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

CrO (Basic oxide): - In CrO, chromium exists as Cr$^{2+}$. - Spin-only magnetic moment ($\mu$):
$\mu = \sqrt{n(n+2)} \, \text{BM}, \quad n = \text{number of unpaired electrons} = 4.$
$\mu = \sqrt{4(4+2)} = 4.90 \, \text{BM}.$
2.Cr$_2$O$_3$ (Amphoteric oxide): - In Cr$_2$O$_3$, chromium exists as Cr$^{3+}$. - Spin-only magnetic moment ($\mu$):
$\mu = \sqrt{n(n+2)} \, \text{BM}, \quad n = 3.$
$\mu = \sqrt{3(3+2)} = 3.87 \, \text{BM}.$
3.CrO$_3$ (Acidic oxide): - In CrO$_3$, chromium exists as Cr$^{6+}$. - No unpaired electrons ($n = 0$), so $\mu = 0$.
Sum of spin-only magnetic moments of basic and amphoteric oxides:
$\mu_{\text{total}} = 4.90 + 3.87 = 8.77 \, \text{BM}.$
Expressing in terms of $10^{-2}$ BM:
$\mu_{\text{only}} = 8.77 \times 10^{-2} \, \text{BM}.$
Answer: 877

Answer 2

CrO (Basic oxide): - In CrO, chromium exists as Cr$^{2+}$. - Spin-only magnetic moment ($\mu$):
$\mu = \sqrt{n(n+2)} \, \text{BM}, \quad n = \text{number of unpaired electrons} = 4.$
$\mu = \sqrt{4(4+2)} = 4.90 \, \text{BM}.$
2.Cr$_2$O$_3$ (Amphoteric oxide): - In Cr$_2$O$_3$, chromium exists as Cr$^{3+}$. - Spin-only magnetic moment ($\mu$):
$\mu = \sqrt{n(n+2)} \, \text{BM}, \quad n = 3.$
$\mu = \sqrt{3(3+2)} = 3.87 \, \text{BM}.$
3.CrO$_3$ (Acidic oxide): - In CrO$_3$, chromium exists as Cr$^{6+}$. - No unpaired electrons ($n = 0$), so $\mu = 0$.
Sum of spin-only magnetic moments of basic and amphoteric oxides:
$\mu_{\text{total}} = 4.90 + 3.87 = 8.77 \, \text{BM}.$
Expressing in terms of $10^{-2}$ BM:
$\mu_{\text{only}} = 8.77 \times 10^{-2} \, \text{BM}.$
Answer: 877