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JEE Advanced levelSingle correct optionDifficulty 3 of 3as212

A 3d-series metal M forms an ion $M X^{3 +}$ with spin-only magnetic moment $5.92 BM$ . The number of unpaired electrons in the neutral ground-state atom M is

A5
B4
C6
D3

Answer: B

n (n + 2) = 35

gives

n = 5

, so

M X^{3 +}

is

3 d^{5}

. Adding back three electrons (

4 s^{2}

and one

3 d

) gives

3 d^{6} 4 s^{2}

, i.e. Fe. Fe has 6

d

electrons: 4 unpaired.

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