Match List-I with List-II based on the bond o | vedclass

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(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$

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$(a)$ $\rightarrow (iii), (b)$ $\rightarrow (iv), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$

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(A) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$(a) \ Ne_2$ $(20 \ e^-)$: Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2, \sigma^* 2p_z^2$. $B.O. = \frac{10-10}{2} = 0$.$(b) \ N_2$ $(14 \ e^-)$: $B.O. = \frac{10-4}{2} = 3$.$(c) \ F_2$ $(18 \ e^-)$: $B.O. = \frac{10-8}{2} = 1$.$(d) \ O_2$ $(16 \ e^-)$: $B.O. = \frac{10-6}{2} = 2$.Thus,the correct matching is: $(a)$ $\rightarrow (iii), (b)$ $\rightarrow (iv), (c)$ $\rightarrow (i), (d)$ $\rightarrow (ii)$.

List-$I$	List-$II$
$(a)$ $Ne_2$	$(i)$ $1$
$(b)$ $N_2$	$(ii)$ $2$
$(c)$ $F_2$	$(iii)$ $0$
$(d)$ $O_2$	$(iv)$ $3$

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:

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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: