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:

  1. 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

$$ Sc^{3+}<Ti^{3+}<Ti^{2+}<V^{2+} $$



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