Question

The hybridisation of the central atom of \( BF_3 \), \( SnCl_2 \), \( HgCl_2 \), respectively is?

• A. \( sp^2, sp^2, sp \)
• B. \( sp^3, sp^2, sp^2 \)
• C. \( sp^3, sp, sp^2 \)
• D. \( sp^3, sp, sp \)

Hint:

The hybridization of an atom can be determined using the steric number formula: \[ \text{Steric Number} = \text{Number of bonded atoms} + \text{Number of lone pairs} \] A steric number of 2 corresponds to \( sp \), 3 to \( sp^2 \), and 4 to \( sp^3 \).

Answer 1

The Correct Option is A

The hybridization of a molecule can be determined using the steric number, which is given by: \[ \text{Steric Number} = \text{Number of bonded atoms} + \text{Number of lone pairs} \] Step 1: Hybridization of \( BF_3 \)
- Boron (B) has 3 valence electrons and forms 3 sigma bonds with fluorine atoms.
- No lone pairs on boron.
- Steric number = 3, which corresponds to \( sp^2 \) hybridization.
- Molecular geometry: Trigonal planar.
Thus, the hybridization of \( BF_3 \) is \( sp^2 \).
Step 2: Hybridization of \( SnCl_2 \)
- Tin (Sn) has 4 valence electrons and forms 2 sigma bonds with chlorine atoms.
- It has one lone pair.
- Steric number = 3, which corresponds to \( sp^2 \) hybridization.
- Molecular geometry: Bent (V-shape).
Thus, the hybridization of \( SnCl_2 \) is \( sp^2 \).
Step 3: Hybridization of \( HgCl_2 \)
- Mercury (Hg) has 2 valence electrons and forms 2 sigma bonds with chlorine atoms.
- No lone pairs on mercury in this case.
- Steric number = 2, which corresponds to \( sp \) hybridization.
- Molecular geometry: Linear.
Thus, the hybridization of \( HgCl_2 \) is \( sp \).
Step 4: Verify the Correct Answer
From our calculations, the hybridizations are \( sp^2, sp^2, sp \), which matches Option (1). \bigskip