Two statements are given below.Statement I: S | vedclass

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(C) In group $14$,most $MX_4$ halides are covalent. However,$SnF_4$ and $PbF_4$ are ionic due to the large electronegativity difference between the me

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Statement $I$ is correct,but statement $II$ is not correct.

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(C) In group $14$,most $MX_4$ halides are covalent. However,$SnF_4$ and $PbF_4$ are ionic due to the large electronegativity difference between the metal and fluorine. Thus,Statement $I$ is correct.For group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect. $Ge$ is at the top of the group,so the $+4$ oxidation state is more stable than the $+2$ oxidation state. Therefore,$GeCl_4$ is more stable than $GeCl_2$. Thus,Statement $II$ is incorrect.

Element	Electronic configuration
$X$	$1s^2, 2s^2, 2p^2$
$Y$	$1s^2, 2s^2, 2p^6, 3s^1$
$Z$	$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^5$

Two statements are given below.
Statement $I$: $SnF_4$ and $PbF_4$ are ionic in nature.
Statement $II$: $GeCl_2$ is more stable than $GeCl_4$.
The correct answer is:

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Two statements are given below.Statement $I$: $SnF_4$ and $PbF_4$ are ionic in nature.Statement $II$: $GeCl_2$ is more stable than $GeCl_4$.The correct answer is: