The maximum oxidation state that can be exhibited by a halogen in its second excited state 

Options

A. `+1`
B. `+3`
C. `+5`
D. `+7`

Answer

To determine the maximum oxidation state that can be exhibited by a halogen in its second excited state, we will follow these steps:

### Step 1: Understand the Electronic Configuration of Halogens
Halogens belong to Group 17 of the periodic table. Their general electronic configuration is given as:
\[ \text{ns}^2 \text{np}^5 \text{nd}^0 \]
Where \( n \) represents the principal quantum number corresponding to the period of the halogen.

### Step 2: Identify the First Excited State
In the first excited state, one of the electrons from the \( p \) orbital will be promoted to the \( d \) orbital. The electronic configuration in the first excited state will be:
\[ \text{ns}^2 \text{np}^4 \text{nd}^1 \]
This configuration allows for an oxidation state of +3, as the halogen can lose three electrons (two from \( s \) and one from \( p \)).

### Step 3: Identify the Second Excited State
In the second excited state, another electron from the \( p \) orbital is promoted to the \( d \) orbital. The electronic configuration in the second excited state will be:
\[ \text{ns}^2 \text{np}^3 \text{nd}^2 \]
This configuration allows for an oxidation state of +5, as the halogen can lose five electrons (two from \( s \) and three from \( p \)).

### Step 4: Determine the Maximum Oxidation State
In the second excited state, the halogen can exhibit an oxidation state of +5. This is the maximum oxidation state that can be achieved by halogens in their second excited state.

### Conclusion
The maximum oxidation state that can be exhibited by a halogen in its second excited state is:
\[ \text{+5} \]