According to molecular orbital theory, which of the following statement is not correct?
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Remember, the presence of unpaired electrons in molecular orbitals indicates paramagnetism, while paired electrons indicate diamagnetism. Additionally, molecular orbital theory helps in determining bond orders by counting bonding and antibonding electrons.
Molecular orbital theory helps explain the electronic structure and magnetic properties of molecules. Let’s analyze each option for the \( C_2 \) molecule:
Statement (1): \( C_2 \) molecule is diamagnetic in nature. According to molecular orbital theory, \( C_2 \) has two unpaired electrons in the \( \pi \)-orbitals, making it paramagnetic, not diamagnetic. Hence, this statement is incorrect.
Statement (2): Bond order of \( C_2 \) molecule is 2. The bond order is calculated using the formula: \[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \] There are 6 bonding electrons and 2 antibonding electrons, leading to a bond order of: \[ \text{Bond Order} = \frac{6 - 2}{2} = 2 \] Hence, this statement is correct.
Statement (3): \( C_2^- \) ion is paramagnetic in nature. The \( C_2^- \) ion has one additional electron, which occupies an antibonding \( \pi^* \) orbital. This results in an unpaired electron, making the \( C_2^- \) ion paramagnetic. Thus, this statement is also correct.
Statement (4): \( C_2 \) consists of 1 sigma and 1 pi bond. \( C_2 \) consists of both sigma and pi bonds. However, there are two pi bonds formed by the sideways overlap of the p-orbitals, in addition to the sigma bond. Therefore, this statement is incorrect, as it underestimates the number of pi bonds in the \( C_2 \) molecule.
\(\boxed{ \text{Final Answer: Statement (2) and (3) are correct.} }\)
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