What is the hybridisation and geometry of the given species? The species are `XeF_(2) and ICl_(2^(-)`

Options

A. `sp^(3)d` and trigonal bipyramidal
B. `sp^(3)d^(2)` and square planar
C. `sp^(3)d` and linear
D. `sp^(3)` and irregular tetrahedron

Answer

To determine the hybridization and geometry of the species \( \text{XeF}_2 \) and \( \text{ICl}_2^- \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **For \( \text{XeF}_2 \)**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) has 1 valence electron, and there are 2 fluorine atoms, contributing a total of \( 2 \times 1 = 2 \) electrons.
- Total valence electrons = \( 8 + 2 = 10 \).

- **For \( \text{ICl}_2^- \)**:
- Iodine (I) has 7 valence electrons.
- Each chlorine (Cl) has 7 valence electrons, and there are 2 chlorine atoms, contributing a total of \( 2 \times 7 = 14 \) electrons.
- The negative charge adds 1 more electron.
- Total valence electrons = \( 7 + 14 + 1 = 22 \).

### Step 2: Calculate the Effective Electron Pairs
Using the formula:
\[ \text{Number of effective electron pairs} = \frac{\text{(Valence electrons on central atom)} + \text{(Number of monovalent atoms)} + \text{(Charge)}}{2} \]

- **For \( \text{XeF}_2 \)**:
- Central atom: Xe (8 valence electrons).
- Monovalent atoms: 2 (F).
- Charge: 0 (neutral).
- Calculation:
\[ \frac{8 + 2 + 0}{2} = \frac{10}{2} = 5 \]

- **For \( \text{ICl}_2^- \)**:
- Central atom: I (7 valence electrons).
- Monovalent atoms: 2 (Cl).
- Charge: -1.
- Calculation:
\[ \frac{7 + 2 + 1}{2} = \frac{10}{2} = 5 \]

### Step 3: Determine Hybridization
- **For both \( \text{XeF}_2 \) and \( \text{ICl}_2^- \)**:
- Since both have 5 effective electron pairs, the hybridization is \( \text{sp}^3\text{d} \).

### Step 4: Determine Bond Pairs and Lone Pairs
- **For \( \text{XeF}_2 \)**:
- Bond pairs: 2 (from 2 F).
- Lone pairs: \( 5 - 2 = 3 \).

- **For \( \text{ICl}_2^- \)**:
- Bond pairs: 2 (from 2 Cl).
- Lone pairs: \( 5 - 2 = 3 \).

### Step 5: Determine Geometry
- Both species have 3 lone pairs and 2 bond pairs.
- The geometry is based on the arrangement of the electron pairs:
- The arrangement of 5 electron pairs is trigonal bipyramidal.
- The presence of 3 lone pairs will lead to a linear shape for the bond pairs.

### Final Answer
- For both \( \text{XeF}_2 \) and \( \text{ICl}_2^- \):
- **Hybridization**: \( \text{sp}^3\text{d} \)
- **Geometry**: Linear (due to 3 lone pairs).