Which of the following orders is correct?

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(C) $1$. Option $A$ is incorrect because the bond energy order for group $14$ elements is $C-C > Si-Si > Ge-Ge$ due to the decrease in atomic size and

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$Ge < Sn < Pb$ (ability of $ns^2$ electrons to participate in bonding)

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(C) $1$. Option $A$ is incorrect because the bond energy order for group $14$ elements is $C-C > Si-Si > Ge-Ge$ due to the decrease in atomic size and increase in bond length.$2$. Option $B$ is incorrect because the bond energy of $H-H$ is $436 \ kJ/mol$,$C-C$ is $348 \ kJ/mol$,and $F-F$ is $158 \ kJ/mol$. The correct order is $H-H > C-C > F-F$.$3$. Option $C$ is correct. Due to the inert pair effect,the stability of the $+2$ oxidation state increases down the group $(Ge  Sn > Pb$. Therefore,option $C$ as written is incorrect.$4$. Option $D$ is incorrect. $CH_4$ does not undergo hydrolysis. The ease of hydrolysis depends on the availability of vacant $d$-orbitals. $SiH_4$ hydrolyzes easily,while $CH_4$ does not.$5$. Re-evaluating the question: It seems there might be a typo in the provided options. However,based on standard chemistry,none of these are strictly correct as written. If we look for the most chemically sound statement,the inert pair effect is a fundamental concept. Let's re-check the options. Actually,the ability of $ns^2$ electrons to participate in bonding is $Ge > Sn > Pb$. The option $C$ says $Ge  C-C$. Let's select the best fit or correct the logic.

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