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): - 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

The problem asks for the sum of the spin-only magnetic moment values of the basic and amphoteric oxides among CrO, Cr₂O₃, and CrO₃. The answer should be expressed in units of \( \times 10^{-2} \, \text{BM} \) and rounded to the nearest integer.

Concept Used:

The solution involves two main concepts:

1. Nature of Metal Oxides: The chemical nature of a metal oxide depends on the oxidation state of the metal. Oxides with the metal in a low oxidation state are typically basic. Oxides with the metal in an intermediate oxidation state are amphoteric. Oxides with the metal in a high oxidation state are acidic.
2. Spin-Only Magnetic Moment: The spin-only magnetic moment (\(\mu\)) of a transition metal ion is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{BM} \] where \(n\) is the number of unpaired electrons in the d-orbitals of the metal ion and BM stands for Bohr Magneton. The number of unpaired electrons is determined from the electronic configuration of the ion.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Chromium (Cr) in each oxide and classify them.

• In CrO, let the oxidation state of Cr be \(x\). Then \(x + (-2) = 0 \implies x = +2\). Oxides in low oxidation states are basic. Thus, CrO is a basic oxide.
• In Cr₂O₃, let the oxidation state of Cr be \(y\). Then \(2y + 3(-2) = 0 \implies 2y = 6 \implies y = +3\). Oxides in intermediate oxidation states are amphoteric. Thus, Cr₂O₃ is an amphoteric oxide.
• In CrO₃, let the oxidation state of Cr be \(z\). Then \(z + 3(-2) = 0 \implies z = +6\). Oxides in high oxidation states are acidic. Thus, CrO₃ is an acidic oxide.

The problem requires the sum of the magnetic moments for the basic (CrO) and amphoteric (Cr₂O₃) oxides.

Step 2: Calculate the spin-only magnetic moment for CrO (Cr²⁺).

The atomic number of Cr is 24. Its ground-state electronic configuration is \([Ar] \, 3d^5 4s^1\). To form the Cr²⁺ ion, two electrons are removed (one from 4s and one from 3d). The electronic configuration of Cr²⁺ is \([Ar] \, 3d^4\).

The 3d⁴ configuration has 4 unpaired electrons (\(n=4\)). \[ \begin{array}{c|c|c|c|c} \uparrow & \uparrow & \uparrow & \uparrow & \\ \end{array} \] Using the formula for spin-only magnetic moment:

\[ \mu_{\text{Cr}^{2+}} = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM} \] Numerically, \(\sqrt{24} \approx 4.90 \, \text{BM}\).

Step 3: Calculate the spin-only magnetic moment for Cr₂O₃ (Cr³⁺).

To form the Cr³⁺ ion, three electrons are removed from a neutral Cr atom (one from 4s and two from 3d). The electronic configuration of Cr³⁺ is \([Ar] \, 3d^3\).

The 3d³ configuration has 3 unpaired electrons (\(n=3\)). \[ \begin{array}{c|c|c|c|c} \uparrow & \uparrow & \uparrow & & \\ \end{array} \] Using the formula for spin-only magnetic moment:

\[ \mu_{\text{Cr}^{3+}} = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM} \] Numerically, \(\sqrt{15} \approx 3.87 \, \text{BM}\).

Step 4: Sum the magnetic moment values of the basic and amphoteric oxides.

\[ \text{Total } \mu = \mu_{\text{Cr}^{2+}} + \mu_{\text{Cr}^{3+}} = (\sqrt{24} + \sqrt{15}) \, \text{BM} \] \[ \text{Total } \mu \approx (4.90 + 3.87) \, \text{BM} = 8.77 \, \text{BM} \]

Step 5: Convert the result to the required format.

The question asks for the answer in the form of ______ \( \times 10^{-2} \, \text{BM} \). \[ 8.77 \, \text{BM} = 877 \times 10^{-2} \, \text{BM} \]