The $H.O.M.O.$ (Highest Occupied Molecular Orbital) of the $CO$ molecule is:
- ANon-bonding $M.O.$ with slight antibonding character
- BNon-bonding $M.O.$ with slight bonding character
- CPure non-bonding $M.O.$
- DNone of the above
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Similar Questions
According to "molecular orbital theory",which among the following diatomic molecules only has $\pi$-bonds?
Match List-$I$ with List-$II$ based on the bond order of the molecules.
List-$I$ List-$II$ $(a)$ $Ne_2$ $(i)$ $1$ $(b)$ $N_2$ $(ii)$ $2$ $(c)$ $F_2$ $(iii)$ $0$ $(d)$ $O_2$ $(iv)$ $3$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(a)$ $Ne_2$ | $(i)$ $1$ |
| $(b)$ $N_2$ | $(ii)$ $2$ |
| $(c)$ $F_2$ | $(iii)$ $0$ |
| $(d)$ $O_2$ | $(iv)$ $3$ |
Choose the correct answer from the options given below:
Although $CN^{-}$ ion and $N_2$ molecule are isoelectronic,yet $N_2$ molecule is chemically inert because of
DifficultAIEEE 2012
View SolutionThe correct molecular orbital diagram for $F_2$ molecule in the ground state is
In which change do both the bond order and the number of unpaired electrons increase?
Difficult
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