Question

Among \(\text{CrO}\), \(\text{Cr}_2\text{O}_3\), and \(\text{CrO}_3\), the sum of spin-only magnetic moment values of basic and amphoteric oxides is ______ \( \times 10^{-2} \, \text{BM} \) (nearest integer).
Given: Atomic number of Cr is 24.

Answer 1

Correct Answer: 877

CrO (Basic oxide): Chromium is Cr$^{2+}$ in CrO. Spin-only magnetic moment ($\mu$): $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons, which is 4. Therefore, $\mu = \sqrt{4(4+2)} = 4.90$ BM. 2. Cr$_2$O$_3$ (Amphoteric oxide): Chromium is Cr$^{3+}$ in Cr$_2$O$_3$. Spin-only magnetic moment ($\mu$): $\mu = \sqrt{n(n+2)}$ BM, with $n = 3$. Therefore, $\mu = \sqrt{3(3+2)} = 3.87$ BM. 3. CrO$_3$ (Acidic oxide): Chromium is Cr$^{6+}$ in CrO$_3$. There are no unpaired electrons ($n = 0$), so $\mu = 0$. Sum of spin-only magnetic moments of basic and amphoteric oxides: $\mu_{\text{total}} = 4.90 + 3.87 = 8.77$ BM. Expressed in $10^{-2}$ BM: $\mu_{\text{only}} = 8.77 \times 10^{-2}$ BM. Answer: 877