Transition and InnerTransition Elements - Result Question 5
####5. Consider the hydrated ions of $Ti^{2+}, V^{2+}, Ti^{3+}$ and $Sc^{3+}$. The correct order of their spin-only magnetic moment is
(2019 Main, 10 April I)
(a) $Sc^{3+}<Ti^{3+}<Ti^{2+}<V^{2+}$
(b) $Sc^{3+}<Ti^{3+}<V^{2+}<Ti^{2+}$
(c) $Ti^{3+}<Ti^{2+}<Sc^{3+}<V^{2+}$
(d) $V^{2+}<Ti^{2+}<Ti^{3+}<Sc^{3+}$
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Solution:
- The spin only magnetic moment $(\mu)$ of each ion can be calculated as :
$$ \mu=\sqrt{n(n+2)} BM $$
$[\because n=$ No. of unpaired electron(s) $] \Rightarrow \mu \propto n$, i.e. higher the number of unpaired electron, higher will be the value of $\mu$.
| Metal ion | $\boldsymbol{Z}$ | $\boldsymbol{n}$ (for metal ion) | $\boldsymbol{M}$ (BM) | Nature |
|---|---|---|---|---|
| $Ti^{2+}$ | 22 | $2\left(3 d^{2}\right)$ | $\sqrt{8}$ | Paramagnetic |
| $V^{2+}$ | 23 | $3\left(3 d^{3}\right)$ | $\sqrt{15}$ | Paramagnetic |
| $Ti^{3+}$ | 22 | $1\left(3 d^{1}\right)$ | $\sqrt{3}$ | Paramagnetic |
| $Sc^{3+}$ | 21 | $0\left(3 d^{0}\right)$ | 0 | Diamagnetic |
Thus, the correct order of spin only magnetic moments of given hydrated ions will be
$$ Sc^{3+}<Ti^{3+}<Ti^{2+}<V^{2+} $$
