Question

\(AB_3\) is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is _________. (Integer answer)

Hint:

For halogens, the formula \(LP = \frac{7 - n}{2}\) where \(n\) is the number of bonds (assuming all halogens) works perfectly. For \(AB_3\), \(LP = \frac{7 - 3}{2} = 2\).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
Interhalogen molecules of the type \(AB_3\) are formed by two different halogens. The geometry and shape are determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory based on the total number of electron pairs around the central atom \(A\).
Step 2: Key Formula or Approach:
Steric Number = \(\frac{1}{2} [\text{Valence } e^- \text{ on central atom} + \text{Monovalent atoms} - \text{Charge}]\)
Step 3: Detailed Explanation:
1. Central atom \(A\) is a halogen, which has 7 valence electrons.
2. In \(AB_3\), there are 3 monovalent \(B\) atoms bonded to \(A\).
3. Total valence electrons involved = 7 (from \(A\)) + 3 (from bonds) = 10 electrons (or 5 pairs).
4. Steric Number = 5. This corresponds to a trigonal bipyramidal (\(sp^3d\)) hybridization.
5. Out of the 5 electron pairs, 3 are bond pairs (BPs) with atoms \(B\).
6. Number of Lone Pairs (LPs) = \(\text{Steric Number} - \text{Bond Pairs} = 5 - 3 = 2\).
7. According to VSEPR, 2 LPs and 3 BPs result in a "T-shaped" geometry to minimize repulsion.
Step 4: Final Answer:
The number of lone pairs on the central atom \(A\) is 2.