Question

Which of the following molecule doesn't have $d\pi -p\pi$ bonding?

• A. $S{O}_2$
• B. $P_4{O}_{10}$
• C. $H_{3}P{O}_4$
• D. $B_3{N}_3{H}_6$

Answer 1

Answer: (D)

The structural study of $B_3{N}_3{H}_6$ shows that it has hexagonal ring with alternating B and N atom along with H atom. Each B and N are $s{p}^2$ hybridized and form three sigma bond one to terminal H atom and two to the adjacent N or B atom. There is a p$\pi$-p$\pi$ donation of filled nitrogen p${}_z$ atomic orbital to adjacent Boron p${}_z$ atomic orbitals. So, borazine contain 3 pi and 12 sigma bonds. Hence, there is no involvement of any d-orbital in hybridization of $B_3{N}_3{H}_6$.