In molecular orbital theory, molecular orbitals form from the linear combination of atomic orbitals. When dealing with diatomic molecules, these combinations form bonding and antibonding orbitals.
The symbols \(\psi_A\) and \(\psi_B\) represent the wave functions of atomic orbitals from atoms A and B, respectively.
To form molecular orbitals, atomic orbitals can combine constructively or destructively:
The antibonding molecular orbital, \(\sigma^*\), is formed when the wave functions interfere destructively, leading to a nodal plane between the nuclei where the electron density is low.
Given these explanations, the correct representation of \(\sigma^*\) is:
Correct Answer: \(\psi_A - \psi_B\)
Let's rule out the other options:
The bonding ($\sigma$) and anti-bonding ($\sigma^*$) molecular orbitals are formed by the constructive and destructive interference of atomic orbitals' wave functions.
For an anti-bonding molecular orbital ($\sigma^*$):
\[ \psi_{\sigma^*} = \psi_A - \psi_B. \]
This occurs due to the out-of-phase overlap of the wave functions, leading to a node between the nuclei and a higher energy state.
Thus, $\sigma^*$ is represented by:
\[ \psi_A - \psi_B. \]
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
