A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula M_2 O_n (where n=3,4,5 ). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is \underline\hspace2cmBM (near integer) (Given atomic number: Sc: 21, Ti: 22, ~V: 23, Cr: 24, Mn: 25, Fe: 26, Co: 27, Ni: 28, Cu: 29, Zn: 30 )

D and F Block Elements PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q30 with Solution

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Correct answer: 2

Identify the first-row transition metal with highest enthalpy of atomisation

Among the first transition series, the metal with the highest enthalpy of atomisation is vanadium, $\mathrm{V}$ .

Check the oxides formed of type $\mathrm{M}_2\mathrm{O}_n$ where $n=3,4,5$

For vanadium, the relevant oxides are:

\mathrm{V}_2\mathrm{O}_3,\quad \mathrm{V}_2\mathrm{O}_4\;(=2\,\mathrm{VO}_2),\quad \mathrm{V}_2\mathrm{O}_5

These correspond to oxidation states:

In $\mathrm{V}_2\mathrm{O}_3$ : $\mathrm{V}$ is $+3$
In $\mathrm{V}_2\mathrm{O}_4$ : $\mathrm{V}$ is $+4$
In $\mathrm{V}_2\mathrm{O}_5$ : $\mathrm{V}$ is $+5$

Identify the amphoteric oxide

Among vanadium oxides:

$\mathrm{V}_2\mathrm{O}_3$ is basic
$\mathrm{VO}_2$ (or $\mathrm{V}_2\mathrm{O}_4$ ) is amphoteric
$\mathrm{V}_2\mathrm{O}_5$ is acidic

So the amphoteric oxide is:

\mathrm{V}_2\mathrm{O}_4\;(\text{or }\mathrm{VO}_2)

Find the electronic configuration of the metal ion in the amphoteric oxide

In $\mathrm{VO}_2$ , vanadium is in the $+4$ oxidation state.

Vanadium atom ( $Z=23$ ):

\mathrm{V}: [\mathrm{Ar}]\,3d^3 4s^2

Therefore,

\mathrm{V}^{4+}: [\mathrm{Ar}]\,3d^1

So there is 1 unpaired electron.

Calculate spin-only magnetic moment

The spin-only magnetic moment is:

\mu = \sqrt{n(n+2)}\;\text{BM}

where $n$ is the number of unpaired electrons.

Here, $n=1$ , so

\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\;\text{BM}

Near integer value:

\boxed{2}

Compare with stored correct answer

Stored answer is $0$ , but the derived answer is $2$ .

Hence, the stored answer appears incorrect.

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View written solutionFree

Correct answer: 2

Identify the first-row transition metal with highest enthalpy of atomisation

Among the first transition series, the metal with the highest enthalpy of atomisation is vanadium, $\mathrm{V}$ .

Check the oxides formed of type $\mathrm{M}_2\mathrm{O}_n$ where $n=3,4,5$

For vanadium, the relevant oxides are:

\mathrm{V}_2\mathrm{O}_3,\quad \mathrm{V}_2\mathrm{O}_4\;(=2\,\mathrm{VO}_2),\quad \mathrm{V}_2\mathrm{O}_5

These correspond to oxidation states:

In $\mathrm{V}_2\mathrm{O}_3$ : $\mathrm{V}$ is $+3$
In $\mathrm{V}_2\mathrm{O}_4$ : $\mathrm{V}$ is $+4$
In $\mathrm{V}_2\mathrm{O}_5$ : $\mathrm{V}$ is $+5$

Identify the amphoteric oxide

Among vanadium oxides:

$\mathrm{V}_2\mathrm{O}_3$ is basic
$\mathrm{VO}_2$ (or $\mathrm{V}_2\mathrm{O}_4$ ) is amphoteric
$\mathrm{V}_2\mathrm{O}_5$ is acidic

So the amphoteric oxide is:

\mathrm{V}_2\mathrm{O}_4\;(\text{or }\mathrm{VO}_2)

Find the electronic configuration of the metal ion in the amphoteric oxide

In $\mathrm{VO}_2$ , vanadium is in the $+4$ oxidation state.

Vanadium atom ( $Z=23$ ):

\mathrm{V}: [\mathrm{Ar}]\,3d^3 4s^2

Therefore,

\mathrm{V}^{4+}: [\mathrm{Ar}]\,3d^1

So there is 1 unpaired electron.

Calculate spin-only magnetic moment

The spin-only magnetic moment is:

\mu = \sqrt{n(n+2)}\;\text{BM}

where $n$ is the number of unpaired electrons.

Here, $n=1$ , so

\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\;\text{BM}

Near integer value:

\boxed{2}

Compare with stored correct answer

Stored answer is $0$ , but the derived answer is $2$ .

Hence, the stored answer appears incorrect.

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