Question

# Which of the following options represents the correct bond order?

A
O2>O2<O+2
B
O2<O2>O+2
C
O2>O2>O+2
D
O2<O2<O+2

Solution

## The correct option is D $$O_2{^-} < O_2 < O_2^+$$$$Bond\ Order = \dfrac { Bonding\quad Orbital\quad electrons\quad -\quad Antibonding\quad Orbital\quad electrons }{ 2 }$$The electronic configuration of the $$O_2$$ containing 16 electrons can be written as: $$1s^2 *1s^2 2s^2 *2s^2 2pz^2 2px^2 2py^2 *2px^1 *2py^1$$Bond Order = $$\dfrac { 8-4 }{ 2 }$$ = 2The electronic configuration of the $$O_2^+$$ containing 15 electrons can be written as: $$1s^2 *1s^2 2s^2 *2s^2 2pz^2 2px^2 2py^2 *2px^1$$Bond Order = $$\dfrac { 8-3 }{ 2 }$$ = 2.5The electronic configuration of the $$O_2^-$$ ion containing 17 electrons can be written as:$$1s^2 *1s^2 2s^2 *2s^2 2pz^2 2px^2 2py^2 *2px^2 *2py^1$$Bond Order = $$\dfrac { 8-5 }{ 2 }$$ = 1.5  $$O_2^-$$contains 3 electrons in the anti-bonding orbital. It requires less energy to remove an electron from anti-bonding orbital.Hence, the correct bond order in the following species is $$O_2^{2+} > O_2 >O_2^-$$.Therefore option D is correct.Chemistry

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