The incorrect statement among the following is

Options

A. The bond order of the `CO^+` molecule is unusually 2.5
B. When a magnet is dipped into liquid oxygen, it sticks to it
C. Out of `H_(2)^(+) and H_(2)^(-)` the latter is more stable due to more number of bonding electrons.
D. Bond angle is `PH_3` is less than that in `NH_3`.

Answer

To determine the incorrect statement among the options provided, we need to analyze each statement based on our understanding of molecular orbital theory, bond order, and molecular geometry.

### Step-by-Step Solution

1. **Understanding Bond Order**:
- Bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
- For \(H_2^+\) and \(H_2^-\):
- \(H_2^+\) has 1 bonding electron and 0 antibonding electrons, so:
\[ \text{Bond Order of } H_2^+ = \frac{(1 - 0)}{2} = 0.5 \]
- \(H_2^-\) has 2 bonding electrons and 1 antibonding electron, so:
\[ \text{Bond Order of } H_2^- = \frac{(2 - 1)}{2} = 0.5 \]
- Both have the same bond order of 0.5, indicating they have the same stability.

2. **Analyzing Stability**:
- The statement claims that \(H_2^-\) is more stable due to having more bonding electrons. However, since both species have the same bond order, they are equally stable. This statement is incorrect.

3. **Bond Angles in \(NH_3\) and \(PH_3\)**:
- The bond angle in \(NH_3\) is approximately 107 degrees due to the presence of a lone pair which causes repulsion.
- The bond angle in \(PH_3\) is around 90 degrees because the lone pair in \(PH_3\) is in a pure s orbital and does not contribute to bond angle repulsion in the same way as in \(NH_3\).
- Thus, the statement that the bond angle in \(PH_3\) is less than in \(NH_3\) is correct.

4. **Magnetic Properties of \(O_2\)**:
- \(O_2\) has two unpaired electrons in its molecular orbital configuration, making it paramagnetic. Therefore, the statement about a magnet sticking to liquid oxygen is also correct.

5. **Conclusion**:
- The incorrect statement among the options provided is that \(H_2^-\) is more stable due to having more bonding electrons. Both \(H_2^+\) and \(H_2^-\) have the same bond order and thus the same stability.

### Final Answer
The incorrect statement is: "Out of \(H_2^+\) and \(H_2^-\), the latter is more stable due to more number of bonding electrons."