Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Hybridization is the intermixing of the atomic orbitals that have similar or comparable energies.

Hybridisation can be calculated by the electron pair number method.

$$\begin{gathered} \mathrm{Electron} \mathrm{Pair} \mathrm{Hybridization} \\ 2 \mathrm{sp} \\ 3 {\mathrm{sp}}^2 \\ 4 {\mathrm{sp}}^3 \\ 5 {\mathrm{sp}}^3\mathrm{d} \\ 6 {\mathrm{sp}}^3{\mathrm{d}}^2 \\ 7 {\mathrm{sp}}^3{\mathrm{d}}^3 \end{gathered}$$

Electron pair can be calculated as the given below.

$\mathrm{E}.\mathrm{P}= \frac{\mathrm{valence} \mathrm{electron} \mathrm{of} \mathrm{central} \mathrm{atom} + \mathrm{monovalent} \mathrm{atom} + \mathrm{negative} \mathrm{charge}- \mathrm{positve} \mathrm{charge}}{2}$

For ${\mathrm{CO}}_3^{2-}$

$\mathrm{E}.\mathrm{P} = \frac{4+0+2+0}{2}=3 ={\mathrm{sp}}^2$

For ${\mathrm{XeF}}_4$

$\mathrm{E}.\mathrm{P} = \frac{8+4+2+0}{2}=6 ={\mathrm{sp}}^3{\mathrm{d}}^2$

For ${\mathrm{PCl}}_3$

$\mathrm{E}.\mathrm{P} = \frac{5+3+2+0}{2}=4 ={\mathrm{sp}}^3$

For ${\mathrm{I}}_3^{-}$

$\mathrm{E}.\mathrm{P} = \frac{7+2+1+0}{2}=5 ={\mathrm{sp}}^3\mathrm{d}$

Therefore the same hybridization as carbon in the above ethene molecule is only with the carbonate ion.