Question

Match the molecule with the hybridisation of its central atom.

• A. sp hybridisation
• B. $s{p}^2$ hybridisation
• C. $s{p}^3$ hybridisation

Answer 1

$\bullet$ Electronic configuration of beryllium: $1s^2$ $2s^2$
Excited state electronic configuration: $1s^2$$2s^1$$2p^1$
The valence 2s and 2p orbitals in beryllium hybridise to form two sp hybrid orbitals with one unpaired electron each. These hybrid orbitals overlap with $2p$ orbital of two fluorine atoms to form beryllium fluoride.

$\bullet$ Electronic configuration of boron: $1s^2$ $2s^2$$2p^1$
Excited state electronic configuration: $1s^2$ $2s^1$ $2p_x^1$ $2p_y^1$.
The one 2s and two 2p orbitals (valence orbitals) in boron hybridise to form three $s{p}^2$ hybrid orbitals. These hybrid orbitals overlap with $3p$ orbital of three chlorine atoms to form boron trichloride.

$\bullet$ Electronic configuration of nitrogen: $1s^2$ $2s^2$$2p^3$
The one 2s and three 2p orbitals (valence orbitals) in nitrogen hybridise to form four $s{p}^3$ hybrid orbitals to form ammonia. Three hybrid overlap with $1s$ orbital of three hydrogen atoms to form ammonia.