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4.Chemical Bonding and Molecular Structure
medium
The bond length the species ${O_2},O_2^ + $ and $O_2^ - $ are in the order of
A
$O_2^ - > {O_2} > O_2^ + $
B
$O_2^ + > O_2^ - > {O_2}$
C
${O_2} > O_2^ + > O_2^ - $
D
$O_2^ + > {O_2} > O_2^ - $
Solution
(a) As bond order increase bond length decrease the bond order of species are
$ = \frac{{{\rm{number \,of\, bonding \,electron – Number \,of \,}}a{\rm{.}}b{\rm{. electron}}}}{2}$
For ${O_2} = \frac{{10 – 6}}{2} = 2$ ;
$O_2^ + = \frac{{10 – 5}}{2} = 2.5$
$O_2^ – = \frac{{10 – 7}}{2} = 1.5$
So, bond order $O_2^ + > {O_2} > O_2^ – $
Standard 11
Chemistry
Similar Questions
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
