Manipal Entrance Test 2014 Chemistry Paper

The molecular orbital electronic configuration of CF(15e) is σ1s2,σ*1s2α2s2,σ*2s2σ2Pz2 π2px2≈π2py2,π2px1
Bond order (BO) =

10−5

2

=2.5
In CF+, there is one electron less in antibonding MO therefore, BO=

10−4

2

=3
In CF−, there is one electron more inantibonding MO therefore, BO=

10−6

2

=2
Bond length ∝

1

Bond order

Hence, the order of bond length in given series is CF+<CF<CF−