Question

The hybridization of central atom of ClF$_3$, NH$_3$, SO$_3$ are respectively

• A. sp$^2$, sp$^2$, sp$^2$
• B. sp$^3$d, sp$^3$, sp$^2$
• C. sp$^2$, sp$^3$, sp$^3$d
• D. sp$^3$d, sp$^3$, sp$^3$

Hint:

Hybridization can be found by counting the number of sigma bonds and lone pairs on the central atom: 2 regions = sp, 3 = sp$^2$, 4 = sp$^3$, 5 = sp$^3$d, 6 = sp$^3$d$^2$.

Answer 1

The Correct Option is B

Step 1: Determine the hybridization of the central atom in ClF$_3$.
Central atom: Cl
Valence electrons in Cl: 7
Number of F atoms attached: 3
Lone pairs on Cl: $7 - 3 = 4$ electrons = 2 lone pairs
Total regions of electron density (bonds + lone pairs): $3 + 2 = 5$ Hybridization: sp$^3$d

Step 2: Determine the hybridization of the central atom in NH$_3$. Central atom: N
Valence electrons in N: 5
Number of H atoms attached: 3
Lone pairs on N: $5 - 3 = 2$ electrons = 1 lone pair
Total regions of electron density: $3 + 1 = 4$
Hybridization: sp$^3$

Step 3: Determine the hybridization of the central atom in SO$_3$. Central atom: S
Valence electrons in S: 6
Number of O atoms attached: 3
Lone pairs on S: $6 - 3 \times 2 = 0$ (since each S=O double bond uses 2 electrons from S, but in resonance, all are equivalent)
Total regions of electron density: 3 (for the three double bonds)
Hybridization: sp$^2$
Final Answer: \[ \boxed{\text{sp}^3\text{d},\ \text{sp}^3,\ \text{sp}^2} \]