Which of the following order of boiling point | vedclass

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Question

(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.

List-$I$ (Molecule Species)	List-$II$ (Property/Shape)
$A$. $SO_2Cl_2$	$I$. Paramagnetic
$B$. $NO$	$II$. Diamagnetic
$C$. $NO_2^-$	$III$. Tetrahedral
$D$. $I_3^-$	$IV$. Linear

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

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Match List-$I$ with List-$II$ :

List-$I$ (Molecule Species) List-$II$ (Property/Shape)

$A$. $SO_2Cl_2$ $I$. Paramagnetic

$B$. $NO$ $II$. Diamagnetic

$C$. $NO_2^-$ $III$. Tetrahedral

$D$. $I_3^-$ $IV$. Linear

Choose the correct answer from the options given below :

$X-O-X$ linkage is present in $[X = P/S/Cl/Mn]$:
$(a) P_2O_7^{4-}$
$(b) S_2O_5^{2-}$
$(c) S_2O_7^{2-}$
$(d) Cl_2O_7$
$(e) Mn_2O_7$
$(f) S_2O_8^{2-}$

Which of the following is not a Lewis acid?

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