Consider the hydrated ions of Ti^2+,… - JEE Main 2019 Chemistry PYQ | PrepForBharat

JEE Main 2019 — Chemistry Inorganic Chemistry

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Consider the hydrated ions of $T{i}^{2+}, {V}^{2+}, T{i}^{3+}$ and $S{c}^{3+}$ . The correct order of their spin-only magnetic moments is:

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

A
$S{c}^{3+}<T{i}^{3+}<{V}^{2+}<T{i}^{2+}$
B
$T{i}^{3+}<T{i}^{2+}<S{c}^{3+}<{V}^{2+}$
C
$S{c}^{3+}<T{i}^{3+}<T{i}^{2+}<{V}^{2+}$
D
${V}^{2+}<T{i}^{2+}<T{i}^{3+}<S{c}^{3+}$

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Solution

Magnetic moment $(\mu )=\sqrt{n(n+2)}$

More than value of n, more will be magnetic moment.

$T{i}^{+2}$ electronic configuration;

$3{d}^{2}$ JEE Main 2019 Chemistry, d and f Block Elements — solution figure 1 $\mu =\sqrt{2(2+2)}=2.82$

${\text{Ti}}^{+3}\Rightarrow 3{d}^{1}$ JEE Main 2019 Chemistry, d and f Block Elements — solution figure 2 $\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73$

${V}^{+2}\Rightarrow 3{d}^{3}$ JEE Main 2019 Chemistry, d and f Block Elements — solution figure 3 $\mu =\sqrt{3(3+2)}=3.87$

$S{c}^{3+}\Rightarrow [Ar]$ no unpaired electron. $\rightarrow \mu =0$

Hence Order of Magnetic Moment is ${V}^{+2}>T{i}^{+2}>T{i}^{+3}>S{c}^{+3}$

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