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4.Chemical Bonding and Molecular Structure
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The bond order of $O_2^ + $ is the same as in
A
$N_2^ + $
B
$C{N^ - }$
C
$CO$
D
$N{O^ + }$
Solution
(a) $O_2^ + (15{e^ – }) = K{K^*}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_x})^2}$
${(\pi 2{p_y})^2}{(\pi 2{p_z})^2}{({\pi ^*}2{p_y})^1}{({\pi ^*}2{p_z})^0}$
Hence, bond order $ = \frac{1}{2}(10 – 5) = 2.5$$N_2^ + (13{e^ – }) = K{K^*}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_x})^2}$
${(\pi \,2{p_y})^2}{(\pi \,2{p_z})^1}$
Hence, bond order $ = \frac{1}{2}(9 – 4) = 2.5$.
Standard 11
Chemistry
Similar Questions
Match each of the diatomic molecules in Column $I$ with its property / properties in Column $II$.
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{B}_2$ | $(p)$ Paramagnetic |
| $(B)$ $\mathrm{N}_2$ | $(q)$ Undergoes oxidation |
| $(C)$ $\mathrm{O}_2^{-}$ | $(r)$ Undergoes reduction |
| $(D)$ $\mathrm{O}_2$ | $(s)$ Bond order $\geq 2$ |
| $(t)$ Mixing of ' $\mathrm{s}$ ' and ' $\mathrm{p}$ ' orbitals |
