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Question : 84 of 180
Marks:
+1,
-0
Solution:
N2[14 electrons
]=σ1s2,σ*1s2,σ2s2,σ*2s2,π2px2=π2py2,σ2pz2Bond order
=‌[| Nb‌ and ‌Na‌ are bonding and ‌ |
| ‌ anti bonding molecular ‌ |
| ‌ orbital ‌s‌ - electron ‌ |
] =‌=3∴N2 has highest bond order i.e. 3 .
While
O2,He2 and
H2 have 2,0 and
1∕2 bond order respectively.
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