Coordination Compounds Part 4

Concept:Spin-only magnetic moment depends on the number of unpaired electrons ($n$) in the central metal ion.It is calculated as $\mu = \sqrt{n(n+2)}$ BM.Explanation:Determine the oxidation state and $d$ configuration for each complex.Find unpaired electrons based on ligand field strength and geometry.For $[\mathrm{MnBr}_4]^{2-}$ (A): Mn is in $+2$ state, configuration $3d^5$. $\mathrm{Br}^-$ is weak field; tetrahedral geometry → high spin. All five $d$ electrons remain unpaired: $n=5$. For $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ (B): Cu is in $+2$ state, configuration $3d^9$. Octahedral complex with weak field ($\mathrm{H}_2\mathrm{O}$). $d^9$ always has one unpaired electron: $n=1$. For $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ (C): Ni is in $+2$ state, configuration $3d^8$. $\mathrm{CN}^-$ is strong field → square planar geometry. All electrons pair up: $n=0$. For $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ (D): Ni is in $+2$ state, $3d^8$. Octahedral with weak field ($\mathrm{H}_2\mathrm{O}$) → high spin. Two unpaired electrons: $n=2$. Order of unpaired electrons: $n(C)=0 < n(B)=1 < n(D)=2 < n(A)=5$. Same order applies for $\mu$ values. Shortcut:For $d^4$–$d^7$, weak field = high spin, strong field = low spin.$d^9$ always gives 1 unpaired, $d^8$ square planar with strong field gives 0 unpaired.Answer:C $<$ B $<$ D $<$ A, which corresponds to option C.