Question

The bond order of HeH$^+$ is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

MO bond order rule: \begin{itemize} \item Remove electrons for positive charge \item Add for negative charge \item Use $(N_b - N_a)/2$ \end{itemize}

Answer 1

The Correct Option is A

Concept: Bond order using Molecular Orbital Theory: \[ \text{Bond order} = \frac{N_b - N_a}{2} \] Step 1: Electron count He = 2 electrons H = 1 electron HeH$^+$ loses one electron: \[ 2 + 1 - 1 = 2 \text{ electrons} \] Step 2: MO filling Two electrons occupy bonding $\sigma$ orbital. \[ N_b = 2, \quad N_a = 0 \] Step 3: Bond order \[ \frac{2 - 0}{2} = 1 \]