The shapes of `XeF_(4),XeF_(5)^(-) and SnCl_(2)` are :

Options

A. octahedral, trigonal bipyramidal and bent
B. square pyramidal, pentagonal planar and linear
C. square planar, pentagonal planar and angular
D. see-saw, T-shaped and linear

Answer

To determine the shapes of `XeF_(4)`, `XeF_(5)^(-)`, and `SnCl_(2)`, we will follow a systematic approach using the VSEPR (Valence Shell Electron Pair Repulsion) theory.

### Step 1: Determine the number of valence electrons for each molecule.

1. **For `XeF_(4)`**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are 4 fluorine atoms: \( 4 \times 7 = 28 \) electrons.
- Total valence electrons = \( 8 + 28 = 36 \) electrons.

2. **For `XeF_(5)^(-)`**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are 5 fluorine atoms: \( 5 \times 7 = 35 \) electrons.
- The negative charge adds 1 more electron: \( 8 + 35 + 1 = 44 \) electrons.

3. **For `SnCl_(2)`**:
- Tin (Sn) has 4 valence electrons.
- Each chlorine (Cl) has 7 valence electrons, and there are 2 chlorine atoms: \( 2 \times 7 = 14 \) electrons.
- Total valence electrons = \( 4 + 14 = 18 \) electrons.

### Step 2: Determine the number of bonding pairs and lone pairs.

1. **For `XeF_(4)`**:
- Bonding pairs: 4 (from 4 Xe-F bonds).
- Lone pairs: Total valence electrons (36) - (2 electrons per bond) = \( 36 - 8 = 28 \) electrons left for lone pairs. Since each lone pair consists of 2 electrons, \( 28 / 2 = 14 \) lone pairs. However, in the context of VSEPR, we only consider the lone pairs that affect the shape, which are 2 lone pairs.
- Steric number = 4 (bonding pairs) + 2 (lone pairs) = 6.
- Hybridization: \( sp^3d^2 \).
- Shape: Square planar.

2. **For `XeF_(5)^(-)`**:
- Bonding pairs: 5 (from 5 Xe-F bonds).
- Lone pairs: Total valence electrons (44) - (10 electrons for 5 bonds) = \( 44 - 10 = 34 \) electrons left for lone pairs. Thus, \( 34 / 2 = 17 \) lone pairs. But we consider only 2 lone pairs for the shape.
- Steric number = 5 (bonding pairs) + 2 (lone pairs) = 7.
- Hybridization: \( sp^3d^3 \).
- Shape: Pentagonal planar.

3. **For `SnCl_(2)`**:
- Bonding pairs: 2 (from 2 Sn-Cl bonds).
- Lone pairs: Total valence electrons (18) - (4 electrons for 2 bonds) = \( 18 - 4 = 14 \) electrons left for lone pairs. Thus, \( 14 / 2 = 7 \) lone pairs. We consider only 1 lone pair for the shape.
- Steric number = 2 (bonding pairs) + 1 (lone pair) = 3.
- Hybridization: \( sp^2 \).
- Shape: Angular.

### Final Shapes:
- `XeF_(4)` - Square planar
- `XeF_(5)^(-)` - Pentagonal planar
- `SnCl_(2)` - Angular