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JEE Main levelSingle correct optionDifficulty 2 of 3as200

An ion of a 3d-series element has a spin-only magnetic moment of $3.87 BM$ . Which ion could it be?

A $Cr X^{3 +}$
B $Ti X^{3 +}$
C $Mn X^{3 +}$
D $Fe X^{3 +}$

Answer: A

n (n + 2) = 3.87

gives

n (n + 2) = 15

, so

n = 3

.

Cr X^{3 +}

is

3 d^{3}

(3 unpaired).

Ti X^{3 +}

is

3 d^{1}

(1),

Mn X^{3 +}

is

3 d^{4}

(4),

Fe X^{3 +}

is

3 d^{5}

(5).

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