write molecular orbital electronic configuration of N2²- and N2+ .Compare their relative bond dissociation energies and bond lengths
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Bond order =
2
1
[N
b
−N
a
]
N
b
= number of bonding electrons
N
a
= number of anti-bonding electrons
→ Bond order of N
2
is 3
→ Bond order of N
2
−
is 2.5
→ Bond order of N
2
+
is 2.5
Bond order α stability α
bondlength
1
order of bond length
Ans :- N
2
+
=N
2
−
>N
2
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