Stability order of $+ 3$ and $+ 1$ states of boron family elements is:

{G a}^{3 +} < {I n}^{3 +} < {T l}^{3 +}

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{G a}^{3 +} > {I n}^{3 +} {> T l}^{3 +}

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{G a}^{+} < {I n}^{+} < {T l}^{+}

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{G a}^{3 +} < {G a}^{+}

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Solution

The correct options are
B ${G a}^{3 +} > {I n}^{3 +} {> T l}^{3 +}$
C ${G a}^{+} < {I n}^{+} < {T l}^{+}$
Due to Inert pair effect, in the elements which are at a lower position in a group, there are d and f orbitals which have a poor shielding effect. So outermost s electrons feel more attractions with the nucleus and become more difficult to lose. So as 2 electrons are not getting lost, oxidation state decreases from +3 to +1.
Hence, $T l^{+} > I n^{+} > G a^{+}$ and $G a^{3 +} > I n^{3 +} > T l^{3 +}$

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G a^{3 +} > I n^{3 +} > T l^{3 +}

G a^{3 +} < I n^{3 +} < T l^{3 +}

G e^{+ 4} > S n^{+ 4} > P b^{+ 4}

G e^{+ 2} < S n^{2 +} < P b^{2 +}

A s^{3 +} < S b^{3 +} < B i^{3 +}

The relative stability of the different oxidation states,

$T I^{+} > T I^{3 +}, > G a^{3 +} > G a^{+} T I^{+} > I n^{+} > G a^{+} is because of$

T l^{3 +} / T l

T l^{3 +} + 2 e^{-} \to T l^{+}; E^{\circ} = 1.26 v o l t

T l^{+} + e^{-} \to T l; E^{\circ} = - 0.336 v o l t

T l^{+}

T l^{3 +}

What is the standard potential of the $T l^{3 +} / T l$ electrode?

$T l^{3 +} + 2 e^{-} \to T l^{+} E^{0} = 1.26 V T l^{+} + e^{-} \to T l E^{0} = - 0.336 V$

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