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 $CORRECT$?

Similar Questions

Which of the following statements is true?

Which of the following order is incorrect?

In which of the following options, the order of arrangement does not agree with the variation of property indicated against it?
$(a)$ $BF_3 < NF_3 < NH_3$ (Dipole moment)
$(b)$ $HgCl_2 > NH_4^+ > SF_6$ (Bond angle)

Write Lewis dot symbols for atoms of the following elements: $Mg$,$Na$,$B$,$O$,$N$,$Br$.

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