Increasing order of bond length is

$$\mathop {C - H}\limits_{107\,pm} < \mathop {C = C}\limits_{134\,pm} < \mathop {C - O}\limits_{141\,pm} < \mathop {C - C}\limits_{134\,pm} $$

Electronic configuration
O₂ : $$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_y^1{\pi ^*}2p_z^1$$

Bond order = $${1 \over 2}\left( {8 - 4} \right)$$ = 2

O₂⁺ : Bond order = $${1 \over 2}\left( {8 - 3} \right)$$ = 2.5

O₂^- : Bond order = $${1 \over 2}\left( {8 - 5} \right)$$ = 1.5

O₂^2- : Bond order = $${1 \over 2}\left( {8 - 6} \right)$$ = 1

As bond order increases, bond length decreases.

For neutral molecules

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom]

$$ \therefore $$For CH₄, no. of e^- pairs = 4 + 1/2[4 - 4] = 4 (sp³ hybridisation)

For SF₄, no. of e^- pairs = 4 + 1/2[6 - 4] = 5 (sp³d hybridisation)

and For ions,

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom $$ \pm $$ No. of electrons]

$$ \therefore $$For BF₄^-, no. of e^- pairs = 4 + 1/2[3 - 4 + 1] = 4 (sp³ hybridisation)

For NH₄⁺, no. of e^- pairs = 4 + 1/2[5 - 4 - 1] = 4 (sp³ hybridisation)

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom $$ \pm $$ No. of electrons]

For NO₃^- = 3 + 1/2[5 - 6 + 1] = 3 (sp² hybridisation)

For H₃O⁺ = 3 + 1/2[6 - 3 - 1] = 4 (sp³ hybridisation)