`H_(2), Li_(2), B_(2)` each has bond order equal to 1 the order of their stability is

Options

A. `H_(2)=Li_(2)=B_(2)`
B. `H_(2) gt Li_(2) gt B_(2)`
C. `H_(2) gt B_(2) gt Li_(2)`
D. `B_(2) gt Li_(2) gt H_(2)`

Answer

To determine the order of stability for the molecules \(H_2\), \(Li_2\), and \(B_2\), we will analyze their bond orders and the number of electrons in their anti-bonding orbitals. Here's a step-by-step solution:

### Step 1: Understand Bond Order
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of anti-bonding electrons})}{2} \]
A higher bond order generally indicates greater stability.

### Step 2: Analyze \(H_2\)
- **Electronic Configuration**: Each hydrogen atom has 1 electron, so for \(H_2\):
- Bonding electrons: 2 (both in the bonding \(\sigma_{1s}\) orbital)
- Anti-bonding electrons: 0 (none in the \(\sigma^*_{1s}\) orbital)

Using the bond order formula:
\[ \text{Bond Order}_{H_2} = \frac{2 - 0}{2} = 1 \]

### Step 3: Analyze \(Li_2\)
- **Electronic Configuration**: Each lithium atom has 3 electrons (1s² 2s¹), so for \(Li_2\):
- Bonding electrons: 4 (2 in \(\sigma_{1s}\) and 2 in \(\sigma_{2s}\))
- Anti-bonding electrons: 2 (both in \(\sigma^*_{1s}\))

Using the bond order formula:
\[ \text{Bond Order}_{Li_2} = \frac{4 - 2}{2} = 1 \]

### Step 4: Analyze \(B_2\)
- **Electronic Configuration**: Each boron atom has 5 electrons (1s² 2s² 2p¹), so for \(B_2\):
- Bonding electrons: 6 (2 in \(\sigma_{1s}\), 2 in \(\sigma_{2s}\), and 2 in \(\pi_{2p}\))
- Anti-bonding electrons: 2 (in \(\sigma^*_{1s}\))

Using the bond order formula:
\[ \text{Bond Order}_{B_2} = \frac{6 - 2}{2} = 2 \]

### Step 5: Compare Stability
- All three molecules \(H_2\), \(Li_2\), and \(B_2\) have a bond order of 1, but we need to consider the number of anti-bonding electrons:
- \(H_2\): 0 anti-bonding electrons
- \(Li_2\): 2 anti-bonding electrons
- \(B_2\): 2 anti-bonding electrons

Since stability is inversely proportional to the number of anti-bonding electrons, we find:
- \(H_2\) is the most stable (0 anti-bonding electrons)
- \(Li_2\) and \(B_2\) are less stable (2 anti-bonding electrons each)

### Final Order of Stability
The order of stability is:
\[ H_2 > Li_2 = B_2 \]