What will be the maximum value of `Sigma(n+1+m)` for all unpaired electrons of chlorine is its second excited state?

Options

A. 4
B. 20
C. 28
D. 27

Answer

To solve the problem, we need to find the maximum value of \( n + l + m \) for all unpaired electrons of chlorine in its second excited state. Let's break down the steps:

### Step 1: Determine the Ground State Electronic Configuration of Chlorine
The ground state electronic configuration of chlorine (Cl) is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5 \]

### Step 2: Identify the First Excited State
In the first excited state, one of the electrons from the 3p subshell gets excited to the 4s subshell. Thus, the configuration becomes:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \]

### Step 3: Identify the Second Excited State
In the second excited state, one electron from the 3s subshell gets excited to the 4s subshell. The electronic configuration now is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^0 \, 3p^6 \, 4s^1 \]

### Step 4: Identify Unpaired Electrons
In the second excited state, we have:
- 3s: 0 electrons (paired)
- 3p: 6 electrons (fully filled, paired)
- 4s: 1 electron (unpaired)

Thus, the only unpaired electron is in the 4s subshell.

### Step 5: Determine Quantum Numbers for the Unpaired Electron
For the unpaired electron in the 4s subshell:
- The principal quantum number \( n = 4 \)
- The azimuthal quantum number \( l = 0 \) (since it is an s orbital)
- The magnetic quantum number \( m = 0 \) (since \( m \) can range from \(-l\) to \(+l\), and for \( l = 0 \), \( m = 0 \))

### Step 6: Calculate \( n + l + m \)
Now, we calculate:
\[ n + l + m = 4 + 0 + 0 = 4 \]

### Conclusion
The maximum value of \( n + l + m \) for all unpaired electrons of chlorine in its second excited state is **4**.