QJEEM24Chc85b: JEE Main 2024 Chemistry - Class 12 JEEM24Chc85b Solution: Among and , the sum of spin-only magnetic moment values of…

First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.

1. $\mathrm{CrO}$ : In $\mathrm{CrO}$ , the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ . In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $3d^4$ .

To find the spin-only magnetic moment, we use the formula:

$\mu = \sqrt{n(n+2)} \mathrm{BM}$

where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$ , we have 4 unpaired electrons in the 3d orbitals.

Thus,

$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mathrm{BM}$

2. $\mathrm{Cr}_2 \mathrm{O}_3$ : In $\mathrm{Cr}_2 \mathrm{O}_3$ , the oxidation state of chromium is +3. The electronic configuration of $\mathrm{Cr}^{3+}$ is $3d^3$ after losing three electrons.

For $\mathrm{Cr}^{3+}$ , there are 3 unpaired electrons.

Thus,

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$

3. $\mathrm{CrO}_3$ : In $\mathrm{CrO}_3$ , the oxidation state of chromium is +6. The electronic configuration of $\mathrm{Cr}^{6+}$ is $3d^0$ after losing six electrons. There are no unpaired electrons for $\mathrm{Cr}^{6+}$ .

Since $\mathrm{Cr}^{6+}$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.

The magnetic moment values for each compound are:

• $\mathrm{CrO}$ : 4.90 BM (basic oxide)
• $\mathrm{Cr}_2 \mathrm{O}_3$ : 3.87 BM (amphoteric oxide)
• $\mathrm{CrO}_3$ : 0 BM

The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:

$4.90 + 3.87 = 8.77 \mathrm{BM}$

Expressing in terms of $10^{-2} \mathrm{BM}$ ,

$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$

Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $\times 10^{-2} \mathrm{BM}$ (nearest integer).