A) \[He_{2}^{+}\]
B) \[H{{e}_{2}}\]
C) \[H_{2}^{+}\]
D) \[H_{2}^{-}\]
Correct Answer: B
Solution :
\[H{{e}_{2}}(4)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}}\] Bond order \[\frac{Number\text{ }of\text{ }bonding\,electrons-Number\text{ }of\text{ }antibonding\,electrons}{2}\] \[=\frac{2-2}{2}=0\]
You need to login to perform this action.
You will be redirected in
3 sec
