Which of the following order is correct for strength and extent of H-bonding between $H_{2} O$ and $H_{2} O_{2}$ respectively:

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$H_{2} O < H_{2} O_{2}$ and $H_{2} O > H_{2} O_{2}$

$H_{2} O > H_{2} O_{2}$ and $H_{2} O_{2} < H_{2} O$

$H_{2} O > H_{2} O_{2}$ and $H_{2} O_{2} > H_{2} O$

$H_{2} O < H_{2} O_{2}$ and $H_{2} O < H_{2} O_{2}$

answer is C.

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Detailed Solution

The strength of H-bonding is greater in $H_{2} O$ than in $H_{2} {O_{2}}_{}$ because water has more formal negative charge on the oxygen atom than $H_{2} O_{2}$ .
Because of the largest electronegativity difference between H and F, HF has the strongest hydrogen bond. The hydrogen molecule is covalently bonded and does not exhibit hydrogen bonding.
As a result, the order is $HF > H_{2} O > H_{2} O_{2} > H_{2}$ .