The spin only magnetic moments (in BM) for free Ti3+, V2+ and Sc3+ ions respectively are ______. (At. No. Sc: 21, Ti:…

🧠 The Unpaired Spin Peak Magnetic moment depends solely on the number of "lonely" (unpaired) electrons ($n$). We use the formula $\mu = \sqrt{n(n+2)}$ BM.

🗺️ The Ion Audit

1. $\mathrm{Sc^{3+}}$ ($Z=21$): Config: $[\text{Ar}] 4s^2 3d^1$. Removing 3 electrons leaves $[\text{Ar}]$. $n = 0 \implies \mu = \mathbf{0 \text{ BM}}$.
2. $\mathrm{Ti^{3+}}$ ($Z=22$): Config: $[\text{Ar}] 4s^2 3d^2$. Removing 3 electrons ($2s, 1d$) leaves $[\text{Ar}] 3d^1$. $n = 1 \implies \mu = \sqrt{1(3)} = \mathbf{1.73 \text{ BM}}$.
3. $\mathrm{V^{2+}}$ ($Z=23$): Config: $[\text{Ar}] 4s^2 3d^3$. Removing 2 electrons ($2s$) leaves $[\text{Ar}] 3d^3$. $n = 3 \implies \mu = \sqrt{3(5)} = \mathbf{3.87 \text{ BM}}$.

Order: $1.73, 3.87, 0$.

⚡ The "N.Point-Something" Trick If $n=1$, $\mu$ is $1.73$. If $n=3$, $\mu$ is $3.87$. The magnetic moment always starts with the number of unpaired electrons!

⚠️ Common Traps Always remove $4s$ electrons before $3d$. For $\mathrm{Ti^{3+}}$, don't accidentally leave two $d$-electrons.

$\boxed{\text{Answer: (b)}}$