Element $F e$ has two oxidation states $F e^{2 +}$ & $F e^{3 +}$ , the correct order of radii of these ions will be:

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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F e^{3 +} \leq F e^{2 +}

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Solution

The correct option is C $F e^{2 +} > F e^{3 +}$
The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus.
Therefore, the correct order is: $F e^{2 +} > F e^{3 +}$

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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^{\circ}

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O i s 1.23 V F e^{3 +} + e^{-} \to F e^{2 +} i s 0.77 V

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

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