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 will analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement 1:** The bond order of CO positive molecule is unusually 2.5.

- **Bond Order Calculation:**
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of electrons in bonding orbitals}) - (\text{Number of electrons in anti-bonding orbitals})}{2} \]

- **Electronic Configuration:**
- Carbon (C): \(1s^2 2s^2 2p^2\)
- Oxygen (O): \(1s^2 2s^2 2p^4\)
- For CO\(^+\): Total electrons = 13 (1 less than CO, which has 14)

- **Molecular Orbital Configuration for CO\(^+\):**
- Bonding orbitals: \( \sigma_{1s}^2, \sigma_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2 \)
- Anti-bonding orbitals: \( \sigma^*_{2s}^0 \)

- **Calculating Bond Order:**
- Bonding electrons = 10 (from \( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} \))
- Anti-bonding electrons = 3 (from \( \sigma^*_{2s} \) and \( \sigma^*_{2p} \))

\[ \text{Bond Order} = \frac{10 - 3}{2} = \frac{7}{2} = 3.5 \]

- **Conclusion:** The bond order of CO\(^+\) is 3.5, not 2.5. Thus, this statement is incorrect.

### Step 2: Analyze the second statement
**Statement 2:** When a magnet is dipped into liquid oxygen, it sticks to it.

- **Explanation:**
Oxygen is paramagnetic due to the presence of unpaired electrons in its molecular orbital configuration. Paramagnetic substances are attracted to magnetic fields.

- **Conclusion:** This statement is correct.

### Step 3: Analyze the third statement
**Statement 3:** Out of H\(_2^+\) and H\(_2^-\), the latter is more stable due to more number of bonding electrons.

- **Electronic Configuration:**
- H\(_2^+\): \( \sigma_{1s}^1 \) (1 bonding electron)
- H\(_2^-\): \( \sigma_{1s}^2 \sigma^*_{1s}^1 \) (2 bonding electrons, 1 anti-bonding electron)

- **Bond Order Calculation:**
- For H\(_2^+\): Bond order = \( \frac{1 - 0}{2} = 0.5 \)
- For H\(_2^-\): Bond order = \( \frac{2 - 1}{2} = 0.5 \)

- **Stability Comparison:**
H\(_2^+\) is more stable than H\(_2^-\) because it has no electrons in anti-bonding orbitals.

- **Conclusion:** This statement is incorrect.

### Step 4: Analyze the fourth statement
**Statement 4:** The bond angle of PH\(_3\) is less than that in NH\(_3\).

- **Explanation:**
NH\(_3\) has a trigonal pyramidal shape with bond pairs closer together due to the smaller size of nitrogen compared to phosphorus in PH\(_3\). The larger size of phosphorus leads to greater bond pair repulsion in PH\(_3\).

- **Conclusion:** This statement is correct.

### Final Conclusion:
The incorrect statements are:
1. The bond order of CO\(^+\) is unusually 2.5.
2. Out of H\(_2^+\) and H\(_2^-\), the latter is more stable due to more number of bonding electrons.