Coordination Compounds Part 4

Concept:The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ (where $n$ is number of unpaired electrons). Ligands are classified as weak field (high spin) or strong field (low spin), which affects the hybridisation, geometry, and number of unpaired electrons.Explanation:For statement I: $K_3[Co(CO_3)_3]$: $Co$ is in $+3$ oxidation state ($d^6$). $CO_3^{2-}$ is a weak field ligand, causing high spin: $t_{2g}^4 e_g^2$ — 4 unpaired electrons. Hybridisation: $sp^3d^2$, octahedral geometry. Magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM. So statement I is correct.For statement II: $[Ni(CN)_4]^{2-}$: $Ni$ is $+2$ ($d^8$). $CN^-$ is strong field, pairing occurs: square planar $dsp^2$, 0 unpaired electrons → 0 BM. $[MnBr_4]^{2-}$: $Mn$ is $+2$ ($d^5$). $Br^-$ is weak field, high spin: tetrahedral $sp^3$, 5 unpaired electrons → $\sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM. $[CoF_6]^{3-}$: $Co$ is $+3$ ($d^6$). $F^-$ is weak field: octahedral $sp^3d^2$, 4 unpaired electrons → $\sqrt{4(4+2)} = 4.9$ BM. All three sets match the values given in statement II, so statement II is correct.Answer:Both Statement I and Statement II are true. Option C.