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Chemistry Question on Chemical bonding and molecular structure

Identify the correct increasing order of energies of molecular orbitals for F2 molecule.

A

σ1s<σ1s∗<σ2s<σ2s∗σ_{1s}< σ^{*}_{1s} < σ_{2s} < σ^{*}_{2s}

B

σ1s<σ2s<σ1s∗<σ2s∗σ_{1s}< σ_{2s} <σ^{*}_{1s} < σ^{*}_{2s}

C

σ1s<σ1s∗<σ2s∗<σ2sσ_{1s} < σ^{*}_{1s} < σ^{*}_{2s} < σ_{2s}

D

σ1s∗<σ1s<σ2s∗<σ2sσ ^{*}_{1s} < σ_{1s} < σ^{ *}_{2s} < σ_{2s}

Answer

σ1s∗<σ1s<σ2s∗<σ2sσ ^{*}_{1s} < σ_{1s} < σ^{ *}_{2s} < σ_{2s}

Explanation

Solution

Let's break down the options to understand why this is the correct order:

  • In the F2 molecule, the σ1sσ_{1s} bonding orbital is formed by the overlap of the two 1s atomic orbitals from each fluorine atom. This bonding orbital is lower in energy compared to the other molecular orbitals.
  • The σ1s∗σ^{*}_{1s} antibonding orbital is formed by the destructive overlap of the two 1s atomic orbitals. It is higher in energy than the σ1sσ_{1s} bonding orbital.
  • The σ2sσ_{2s} bonding orbital is formed by the overlap of the two 2s atomic orbitals from each fluorine atom. It is lower in energy than the σ* 1s orbital.
  • Finally, the σ2s∗σ^{*}_{2s} antibonding orbital is formed by the destructive overlap of the two 2s atomic orbitals. It is higher in energy than the σ2sσ_{2s} bonding orbital.

Therefore, the correct order is σ1s∗<σ1s<σ2s∗<σ2sσ^{*}_{1s} < σ_{1s} < σ^{*}_{2s} < σ_{2s}, which corresponds to option (D).