Which of the following is correct order of stability.

$T l^{3 +} > B i^{3 +}$

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$P b O_{2} > P b O$

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$B i I_{5} < B i F_{5}$

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$S n^{2 +} = G e^{2 +}$

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Solution

The correct option is C
$B i I_{5} < B i F_{5}$

$B i^{+ 5}$ usually gives only ionic compounds. A smaller anion is easier to fit into the lattice and thus $B i I_{5}$ is less stable than $B i F_{5}$ .

Due to inert pair effect, the $n s^{2}$ electrons are difficult to get excited and thus tend not to take part in bond formation.

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B F_{3} < B C l_{3} < B B r_{3} < B I_{3}

The correct order of Lewis acidity of $B X_{3}$ molecule is

A^{o}

A s^{3 +}, S b^{3 +}

B i^{3 +}

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 +}

{B i}^{3 +} (a q) ⟶ B i O_{3}^{-} (a q)

{B i}^{3 +}

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