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Isoelectronic Species

What is corre...

Question

What is correct order of size in each pair given below ?
$F e^{2 +}$ and $F e^{3 +}$ :

F e^{2 +}

F e^{3 +}

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F e^{2 +}

F e^{3 +}

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F e^{2 +}

F e^{3 +}

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None of these

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Solution

The correct option is C $F e^{2 +}$ > $F e^{3 +}$
Size (ionic radius) of $F e^{2 +}$ is more than the size (ionic radius) of $F e^{3 +}$ . 3 electrons are removed from $F e$ to form $F e^{3 +}$ and 2 electrons to form $F e^{2 +}$ , The ratio of the number of protons to the number of electrons is higher for $F e^{3 +}$ than for $F e^{2 +}$ . Hence due to higher effective nuclear charge in $F e^{3 +}$ , the nucleus pulls the electrons to itself to greater extent than that in $F e^{2 +}$ which results in a smaller ionic radius for $F e^{3 +}$ . Option $C$ .

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Similar questions

F e

F e^{2 +}

F e^{3 +}

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

E_{F e^{2 +} / F e}^{\circ} = - 0.44 V and E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V

E_{F e^{3 +} / F e}^{\circ}

F e^{2 +} ⇌ F e; E^{\circ} = - 0.44 V

F e^{3 +} ⇌ F e^{2 +}; E^{\circ} = + 0.77 V

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

F e

(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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