Question
What is the order of stability of $${N_2}$$ and its ions?
A.
$${N_2} > N_2^ + = N_2^ - > N_2^{2 - }$$
B.
$$N_2^ + > N_2^ - > {N_2} > N_2^{2 - }$$
C.
$$N_2^ - > N_2^ + > {N_2} > N_2^{2 - }$$
D.
$$N_2^{2 - } > N_2^ - = N_2^ + > {N_2}$$
Answer :
$${N_2} > N_2^ + = N_2^ - > N_2^{2 - }$$
Solution :
Bond order of $${N_2} = 3,N_2^ + = 2.5,N_2^ - = 2.5$$ and $$N_2^{2 - }$$ is 2. Higher the bond order, more is the stability.