1. Identify nature of the oxides

For chromium oxides:

• $\mathrm{CrO}$ : chromium is in $+2$ oxidation state, and this oxide is basic.
• $\mathrm{Cr_2O_3}$ : chromium is in $+3$ oxidation state, and this oxide is amphoteric.
• $\mathrm{CrO_3}$ : chromium is in $+6$ oxidation state, and this oxide is acidic.

So we need the sum of spin-only magnetic moments of:

• $\mathrm{CrO}$
• $\mathrm{Cr_2O_3}$

2. Find electronic configuration of chromium and its ions

Atomic number of Cr = 24

Ground state of Cr: $\mathrm{Cr} = [\mathrm{Ar}]\,3d^5 4s^1$

(a) In $\mathrm{CrO}$

Oxygen is $-2$, so chromium is $+2$.

$\mathrm{Cr^{2+}} = [\mathrm{Ar}]\,3d^4$

Thus number of unpaired electrons: $n = 4$

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899\ \mathrm{BM}$

(b) In $\mathrm{Cr_2O_3}$

Let oxidation state of Cr be $x$: $2x + 3(-2) = 0$ $2x - 6 = 0$ $x = +3$

So chromium is $\mathrm{Cr^{3+}}$.

$\mathrm{Cr^{3+}} = [\mathrm{Ar}]\,3d^3$

Thus number of unpaired electrons: $n = 3$

Spin-only magnetic moment: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.873\ \mathrm{BM}$

3. Add the required magnetic moments

$\mu_{\text{total}} = \sqrt{24} + \sqrt{15}$ $\mu_{\text{total}} \approx 4.899 + 3.873 = 8.772\ \mathrm{BM}$

4. Convert into the asked form

The question asks for:

$\underline{\hspace{2cm}} \times 10^{-2}\ \mathrm{BM}$

Since $8.772\ \mathrm{BM} = 877.2 \times 10^{-2}\ \mathrm{BM}$

Nearest integer: $877$

5. Final answer

$\boxed{877}$

This matches the stored correct answer.