Q

# Answer please! - Chemical Bonding and Molecular Structure - JEE Main

During the change of $O_{2}$ to $O_{2}^{-}$, the incoming electron goes to the orbital :

• Option 1)

$\pi 2 p_{y}$

• Option 2)

$\sigma ^{*}2p_{z}$

• Option 3)

$\pi^{*}2p_{x}$

• Option 4)

$\pi 2 p_{x}$

Views

For $O_{2}$ molecule , the molecule orbital sequence  follows the given order :

$\sigma 1 s^{2}$ $\sigma^{*} 1 s^{2}$ $\sigma 2 s^{2}$ $\sigma^{*} 2 s^{2}$ $\sigma 2 p_{z}^{2}$ $\pi 2 p_{x}^{2}$ = $\pi 2 p_{y}^{2}$  $\pi^{*} 2 p_{x}^{1}$ = $\pi^{*} 2 p_{y}^{1}$

Now, from $O_{2}$ to $O_{2}^{-}$ , the incoming electron will go into ($\pi^{*}2p_{z}$) orbital and thus

the final sequence will be

$\sigma 1 s^{2}$ $\sigma^{*} 1 s^{2}$ $\sigma 2 s^{2}$ $\sigma^{*} 2 s^{2}$ $\sigma 2 p_{z}^{2}$ $\pi 2 p_{x}^{2}$ = $\pi 2 p_{y}^{2}$  $\pi^{*} 2 p_{x}^{1}$ = $\pi^{*} 2 p_{y}^{1}$

$\therefore$ option (3) is correct.

Option 1)

$\pi 2 p_{y}$

Option 2)

$\sigma ^{*}2p_{z}$

Option 3)

$\pi^{*}2p_{x}$

Option 4)

$\pi 2 p_{x}$

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