JEE Main 2019 — Chemistry Inorganic Chemistry

2019mcqmedium

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
${V}^{2+}<T{i}^{2+}<T{i}^{3+}<S{c}^{3+}$

Did you get this right?

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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JEE Main Inorganic Chemistry in other years

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JEE Main 2019 — Chemistry Inorganic Chemistry

2019mcqmedium

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
${V}^{2+}<T{i}^{2+}<T{i}^{3+}<S{c}^{3+}$

Did you get this right?

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 ${V}^{+2}>T{i}^{+2}>T{i}^{+3}>S{c}^{+3}$

Your note

JEE Main Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2019 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

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Physical Chemistry

Organic Chemistry

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JEE Main 2019 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry