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 bond lengths of diatomic molecules of elements $X$,$Y$ and $Z$ respectively are $143 \text{ pm}$,$110 \text{ pm}$ and $121 \text{ pm}$. The atomic numbers of $X$,$Y$ and $Z$ respectively are:
For $F_2$ and $OF$ molecules,consider the following statements:
$(a)$ Bond order for both is one
$(b)$ $OF$ is paramagnetic but $F_2$ is diamagnetic
$(c)$ $F_2$ is more likely to dissociate into atoms than $OF$
$(d)$ Both have greater number of electrons in $BMO$ than that in $ABMO$
Which of the following statements are true $(T)$ or false $(F)$ respectively?
$(a)$ Bond order for both is one
$(b)$ $OF$ is paramagnetic but $F_2$ is diamagnetic
$(c)$ $F_2$ is more likely to dissociate into atoms than $OF$
$(d)$ Both have greater number of electrons in $BMO$ than that in $ABMO$
Which of the following statements are true $(T)$ or false $(F)$ respectively?
Difficult
View SolutionOxygen molecule is paramagnetic because
The bond order is maximum in
MediumAIIMS 1985
View SolutionFind the odd-electron molecules from the following:
$(i) C_2, (ii) H_2, (iii) SCl_2, (iv) NO, (v) NO_2$
$(i) C_2, (ii) H_2, (iii) SCl_2, (iv) NO, (v) NO_2$
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