Which of the following has the maximum number of unpaired electrons ? $(A t o m i c n u m b e r o f F e = 26)$

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F e^{2 +}

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F e^{3 +}

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F e^{4 +}

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Solution

The correct option is C $F e^{3 +}$
Ground state electronic configuration of $F e = [A r] 3 d^{6} 4 s^{2}$
So, Fe has 4 unpaired electrons.
$F e^{2 +} = [A r] 3 d^{6} 4 s^{0}$
So, $F e^{2 +}$ has 4 unpaired electrons.
$F e^{3 +} = [A r] 3 d^{5} 4 s^{0}$
So, $F e^{3 +}$ has 5 unpaired electrons.
$F e^{4 +} = [A r] 3 d^{4} 4 s^{0}$
So, $F e^{4 +}$ has 4 unpaired electrons.

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