Single Choice

Using MO theory, predict which of the following species has the shortest bond length?

A$$\mathrm{O}_{2}^{2+}$$
B$$\mathrm{O}_{2}^{+}$$
C$$\mathrm{O}_{2}^{-}$$
D$$\mathrm{O}_{2}^{2-}$$
Correct Answer

Solution

1. $$O_2^{ 2+ }$$:

Valence electrons $$=6+6-2=10$$

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

BO $$=\dfrac12 \times (6-0)=3$$;

2. $$O_2^+$$:

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

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

BO $$=\dfrac12 \times (6-1)=2.5$$;

3. $$O_2^-$$:

Valence electrons $$=6+6+1=13$$

$${ \left( \sigma 2s \right) }^{ 2 }{ \left( \sigma *2s \right) }^{ 2 }{ \left( \sigma 2p \right) }^{ 2 }{ \left( \pi 2p \right) }^{ 4 }\left( \pi ^*2p \right) ^{ 3 }$$

BO $$=\dfrac12 \times (6-3)=1.5$$;

4. $$O_2^{ 2- }$$:

Valence electrons $$=6+6+2=14$$

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

BO $$=\dfrac12 \times (6-4)=1$$;

Bond length $$\propto\frac {1}{\text{bond order}}$$

$$\therefore O_2^{ 2- }$$ has the highest bond order and smallest bond length.

Hence, the correct option is $$\text{D}$$

SIMILAR QUESTIONS

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Chemical Bonding

During the change of $$O_2$$ to $$O_2^-$$, the incoming electrons goes to the orbital:

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