### Related Formula mu = sqrtn(n+2) text B.M.$\mu = \sqrt{n(n+2)} \text{ B.M.}$ where n$n$ represents the number of unpaired electrons. ### Core Logic Let's calculate the number of unpaired d-electrons (n$n$) and the resulting spin-only magnetic moment for each transition metal ion: * (A) mathrmTi^3+$\mathrm{Ti}^{3+}$: Electronic configuration = [Ar] 3d^1 ightarrow n = 1$= [Ar] 3d^1 ightarrow n = 1$ mu = sqrt1(1+2) = sqrt3 approx 1.73text B.M. ightarrow text(III)$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ B.M.} ightarrow \text{(III)}$ * (B) mathrmV^2+$\mathrm{V}^{2+}$: Electronic configuration = [Ar] 3d^3 ightarrow n = 3$= [Ar] 3d^3 ightarrow n = 3$ mu = sqrt3(3+2) = sqrt15 approx 3.87text B.M. ightarrow text(I)$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ B.M.} ightarrow \text{(I)}$ * (C) mathrmNi^2+$\mathrm{Ni}^{2+}$: Electronic configuration = [Ar] 3d^8$= [Ar] 3d^8$. The 3d$3d$ subshell has 3 paired orbitals and 2 unpaired orbitals ightarrow n = 2$ ightarrow n = 2$ mu = sqrt2(2+2) = sqrt8 approx 2.84text B.M. ightarrow text(IV)$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84\text{ B.M.} ightarrow \text{(IV)}$ * (D) mathrmSc^3+$\mathrm{Sc}^{3+}$: Electronic configuration = [Ar] 3d^0 ightarrow n = 0$= [Ar] 3d^0 ightarrow n = 0$ mu = 0.00text B.M. ightarrow text(II)$\mu = 0.00\text{ B.M.} ightarrow \text{(II)}$ Matching these values yields the sequence: (A)-(III), (B)-(I), (C)-(IV), (D)-(II). ### Pattern Recognition Shortcut: The digit before the decimal point in a spin-only magnetic moment matches the number of unpaired electrons (n$n$). For example, a value of 3.87text B.M.$3.87\text{ B.M.}$ means there are exactly 3$3$ unpaired electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements