The wave functions of $1 s$-orbitals of two hydrogen atoms are $\psi_A$ and $\psi_B . \psi_A$ and $\psi_B$…
The wave functions of $1 s$-orbitals of two hydrogen atoms are $\psi_A$ and $\psi_B . \psi_A$ and $\psi_B$ are linearly combined to form two molecular orbitals $\left(\sigma\right.$ and $\left.\sigma^*\right)$. Which of the following statements are correct?
I. $\sigma^*$ is equal to $\left(\psi_A-\psi_B\right)$.
II. In $\sigma$-orbital, one nodal plane is present in between two nuclei.
III. The energy of $\sigma$-orbital is lower than the energy of $\sigma^*$-orbital.
I, II, III
I, II only
II, III only
I, III only
Solution
According to molecular orbital theory:
I. $\sigma^*$-orbital are formed by the substraction of two wave functions say $\psi_A$ and $\psi_B$ and therefore $\sigma^*=\psi_A-\psi_B$.
II. $\sigma$-orbital are formed when two orbitals are in same phase and thus, do not have nodal plane in between two nuclei.
III. Combination of two sigma $(\sigma)$ atomic orbitals given two molecular orbitals, out of which one is of lower energy called $\sigma$ bonding orbital and other is of higher energy called $\sigma^*$ anti-bonding orbital.
Hence, (i) and (iii) are the correct statements and option (d) is the correct answer.