Question

draw the molecular orbital (mo) electron diagram for the f₂⁺ molecular ion. be sure your diagram contains all of the electrons in the ion, including any core electrons.

Explanation:

Step1: Determine number of electrons

Fluorine has atomic number 9. In $F_2^+$, total electrons = 2×9 - 1=17.

Step2: Write molecular - orbital energy - level order

For second - period homonuclear diatomic molecules with Z > 7, the molecular - orbital energy - level order is $\sigma_{1s}<\sigma_{1s}^*<\sigma_{2s}<\sigma_{2s}^*<\sigma_{2p_z}<\pi_{2p_x}=\pi_{2p_y}<\pi_{2p_x}^*=\pi_{2p_y}^*<\sigma_{2p_z}^*$.

Step3: Fill electrons into orbitals

Fill the 17 electrons into the orbitals following Pauli - exclusion principle (each orbital can hold a maximum of 2 electrons with opposite spins) and Hund's rule (electrons occupy degenerate orbitals singly first before pairing). The electron - configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^1$.

Answer:

Draw an energy - level diagram with the above - mentioned orbitals in order of increasing energy. Place 2 electrons in $\sigma_{1s}$, 2 in $\sigma_{1s}^*$, 2 in $\sigma_{2s}$, 2 in $\sigma_{2s}^*$, 2 in $\sigma_{2p_z}$, 2 in each of $\pi_{2p_x}$ and $\pi_{2p_y}$, 2 in each of $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ and 1 in the remaining $\pi_{2p_y}^*$ (or $\pi_{2p_x}^*$) orbital. Use arrows to represent electrons with opposite spins for paired electrons and a single arrow for the unpaired electron.