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Question

Which of the following options represents the correct bond order?


A
O2>O2<O+2
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B
O2<O2>O+2
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C
O2>O2>O+2
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D
O2<O2<O+2
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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 = $$\dfrac { 8-4 }{ 2 } $$ = 2

The 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 = $$\dfrac { 8-3 }{ 2 } $$ = 2.5

The 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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