solution:
whenever two atomic orbital combine two molecular orbital form.
one is bonding molecular orbital
other is antibonding molecular orbital
bonding molecular orbital has low energy as compared to antibonding molecular orbital.
hence, four molecular orbital are bonding whose name are as follows:
1.) sigma 2s
2.) sigma 2p
rest two are degenerate molecular orbital or both having same energy.
3.) 2p pi
4.) 2p pi
now coming to NO
this is MO configuration of NO
so bond order is calculated as
bond order = ( bonding electron - anti bonding electron) /2
bond order = (6-1)/2 = 2.5
hence bond order of neutral NO = 2.5
now coming on NO+ and NO-
for NO+ 1 electron will be removed from antibonding molecular orbital.
so bond order =( 6-0)/2 = 3
for NO- one electron will be added in antibonding molecular orbital
so bond order = (6 - 2 )/2 = 2
as bond order is directly proportional to bond energy
hence bond order of NO + is high so bond energy will be high
hence bond of NO+ will be stronger as compared to NO-
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3. (cont).(C) (10 pts) Now consider the MO diagram below, which could be used to describe...
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