Write the electronic configuration of `H_(2)^(+)` ion. Calculate the bond order.

Answer

To solve the problem of finding the electronic configuration of the \( H_2^+ \) ion and calculating its bond order, we can follow these steps:

### Step 1: Determine the number of electrons in \( H_2^+ \)
The \( H_2^+ \) ion is derived from the hydrogen molecule \( H_2 \) by removing one electron. The hydrogen molecule \( H_2 \) has 2 electrons (one from each hydrogen atom). Therefore, \( H_2^+ \) has:

\[ 2 - 1 = 1 \text{ electron} \]

### Step 2: Construct the Molecular Orbital (MO) Diagram
In the case of \( H_2^+ \), we need to consider the molecular orbitals formed by the two hydrogen atoms. The two molecular orbitals are:

- **Bonding Molecular Orbital**: \( \sigma_{1s} \)
- **Antibonding Molecular Orbital**: \( \sigma^*_{1s} \)

Since \( H_2^+ \) has only one electron, it will occupy the lowest energy molecular orbital, which is the bonding molecular orbital \( \sigma_{1s} \).

### Step 3: Write the Electronic Configuration
The electronic configuration for \( H_2^+ \) can be written as:

\[ \sigma_{1s}^1 \]

This indicates that there is one electron in the bonding molecular orbital.

### Step 4: Calculate the Bond Order
The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \]

where:
- \( N_b \) = number of electrons in bonding orbitals
- \( N_a \) = number of electrons in antibonding orbitals

For \( H_2^+ \):
- \( N_b = 1 \) (one electron in \( \sigma_{1s} \))
- \( N_a = 0 \) (no electrons in \( \sigma^*_{1s} \))

Substituting these values into the formula gives:

\[ \text{Bond Order} = \frac{(1 - 0)}{2} = \frac{1}{2} = 0.5 \]

### Final Answer
The electronic configuration of \( H_2^+ \) is \( \sigma_{1s}^1 \) and the bond order is \( 0.5 \).

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