what would be the bond order of the hypothetical molecule ar₂?

Explanation:

Step1: Write the molecular - orbital configuration of Ar

Argon has 18 electrons. The molecular - orbital configuration of $Ar_2$ is based on the combination of two Ar atoms. The general molecular - orbital filling order for second - row homonuclear diatomic molecules is: $\sigma_{1s}<\sigma_{1s}^<\sigma_{2s}<\sigma_{2s}^<\sigma_{2p_z}<\pi_{2p_x}=\pi_{2p_y}<\pi_{2p_x}^=\pi_{2p_y}^<\sigma_{2p_z}^$. Each Ar atom has 18 electrons, so $Ar_2$ has 36 electrons. The filled molecular - orbitals are: $(\sigma_{1s})^2(\sigma_{1s}^)^2(\sigma_{2s})^2(\sigma_{2s}^)^2(\sigma_{2p_z})^2(\pi_{2p_x})^4(\pi_{2p_y})^4(\pi_{2p_x}^)^4(\pi_{2p_y}^)^4(\sigma_{2p_z}^)^2$.

Step2: Use the bond - order formula

The bond - order formula is $BO=\frac{1}{2}(N_b - N_a)$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. Counting the electrons: The number of bonding electrons $N_b = 2 + 2+2 + 4+4 = 14$. The number of antibonding electrons $N_a=2 + 2+2 + 4+4 + 2=16$.

Step3: Calculate the bond order

Substitute $N_b = 14$ and $N_a = 16$ into the bond - order formula: $BO=\frac{1}{2}(14 - 16)=\frac{1}{2}\times(- 2)=0$.

Answer:

0