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(C) Bond order is defined as the number of chemical bonds present between two atoms of a molecule.Using Molecular Orbital Theory $(MOT)$:$I. N_2$ ($14

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$I > III > II > IV$

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(C) Bond order is defined as the number of chemical bonds present between two atoms of a molecule.Using Molecular Orbital Theory $(MOT)$:$I. N_2$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order $= \frac{10-4}{2} = 3$.$II. O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order $= \frac{10-6}{2} = 2$.$III. O_2^+$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10-5}{2} = 2.5$.$IV. O_2^-$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10-7}{2} = 1.5$.Comparing the bond orders: $3 (I) > 2.5 (III) > 2 (II) > 1.5 (IV)$.Thus,the correct order is $I > III > II > IV$.

List-$I$	List-$II$
$(1)$ ${\rm{NO}}$	$(A)$ $1.5$
$(2)$ ${\rm{CO}}$	$(B)$ $2.0$
$(3)$ ${\rm{O}}_2^-$	$(C)$ $2.5$
$(4)$ ${\rm{O}}_2$	$(D)$ $3.0$

Arrange the following in the correct order of their bond orders.
$I. N_2, II. O_2, III. O_2^+, IV. O_2^-$

Similar Questions

Match the species given in List-$I$ with their bond orders given in List-$II$.

List-$I$ List-$II$

$(1)$ ${\rm{NO}}$ $(A)$ $1.5$

$(2)$ ${\rm{CO}}$ $(B)$ $2.0$

$(3)$ ${\rm{O}}_2^-$ $(C)$ $2.5$

$(4)$ ${\rm{O}}_2$ $(D)$ $3.0$

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Arrange the following in the correct order of their bond orders.$I. N_2, II. O_2, III. O_2^+, IV. O_2^-$