Identify the most stable ion from the following:

F e^{+}

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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 $F e^{3 +}$
$F e : [A r] 3 d^{6} 4 s^{2}$

$F e^{+} : [A r] 3 d^{6} 4 s^{1}$

$F e^{+ +} : [A r] 3 d^{6} 4 s^{0}$

$F e^{3 +} : [A r] 3 d^{5}$

$F e^{4 +} : [A r] 3 d^{4}$

Half-filled orbitals are quite stable. Hence, $F e^{+ 3}$ is most stable.

Option (C) is correct.

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Similar questions

F e | F e^{2 +} | | F e^{3 +} | F e

F e | F e^{2 +} | | F e^{3 +}, F e^{2 +} | P t

F e | F e^{3 +} | | F e^{3 +}, F e^{2 +} | P t

C u^{+}, F e^{+}, F e^{2 +}, a n d F e^{3 +}

F e^{2 +}, F e^{3 +}

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e 6 - \to F e; E_{F}^{o} e_{F e^{2 +} / F e} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3 +} / F e^{2 +}}^{o} = + 0.77 V

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

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