Chlorine atom, in its third excited state, react with fluorine to form compound X. the formula and shape of X are

Options

A. `ClF_5`, Pentagonal
B. `ClF_4`, Tetrahedral
C. `ClF_4`, Pentagonal bipyramidal
D. `ClF_7`, Pentagonal bipyramidal

Answer

To determine the formula and shape of compound X formed when a chlorine atom in its third excited state reacts with fluorine, we can follow these steps:

### Step 1: Understand the Electronic Configuration of Chlorine
Chlorine (Cl) is in Group 17 of the periodic table and has an atomic number of 17. Its ground state electronic configuration is:
- **Ground State:** 1s² 2s² 2p⁶ 3s² 3p⁵

### Step 2: Determine the Third Excited State Configuration
In the third excited state, electrons from the outermost orbitals can be promoted to higher energy orbitals. The process is as follows:
1. **First Excited State:** One electron from the 3p orbital jumps to the 3d orbital, resulting in:
- 3s² 3p⁴ 3d¹
2. **Second Excited State:** Another electron from the 3p orbital jumps to the 3d orbital:
- 3s² 3p³ 3d²
3. **Third Excited State:** One electron from the 3s orbital jumps to the 3d orbital:
- 3s¹ 3p³ 3d³

### Step 3: Analyze the Hybridization
In the third excited state, chlorine has:
- 1 electron in the 3s orbital
- 3 electrons in the 3p orbitals
- 3 electrons in the 3d orbitals

This results in a total of 7 half-filled orbitals (1 from 3s, 3 from 3p, and 3 from 3d) that can participate in bonding.

### Step 4: Determine the Formula of Compound X
When chlorine reacts with fluorine, it can form a compound where chlorine is bonded to 7 fluorine atoms due to the availability of 7 half-filled orbitals. Thus, the formula of compound X is:
- **Formula:** ClF₇

### Step 5: Determine the Shape of Compound X
The hybridization corresponding to 7 orbitals (1 s, 3 p, and 3 d) is sp³d³. The molecular geometry for sp³d³ hybridization is:
- **Shape:** Pentagonal bipyramidal

### Final Answer
The formula and shape of compound X are:
- **Formula:** ClF₇
- **Shape:** Pentagonal bipyramidal