Analytical ChemistryHard
Question
Bond order of 1.5 is shown by
Options
A.O2+
B.O2-
C.O22-
D.O2
Solution
(a) MO configuration of O2+ ( 8 + 8 - 1 = 15)
where, Nb = number of electrons in bounding molecular orbital
Na = number of electrons in anti-bounding molecular orbital
∴
Similarly,
O2- ( 8 + 8 + 1 = 17 )
so BO
(c)O22- ( 8 + 8 + 2 = 18 )
BO
(d) O2( 8 + 8 = 16 )
BO
Thus,O2- shows the bond order 1.5.
where, Nb = number of electrons in bounding molecular orbital
Na = number of electrons in anti-bounding molecular orbital
∴
Similarly,
O2- ( 8 + 8 + 1 = 17 )
so BO
(c)O22- ( 8 + 8 + 2 = 18 )
BO
(d) O2( 8 + 8 = 16 )
BO
Thus,O2- shows the bond order 1.5.
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