Question

A ${N_2}$, ${N_2}^ + $, ${N_2}^ - $, ${N_2}^{2 + }$, ${N_2}^{2 - }$.
 Compare N-N bond length in all.

Answer 1

Hint: In order to the question, first we have to find the number of bonding electrons in each given compound, then we should calculate the bond order of each compound and compare the bond orders and bond lengths.

Complete step-by-step answer:
In ${N_2}$; It has triple bonding. Bond order of ${N_2}$ is 3. So, it is the longest Bond Order in all Nitrogen Compounds. So, N-N bond length should be shortest in ${N_2}$ because of its triple bonding but ${N_2}$ Bond is more stronger than any other Nitrogen Compound.

In ${N_2}^ + $ and ${N_2}^ - $; in both Nitrogen Compound, bond order is almost same is 2.5.But the Bond Length of ${N_2}^ - $ is longer than the bond length of ${N_2}^ + $ because ${N_2}^ - $ has 5 electrons in antibonding whereas ${N_2}^ + $ has 4 electrons in antibonding. But ${N_2}^ - $ will make a weaker N-N bond than ${N_2}^ + $.
In ${N_2}^{2 + }$ and ${N_2}^{2 - }$; in these both compounds, again the bond order is almost the same as 2.0. So, the Bond length is also the same in these compounds.
In order of Bond Length:-
Longest Bond $({N_2}^{2 + }\,\& \,{N_2}^{2 - }) > ({N_2}^ + \,\& \,{N_2}^ - ) > {N_2}$ Shortest Bond.
In order of Bond Strength:-
Strongest Bond ${N_2} > ({N_2}^ + \,\& \,{N_2}^ - ) > ({N_2}^{2 + }\,\& \,{N_2}^{2 - })$ Weakest Bond.

Note: Bond Order is inversely proportional to Bond Length. If the Bond Order increases then the Bond Length decreases. And if the Bond Length of any compound is longest then the compound is weakest than other compounds.