Single Choice

Which of the following facts regarding change in bond length is correct?

AIncreases in going from $$N_{2}$$ to $$N_{2}^{+}$$, decreases in going from $$O_{2}$$ to $$O_{2}^{+}$$
Correct Answer
BDecreases in going from $$N_{2}$$ to $$N_{2}^{+}$$, increases in going from $$O_{2}$$ to $$O_{2}^{+}$$
CIncreases in going from $$N_{2}$$ to $$N_{2}^{+} $$ and $$O_{2}$$ to $$O_{2}^{+} $$
DDecrease in going from $$N_{2}$$ to $$N_{2}^{+} $$ and $$O_{2}$$ to $$O_{2}^{+} $$

Solution

Bond order = $$1/2 \times $$[(Number of electrons in bonding molecules) - (Number of electrons in antibonding molecules)]

Bond order = $$1/2 \times (N_B-N_{ AB })$$

1. $$N_2$$:

valence electrons=$$5+5=10$$

$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma *2s \right) }^{ 2 }{ \left( \sigma 2p \right) }^{ 2 }{ \left( \pi 2p \right) }^{ 4 }$$

BO=$$1/2 \times (6-0)=3$$;

2. $$N_2^+$$:

valence electrons=$$5+5-1=9$$

$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma *2s \right) }^{ 2 }{ \left( \sigma 2p \right) }^{ 2 }{ \left( \pi 2p \right) }^{ 3 }$$

BO=$$1/2 \times (5-0)=2.5$$;

3.$$O_2$$:

valence electrons=$$6+6=12$$

$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma 2*s \right) }^{ 2 }{ \left( \sigma 2p \right) }^{ 2 }{ \left( \pi 2p \right) }^{ 4 }\left( \pi ^*2p \right) ^{ 2 }$$

BO=$$1/2 \times (6-2)=2$$;

3.$$O_2^+$$:

valence electrons=$$6+6-1=11$$

$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma 2*s \right) }^{ 2 }{ \left( \sigma 2p \right) }^{ 2 }{ \left( \pi 2p \right) }^{ 4 }\left( \pi ^*2p \right) ^{ 1 }$$

BO=$$1/2 \times (6-1)=2.5$$;

Since, bond length $$\propto \dfrac {1}{bond\ order}$$

Order of bond length is: $$N_2$$<$$N_2^+$$ and $$O_2$$>$$O_2^+$$

Hence, option A is correct.

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