What is covalency of chlorine atom in second excited state ?

Answer

To determine the covalency of a chlorine atom in its second excited state, we need to analyze its electronic configuration and how it changes when the atom is excited.

### Step-by-Step Solution:

1. **Identify the Ground State Configuration of Chlorine:**
- The atomic number of chlorine (Cl) is 17.
- Its electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5 \]

2. **Understanding the Ground State:**
- In the ground state, chlorine has 5 electrons in the 3p subshell (3p^5).
- This means it has 7 valence electrons (2 from 3s and 5 from 3p).

3. **First Excited State:**
- In the first excited state, one of the paired electrons in the 3p subshell gets excited to a higher energy level, which can be a 3d orbital.
- The configuration changes to:
\[ 3s^2 \, 3p^4 \, 3d^1 \]
- Here, the 3p subshell has 4 electrons and the 3d subshell has 1 electron.

4. **Second Excited State:**
- In the second excited state, another electron from the 3p subshell gets excited to the 3d subshell.
- The configuration now becomes:
\[ 3s^2 \, 3p^3 \, 3d^2 \]
- This means there are 3 electrons left in the 3p subshell and 2 electrons in the 3d subshell.

5. **Determining Covalency:**
- Covalency refers to the number of covalent bonds an atom can form, which is determined by the number of unpaired electrons available for bonding.
- In the second excited state, chlorine has:
- 3 unpaired electrons in the 3p subshell.
- 2 unpaired electrons in the 3d subshell.
- Therefore, the total number of unpaired electrons is:
\[ 3 (from \, 3p) + 2 (from \, 3d) = 5 \]

6. **Conclusion:**
- The covalency of chlorine in its second excited state is **5**.