The oxidation state(s) shared by each of the elements in group IV $(C, S i, G e, S n$ and $P b)$ is/are

+ 2

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+ 4

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+ 2, + 4

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- 2, + 2

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Solution

The correct option is C $+ 2, + 4$
$C$ , $S i$ , $G e$ , $S n$ and $P b$ can have both the oxidation states $+ 2$ and $+ 4$ .
Example can be given as oxidation state of Carbon in Carbon Monoxide, that is $C O$ is $+ 2$ . Whereas, oxidation state of Carbon in Carbon Dioxide, that is $C O_{2}$ is $+ 4$
Option C is the correct answer.

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Similar questions

Stability of ions of $G e, S n, a n d P b$ will be in the order:

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

C, S, G e, S n

P b

+ 4

{S n}^{2 +} + {2 H g}^{2 +} ⟶ {H g}_{2}^{2 +} + {S n}^{4 +}

{G e}^{2 +} + {P b}^{4 +} ⟶ {p b}^{2 +} + {G e}^{4 +}

{T 1}^{+} + {G a}^{3 +} ⟶ {G a}^{+} + {T 1}^{3 +}

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