Coordination Compounds Part 4

Concept:The complex $[\mathrm{Ni}(\mathrm{PPh}_3)_2\mathrm{Cl}_2]$ is paramagnetic, which implies a tetrahedral geometry for $\mathrm{Ni}^{2+}$ ($3d^8$) because square planar would be diamagnetic.Explanation:$\mathrm{PPh}_3$ is neutral and each $\mathrm{Cl}$ is $-1$, so Ni is in $+2$ oxidation state.$\mathrm{Ni}^{2+}$ has $3d^8$ electron configuration.For tetrahedral $d^8$, the complex has two unpaired electrons.Statement A: In a tetrahedral $\mathrm{MA}_2\mathrm{B}_2$ complex, all positions are equivalent, so no geometrical isomerism exists. Hence A is incorrect.Statement B: Tetrahedral $\mathrm{Ni}^{2+}$ complexes are generally coloured due to $d$–$d$ transitions, not white. So B is incorrect.Statement C: Spin-only magnetic moment for $n=2$: $\mu_\text{so} = \sqrt{n(n+2)} = \sqrt{8} \approx 2.83\ \mathrm{BM}$, close to $2.84\ \mathrm{BM}$. So C is correct.Statement D: For tetrahedral splitting, CFSE is calculated using $\Delta_t$, not $\Delta_o$. For $d^8$ tetrahedral, CFSE = $4(-0.6\Delta_t) + 4(+0.4\Delta_t) = -0.8\Delta_t$. Therefore, stating $-0.8\Delta_o$ is wrong. So D is incorrect.Statement E: $\mathrm{Ni}(\mathrm{CO})_4$ is tetrahedral, and this complex is also tetrahedral, so the geometry is similar. Hence E is correct.Thus the incorrect statements are A, B, and D.Answer:Option B (A, B and D Only)