If true enter 1, else enter 0.
Both $N$ and $C l$ has almost same electronegativity but later is not involved in $H$ -bonding.
1

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Solution

The correct option is A 1
There is no (or very little) orbital mixing between the HOMO of chlorine and the LUMO of hydrogen. This is because chlorine is in row 3. As you move down the rows, the difference in the size between the hydrogen LUMO and the acceptor orbitals becomes larger resulting in poor overlap and less mixing (all the good acceptors are in group 2). Without orbital mixing, there is no hydrogen bond - only a dipole-dipole interaction.

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