Which of the following order of boiling point | vedclass

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(A, B, C, D) $1$. For noble gases $(He, Ne, Ar, Kr, Xe)$,the boiling point increases down the group due to an increase in the magnitude of van der Waa

Vedclass · Accepted Answer

$H_2O_2 > H_2O$

Vedclass · Answer

(A, B, C, D) $1$. For noble gases $(He, Ne, Ar, Kr, Xe)$,the boiling point increases down the group due to an increase in the magnitude of van der Waals forces as the atomic size increases. Thus,$He $2$. For $H_2O$ and $HF$,$H_2O$ can form four hydrogen bonds per molecule,whereas $HF$ can form only two. Therefore,the boiling point of $H_2O$ $(373 \ K)$ is higher than $HF$ $(293 \ K)$. Thus,$H_2O > HF$ is $CORRECT$.$3$. $H_2O_2$ has a higher boiling point $(423 \ K)$ than $H_2O$ $(373 \ K)$ due to stronger and more extensive hydrogen bonding. Thus,$H_2O_2 > H_2O$ is $CORRECT$.$4$. $N(Me)_3$ has a higher boiling point than $NF_3$ because $N(Me)_3$ is more polar and has higher molecular weight/van der Waals forces compared to $NF_3$. Thus,$N(Me)_3 > NF_3$ is $CORRECT$.Since all options are correct,the question implies identifying the correct statements.

Which of the following order of boiling point is $CORRECT$?

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