3). ### Pattern Recognition Nickel complexes provide classic benchmarks: Nickel zero tetracarbonyl is always tetrahedral diamagnetic. Nickel +2$+2$ tetracyanide is square planar diamagnetic due to strong ligand field pairing. Spotting these properties cuts down the problem solving time significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds