Question

the table below lists information about some diatomic molecules or molecular ions. for each molecule in the table: first, decide if the molecule is stable or not. then, if your answer to this question is \yes\: decide whether the molecule would be diamagnetic or paramagnetic. calculate the molecules bond order. molecule stable? diamagnetic or paramagnetic? bond order n₂ yes diamagnetic no paramagnetic c₂⁻ yes diamagnetic no paramagnetic be₂ yes diamagnetic no paramagnetic

Explanation:

Step1: Recall 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.

Step2: Determine bonding and antibonding electrons for $N_2$

For $N_2$, the molecular orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^2$. The number of bonding electrons $N_b = 10$ and the number of antibonding electrons $N_a=4$. Then $BO=\frac{1}{2}(10 - 4)=3$.

Step3: Determine bonding and antibonding electrons for $C_2^-$

The carbon atom has 6 electrons, and $C_2^-$ has $2\times6 + 1=13$ electrons. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^1$. $N_b = 7$ and $N_a = 4$. So $BO=\frac{1}{2}(7 - 4)=1.5$.

Step4: Determine bonding and antibonding electrons for $Be_2$

The beryllium atom has 4 electrons, and $Be_2$ has $2\times4 = 8$ electrons. The molecular - orbital configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2$. $N_b = 4$ and $N_a = 4$. Then $BO=\frac{1}{2}(4 - 4)=0$.

Answer:

molecule	stable?	diamagnetic or paramagnetic?	bond order
$C_2^-$	yes	paramagnetic	1.5
$Be_2$	no	diamagnetic	0