The incorrect statement regarding molecular orbital $(s)$ is
- AIf there is a nodal plane perpendicular to the internuclear axis and lying between the nuclei of bonded atoms then corresponding orbital is antibonding $M.O.$
- BIf a nodal plane lies in the internuclear axis,then corresponding orbital is $\pi$ bonding $M.O.$
- CThe $\sigma$-bonding molecular orbital does not contain nodal planes containing the internuclear axis.
- DThe $\delta$-bonding molecular orbital possesses three nodal planes containing the internuclear axis.
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Similar Questions
The theory that can completely/properly explain the nature of bonding in $[Ni(CO)_4]$ is
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:
What is the number of unpaired electron$(s)$ in the highest occupied molecular orbital $(HOMO)$ of the following species: $N_2$,$N_2^{+}$,$O_2$,$O_2^{+}$?
Two homo-diatomic molecules $X_2$ and $Y_2$ have the same bond order. Then,select the $CORRECT$ statement.
Medium
View SolutionFrom the following,identify the ions with the same bond order.
$I$. $CN^{-}$
$II$. $N_2^{+}$
$III$. $O_2^{2-}$
$IV$. $NO^{+}$
$I$. $CN^{-}$
$II$. $N_2^{+}$
$III$. $O_2^{2-}$
$IV$. $NO^{+}$
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