Question

The correct order of hybridisation of the central atom in the following species is:
$N{{H}_{3}}$,${{[PtC{{l}_{4}}]}^{2-}}$,$PC{{l}_{5}}$,$BC{{l}_{3}}$ is:
(A) $ds{{p}^{2}}$,$ds{{p}^{3}}$,$s{{p}^{2}}$,$s{{p}^{3}}$
(B) $s{{p}^{3}}$,$ds{{p}^{2}}$,$s{{p}^{3}}d$,$s{{p}^{2}}$
(C) $ds{{p}^{2}}$, $s{{p}^{2}}$, $s{{p}^{3}}$, $ds{{p}^{3}}$
(D) $ds{{p}^{2}}$,$s{{p}^{3}}$, $s{{p}^{2}}$, $ds{{p}^{3}}$

Answer 1

Hint: We can find the hybridization of the above given compound and complex using the valence bond theory. Identify the number of sigma bonds and the number of lone pairs on the central atom. In case of metal complex, identify the strength of the ligands to know if back bonding can take place or not.

Complete step-by-step answer:
According to valence bond theory, electrons in a molecule occupy atomic orbitals and not molecular orbitals. The atomic orbitals overlap on the bond formation and the strength of the bond depends on the extent of overlap.
Postulates of Valence Bond Theory:
- Covalent bonds are formed when two valence orbitals belonging to two different atoms overlap onto each other. Due to overlapping the electron density in that region increases, thereby increasing the stability of the molecule thus formed.
- The presence of many unpaired electrons in the valence shell of an atom enables the atoms to form multiple bonds with each other. However, the paired electrons present in the valent shell do not take part in formation of chemical bonds.
- Covalent bonds are directional and parallel to the region corresponding to atomic orbitals that are going to overlap.
- Sigma bonds and pi bonds differ in the pattern that the atomic orbitals overlap in, i.e. sigma bonds undergo head on overlap however pi bonds undergo sideways overlapping.
Let us take one compound at a time and then report its hybridization.
$N{{H}_{3}}$:
Number of bond pairs = 3
Number of lone pairs = 1
Hybridization of steric number 4(3+1) is $s{{p}^{3}}$.
${{[PtC{{l}_{4}}]}^{2-}}$:
Platinum metal lies in the 4d series of d block metals. All ligands are considered to be strong in the 4d and 5d series. Thus, the hybridization of ${{[PtC{{l}_{4}}]}^{2-}}$ will be $ds{{p}^{2}}$ with 4 bonded ligands and not $s{{p}^{3}}$ which is non-planar as well.
$PC{{l}_{5}}$:
Number of bond pairs = 5
Number of lone pairs = 0
Hybridization is thus $s{{p}^{3}}d$.
$BC{{l}_{3}}$:
Number of bond pairs = 3
Number of lone pairs = 0
Hybridization of steric number 4(3+1) is $s{{p}^{3}}$.
Therefore, the correct answer is option (B).

Note: Valence bond theory was successful in determining the hybridization of atoms in a molecule. However, the theory had some limitations ,like:
- Unable to explain the tetravalency of carbon
- No insight or information on the energies of electrons
- Incorrect assumption that electrons are localized in specific areas only
- No distinction between weak and strong ligands ( hybridization of complex compounds)