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.
  1. I, II, III
  2. I, II only
  3. II, III only
  4. 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.

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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