Question

3. (cont).(C) (10 pts) Now consider the MO diagram below, which could be used to describe the bonding. Explain whether Not or NO- would have a stronger bond. 2po* 2pr* 2pr* P2 2pp 2px.py 2p2 2pT 2pr 2po 2so* 2s 2s 2so Which one is stronger? Why? Stronger bond = (d) (10 pts) What bond order would you predict for neutral NO? BO =

Answer 1

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

18 In coulson treatment take and o 152 4 EN diff. an heteronuclear molecule, when Small (i.e. less than 1.5), face . ie hybridisation take place before mio... formation. No consist of w Heckonic configuration of 282, 284 28y' 23' Heckonic configuration of 0 = 182 25² 2px? after coulson treatment En of oN (SP3)? (Spg)'2 PyJ' (2001) after Coulson treatment EN of o (SP3)? (SP3)' (2py)? (22x)' non bonding Bonding mo Antibonding mo If z is internuclear ani's -bond el Total То но will form 2 4 + S-P₂ → Propy ł bond mo configuration will be. (583). (OSP) G) (SP) Car (37) L. som bonding Non bonding

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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