Question

The number of covalent bonds that a halogen atom can form through $s{p}^{3}d$ hybridization is:

• A. $2$
• B. $4$
• C. $5$
• D. $3$

Answer 1

Answer: (D)

$3$

In the first excited state, an electron is excited from $p$ orbital to d orbital. Now, there are $3$ unpaired orbitals available for hybridization. Hence hybridization is $s{p}^{3}d$ hybridisation. And the halogen in this state can form $3$ covalent bonds.