Question

$N$ forms $NC{l}_3$, whereas P can form both $PC{l}_3$ and $PC{l}_5$. Why?

• A. P has d-orbitals which can be used for bonding but $N_2$ does not have
• B. N atom is larger in size than P
• C. P is more reactive towards CI than N
• D. None of the above

Answer 1

The correct option is A P has d-orbitals which can be used for bonding but $N_2$ does not have

Phosphorus has 15 electrons. So its electronic configuration is:

Phosphorus is a period 3 element. Hence it has an empty d-orbital which the period 2 elements do not have.

When phosphorus forms a compound, if there is enough energy available then one electron from one of the orbitals jumps to the vacant d-orbital. This gives phosphorus extra 2 valencies. This is called expanded octet.

As phosphorus initially has valency as 3 it forms $PC{l}_3$, and due to this expanded octet, $PC{l}_5$ can also be formed.

*Note that the expanded octet phenomenon can occur only in period 3 elements.

As Nitrogen lies in period 2, it does not have the empty d-orbital and so does not form $NC{l}_5$

Hence, the correct option is (A).