In the following the correct bond order sequence is :

O_2^ + \gt O_2 \gt O_2^ - \gt O_2^2 -

O_2^2 - \gt O_2^ + \gt O_2^ - \gt O_2

O_2^ + \gt O_2^ - \gt O_2^2 - \gt O_2

O_2 \gt O_2^ - \gt O_2^2 - \gt O_2^ +

Chemical Bonding and Molecular Structure PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q1 with Solution

View written solutionFree

Correct answer: C

Use molecular orbital theory for oxygen species

For $O_2$ and its ions, the relevant bond order is calculated by

\text{Bond order} = \frac{N_b - N_a}{2}

where $N_b$ = number of electrons in bonding MOs and $N_a$ = number of electrons in antibonding MOs.

For oxygen and heavier diatomic molecules, the MO filling order is:

\sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y)

Bond order of $O_2$

$O_2$ has 12 valence electrons. Its MO configuration in valence shell is:

\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2\,\pi^*(2p_x)^1\,\pi^*(2p_y)^1

So,

Bonding electrons: $8$
Antibonding electrons: $4$

Thus,

\text{B.O.}(O_2)=\frac{8-4}{2}=2

Bond order of $O_2^+$

Removing one electron from $O_2$ removes it from the highest occupied antibonding orbital $\pi^*$ .

So antibonding electrons become $3$ .

\text{B.O.}(O_2^+)=\frac{8-3}{2}=2.5

Bond order of $O_2^-$

Adding one electron to $O_2$ adds it to an antibonding $\pi^*$ orbital.

So antibonding electrons become $5$ .

\text{B.O.}(O_2^-)=\frac{8-5}{2}=1.5

Bond order of $O_2^{2-}$

Adding two electrons to $O_2$ adds both to antibonding $\pi^*$ orbitals.

So antibonding electrons become $6$ .

\text{B.O.}(O_2^{2-})=\frac{8-6}{2}=1

Arrange in decreasing order

O_2^+\;(2.5) > O_2\;(2) > O_2^-\;(1.5) > O_2^{2-}\;(1)

So the correct sequence is:

O_2^+ > O_2 > O_2^- > O_2^{2-}

Match with options

This corresponds to Option C.

More from Chemical Bonding and Molecular Structure

View written solutionFree

Correct answer: C

Use molecular orbital theory for oxygen species

For $O_2$ and its ions, the relevant bond order is calculated by

\text{Bond order} = \frac{N_b - N_a}{2}

where $N_b$ = number of electrons in bonding MOs and $N_a$ = number of electrons in antibonding MOs.

For oxygen and heavier diatomic molecules, the MO filling order is:

\sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y)

Bond order of $O_2$

$O_2$ has 12 valence electrons. Its MO configuration in valence shell is:

\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2\,\pi^*(2p_x)^1\,\pi^*(2p_y)^1

So,

Bonding electrons: $8$
Antibonding electrons: $4$

Thus,

\text{B.O.}(O_2)=\frac{8-4}{2}=2

Bond order of $O_2^+$

Removing one electron from $O_2$ removes it from the highest occupied antibonding orbital $\pi^*$ .

So antibonding electrons become $3$ .

\text{B.O.}(O_2^+)=\frac{8-3}{2}=2.5

Bond order of $O_2^-$

Adding one electron to $O_2$ adds it to an antibonding $\pi^*$ orbital.

So antibonding electrons become $5$ .

\text{B.O.}(O_2^-)=\frac{8-5}{2}=1.5

Bond order of $O_2^{2-}$

Adding two electrons to $O_2$ adds both to antibonding $\pi^*$ orbitals.

So antibonding electrons become $6$ .

\text{B.O.}(O_2^{2-})=\frac{8-6}{2}=1

Arrange in decreasing order

O_2^+\;(2.5) > O_2\;(2) > O_2^-\;(1.5) > O_2^{2-}\;(1)

So the correct sequence is:

O_2^+ > O_2 > O_2^- > O_2^{2-}

Match with options

This corresponds to Option C.

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

More from Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

More from Chemical Bonding and Molecular Structure