Question

Hydrogen bonds are formed in many compounds e.g., $H_{2}O,HF,N{H}_3.$ The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is:

• A. $N{H}_3>HF>H_{2}O$
• B. $HF>H_{2}O>N{H}_3$
• C. $N{H}_3>H_{2}O>HF$
• D. $H_{2}O>HF>N{H}_3$

Answer 1

The correct option is D $H_{2}O>HF>N{H}_3$

Strength of hydrogen bonds and boiling point order

$1.$ Boiling point of compounds like $H_{2}O,HF,N{H}_3$ depends on the strength of hydrogen bonds and the number of hydrogen bonds.

$2.$ The strength of hydrogen bond depends on the size and electronegativity of the atoms.

$3.$ Elements with smaller size and greater electronegativity show stronger $H-$bond.

Among the given molecules- $H_{2}O,HF,N{H}_3$ the order of electronegativity of the various atoms is $F>O>N.$ So, due to the least electronegativity of nitrogen here, boiling point of $N{H}_3$ will be less than boiling points of $H_{2}O$ and $HF.$

Moreover, $HF$ molecules can be linked to only two other $HF$ molecules by hydrogen bonding whereas $H_{2}O$ molecules can be linked to four other $H_{2}O$ molecules by hydrogen bonding i.e., $H_{2}O$ will have greater number of $H$ bonds per molecule as compared to $HF.$

Thus, $H_{2}O$ will have more boiling point than $HF.$

So, the correct decreasing order of boiling points will be $H_{2}O>HF>N{H}_3$

Hence, the correct choice is $\left(B\right).$