Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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

4.90 BM

5.92 BM

2.84 BM

3.87 BM

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 $[Ar] 3d^4$ .
Unpaired Electrons: In the $3d^4$ configuration, there are 4 unpaired electrons ( $n = 4$ ) according to Hund's rule.
Calculation: The 'spin-only' magnetic moment ( $\mu$ ) is calculated using the formula: $\mu = \sqrt{n(n+2)} \text{ BM}$ Substituting $n = 4$ : $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$ This value matches the calculated magnetic moment for $Cr^{2+}$ listed in NCERT Table 4.7 .

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