The distribution of electrons among molecular orbitals is called the electronic configuration of the molecule which provides us the following very important informations about the molecule.
(A)Stability of molecule : The molecule is stable if number of bonding moleculer orbital electrons `(N_b)` is greater than the number of antibonding molecular orbital electrons `(N_a)`.
(B) Bond order : Bond order `=1/2(N_b-N_a)`
A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
( C)Nature of the bond : Bond order 1,2 and 3 corresponds to single , double and triple bonds respectively.
(D)Bond length: Bond length decreases as bond order increases.
(E)Magnetic nature: Molecular orbitals in a molecule are doubly occupied , the substance is diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.
Which of the following statements is incorrect ?

Options

A. Among `O_2^(+),O_2` and `O_2^(-)` the bond length decreases as `O_2^(-) gt O_2 gt O_2^(+)`
B. `He_2` molecule does not exist as the bonding and anti-bonding orbitals cancel each other
C. `C_2, O_2^(2+)` and `Li_2` are dimagnetic
D. In `F_2` molecule the energy of `sigma_(2p_z)` is more than `pi_2p_x` and `pi_2p_y`

Explanation

To determine which statement is incorrect regarding the electronic configuration of molecules and their properties, we will analyze each statement based on the information provided.

### Step-by-Step Solution:

1. **Understanding Bond Order and Stability**:
- The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2}(N_b - N_a) \]
- Where \(N_b\) is the number of electrons in bonding molecular orbitals and \(N_a\) is the number of electrons in antibonding molecular orbitals. A positive bond order indicates a stable molecule, while a zero or negative bond order indicates instability.

2. **Analyzing Statement A**:
- Statement A claims that among \(O_2\), \(O_2^+\), and \(O_2^-\), the bond length decreases in the order given.
- For \(O_2\): \(N_b = 6\), \(N_a = 2\) → Bond order = 2.
- For \(O_2^-\): \(N_b = 6\), \(N_a = 3\) → Bond order = 1.5.
- For \(O_2^+\): \(N_b = 6\), \(N_a = 1\) → Bond order = 2.5.
- As bond order increases from \(O_2^-\) to \(O_2\) to \(O_2^+\), bond length decreases. Thus, Statement A is **correct**.

3. **Analyzing Statement B**:
- Statement B states that the helium molecule \(He_2\) does not exist because bonding and antibonding orbitals cancel each other out.
- In \(He_2\), there are 4 electrons: 2 in the bonding orbital and 2 in the antibonding orbital, resulting in a bond order of 0. Therefore, \(He_2\) does not exist. Statement B is **correct**.

4. **Analyzing Statement C**:
- Statement C claims that \(Cu_2\), \(O_2^{2+}\), and \(Li_2^+\) are diamagnetic.
- \(Cu_2\) has 12 electrons and is diamagnetic as all electrons are paired.
- \(O_2^{2+}\) has 14 electrons and is also diamagnetic.
- \(Li_2^+\) has 5 electrons, and since it has unpaired electrons, it is paramagnetic, making Statement C **incorrect**.

5. **Analyzing Statement D**:
- Statement D states that in \(F_2\), the \(\sigma 2p_z\) orbital has more energy than the \(\pi 2p_x\) and \(\pi 2p_y\) orbitals.
- In \(F_2\), the energy levels are such that \(\sigma 2p_z\) is lower in energy than \(\pi 2p_x\) and \(\pi 2p_y\). Therefore, Statement D is **incorrect**.

6. **Conclusion**:
- The incorrect statements are C and D. However, since the question asks for one incorrect statement, the most clearly incorrect one based on the analysis is **Statement C**.

### Final Answer:
The incorrect statement is **C**: \(Cu_2\), \(O_2^{2+}\), and \(Li_2^+\) are diamagnetic.

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