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Magnetic moment of $_{26} {Fe}^{2 +}$ ion is same as:

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_{26} Fe

_{24} {Cr}^{2 +}

_{28} {Ni}^{4 +}

all of these

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detailed solution

Correct option is D

For $_{26} {Fe}^{2 +}$ , n=4.

So magnetic moment,

$μ = \sqrt{4 (4 + 2)} = 4.9 B M$

For $_{28} {Ni}^{4 +}$ , n=4 as,

$_{28} {Ni}^{4 +} = 3 d^{4}$

Now for $_{24} {Cr}^{2 +}$ , n=4 as,

$_{24} {Cr}^{2 +} = 3 d^{4}$

Thus, all options are correct.

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Magnetic moment of 26Fe2+ ion is same as: