Question

The hybridization involved in the central atom of \(PCl_5\) is

• A. \(dsp^3\)
• B. \(sp^3d^2\)
• C. \(sp^2d^2\)
• D. \(d^2sp^3\)
• E. \(sp^3d\)

Hint:

Count the total number of bonds. \(P\) is bonded to 5 \(Cl\) atoms.
To accommodate 5 bonds, we need 5 hybrid orbitals.
Order of orbitals: \(s(1) + p(3) + d(1) = 5\). Thus, \(sp^3d\).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Hybridization is determined by the number of hybrid orbitals, which equals the number of sigma bonds plus the number of lone pairs on the central atom (Steric Number).

Step 2: Key Formula or Approach:
Steric Number (\(n\)) = \(\frac{1}{2} [V + M - C + A]\)
Where:
\(V\) = Valence electrons of central atom.
\(M\) = Number of monovalent surrounding atoms.
\(C, A\) = Cationic and Anionic charges.

Step 3: Detailed Explanation:
For \(PCl_5\):
Central atom is Phosphorus (\(P\)).
Phosphorus belongs to Group 15, so \(V = 5\).
Surrounding atoms are 5 Chlorine (\(Cl\)) atoms, which are monovalent, so \(M = 5\).
Charge is zero, so \(C=0, A=0\).
\[ n = \frac{1}{2} [5 + 5 - 0 + 0] = \frac{10}{2} = 5 \]
A steric number of 5 corresponds to \(sp^3d\) hybridization.
This results in a trigonal bipyramidal geometry.

Step 4: Final Answer:
The hybridization of the central atom in \(PCl_5\) is \(sp^3d\).