Question

a) Describe the hybridization in case of ${PCl}_5$. Why are the axial bonds longer as compared to equatorial bonds?
b) Calculate the bond order of $O_2^2{O}_2^{+}$ and $O_2^{-}{O}_2^{2-}$. Arrange them in increasing order of bond length.

Answer 1

(a) Ground state electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}3d^0$

(Refer to Image 1)

(b) Excited state electronic configuration of $P:1s^{2}2s^{2}2p^{6}3s^{1}3p633d^1$

(Refer to Image 2)

$P$ atom is $s{p}^{3}d$ hybridization in the excited state. These orbitals are filled by the electron pairs donated by $Cl$ atoms as:

(Refer to Image 3)

(Refer to Image 4)

The axial bonds are larger than equatorial bonds because the axial $Cl$ atoms suffer from more repulsion than the equatorial $Cl$ atoms, as a result the axial $cl$ atoms tries to reside far away from the equatorial $Cl$ atoms amd hence axial bonds are larger.

(b) Bond order= $\frac{\text{(Number of bonding electrons)-(number of antibonding electrons)}}{2}$

$O_2:{\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_2}^2{\Pi }_{2p_x}^2{\Pi }_{2p_y}^2{\Pi }_{2p_x}^{\ast 1}{\Pi }_{2p_y}^{\ast 1}$

$\therefore$ Bond order of $O_2^{2+}=\frac{10-4}{2}=3$

Bond order of $O_2^{+}=\frac{10-5}{2}=2.5$

Bond order of $O_2^{-}=\frac{10-7}{2}=1.5$

Bond order of $O_2^{2-}=\frac{10-8}{2}=1$

$\therefore$ Increasing order of Bond order: $O_2^{2-}<O_2^{-}<O_2^{+}<O_2^{2+}$