Which of the following ion has the maximum number of unpaired electrons?

^{+ 2}

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^{+ 3}

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^{+ 3}

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^{+ 1}

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Solution

The correct option is B Fe $^{+ 3}$
Electronic configuration of ions:

$F e^{+ 3}$ - $[A r] 3 d^{5}$
$M n^{+ 3}$ - $[A r] 3 d^{4}$
$C u^{+}$ - $[A r] 3 d^{10}$
$Z n^{+ 2}$ - $[A r] 3 d^{10}$

$F e^{+ 3}$ has 5 unpaired electrons.

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F e^{+ 3}

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V^{3 +}

(8)^{\frac{1}{2}} B M

F e 26

(A t o m i c n u m b e r o f F e = 26)

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