The calculated spin only magnetic moment of $Cr^{2+}$ ion is:

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Accepted Answer

1.  **Electronic Configuration:** The atomic number of Chromium ($Cr$) is 24. Its ground state electronic configuration is $[Ar] 3d^5 4s^1$. To form the $Cr^{2+}$ ion, two electrons are removed (one from $4s$ and one from $3d$), resulting in the configuration $[Ar] 3d^4$ .
2.  **Unpaired Electrons:** In the $3d^4$ configuration, there are 4 unpaired electrons ($n = 4$) according to Hund's rule.
3.  **Calculation:** The 'spin-only' magnetic moment ($\mu$) is calculated using the formula:
    $$\mu = \sqrt{n(n+2)} 	ext{ BM}$$
    Substituting $n = 4$:
    $$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 	ext{ BM}$$
    This value matches the calculated magnetic moment for $Cr^{2+}$ listed in NCERT Table 4.7 .

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