Among $\mathrm{CrO}, \mathrm{Cr}_2 \mathrm{O}_3$ and $\mathrm{CrO}_3$, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _________ $10^{-2} \mathrm{BM}$ (nearest integer).
(Given atomic number of $\mathrm{Cr}$ is 24 )
Answer (integer)
877
Solution
<p>First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.</p>
<p>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 $3d^4$.</p>
<p>To find the spin-only magnetic moment, we use the formula:</p>
<p>$\mu = \sqrt{n(n+2)} \mathrm{BM}$</p>
<p>where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$, we have 4 unpaired electrons in the 3d orbitals.</p>
<p>Thus,</p>
<p>$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mathrm{BM}$</p>
<p>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.</p>
<p>For $\mathrm{Cr}^{3+}$, there are 3 unpaired electrons.</p>
<p>Thus,</p>
<p>$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$</p>
<p>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 $\mathrm{Cr}^{6+}$.</p>
<p>Since $\mathrm{Cr}^{6+}$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.</p>
<p>The magnetic moment values for each compound are:</p>
<ul>
<li>$\mathrm{CrO}$: 4.90 BM (basic oxide)</li>
<li>$\mathrm{Cr}_2 \mathrm{O}_3$: 3.87 BM (amphoteric oxide)</li>
<li>$\mathrm{CrO}_3$: 0 BM</li>
</ul>
<p>The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:</p>
<p>$4.90 + 3.87 = 8.77 \mathrm{BM}$</p>
<p>Expressing in terms of $10^{-2} \mathrm{BM}$,</p>
<p>$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$</p>
<p>Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $\times 10^{-2} \mathrm{BM}$ (nearest integer).</p>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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