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Two pi and half sigma bonds are present in:
(A) $N_2^ + $
(B) ${N_2}$
(C) $O_2^ + $
(D) ${O_2}$

seo-qna
Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: First write down the electronic configurations of the given molecules in the option and see the number of electrons in their bonding and antibonding orbitals. Accordingly, see how many sigma and pi bonds each of the molecules can form. Also, the molecule which has two pi and half sigma bonds will have a bond order of 2.5

Complete step by step solution:
-We all know that bond order basically gives us the number of bonds or the number of electron pairs present between two atoms in a molecule. Like if 2 atoms in a molecule are bonded via a single bond, their bond order will be one; if via a triple bond. Their bond order will be 3 and so on.
Similarly, we will now check for the number of bonds present in the molecule we are talking about in the question. The question says that the molecule has two pi and half sigma bond hence the bond order of the molecule will be 2.5
The Molecular Orbital Theory describes the bond order to be basically the difference between the number of bonding and antibonding electrons, divided by two. Mathematically it can be written as:
$BO =$ $\dfrac{1}{2}$ (Bonding electrons – Anti bonding electrons)--------(1)
-Now coming back to what the question is asking, to find the molecule we need to see which of them has a bond order of 2.5 and forms half sigma and 2 pi bonds. So, we will now write down the electronic configuration of all the given molecules and calculate their bond order to verify.
-For (A) $N_2^ + $: The total number of electrons in it are = 13
Configuration: ${\sigma }$${1}$${s^2}$ ${\sigma ^*}$${1}$${s^2}$ ${\sigma}$${2}$${s^2}$ ${\sigma ^*}{2}{s^2}$ ${\pi}$${2}$${p_x^2}$ ${\pi}$${2}$${p_y^2}$ ${\sigma}$${2}$${p_z^1}$
From this configuration we can tell that the 2 electrons in $\left( {\pi 2{p_x}} \right)$ form a pi bond and the 2 electrons in $\left( {\pi 2{p_y}} \right)$ also form a pi bond, while the 1 electron in $\left( {\sigma 2{p_z}} \right)$ forms a half sigma bond. In total, the ${N_2}$ molecule forms half sigma and 2 pi bonds.
Here we can see that: Number of bonding electrons = 9 and the number of antibonding electrons = 4. Calculate the bond order using equation (1):
$BO =$ $\dfrac{1}{2}(9 - 4)$
$=$ $\dfrac{1}{2} \times 5$
$= 2.5$
So, the bond order for $N_2^ + $ is $2.5$ and it also forms half sigma and $2$ pi bonds.
-For (B) ${N_2}$: The total number of electrons = 14
Configuration: ${\sigma }$${1}$${s^2}$ ${\sigma ^*}$${1}$${s^2}$ ${\sigma}$${2}$${s^2}$ ${\sigma ^*}{2}{s^2}$ ${\pi}$${2}$${p_x^2}$ ${\pi}$${2}$${p_y^2}$ ${\sigma}$${2}$${p_z^2}$
From this configuration we can tell that the 2 electrons in $\left( {\pi 2{p_x}} \right)$ form a pi bond and the 2 electrons in $\left( {\pi 2{p_y}} \right)$ also form a pi bond, while the 2 electrons in $\left( {\sigma 2{p_z}} \right)$ form a sigma bond. In total the ${N_2}$ molecule forms 1 sigma and 2 pi bonds.
-For (C) $O_2^ + $: total number of electrons = 15
Configuration: ${\sigma }$${1}$${s^2}$ ${\sigma ^*}$${1}$${s^2}$ ${\sigma}$${2}$${s^2}$ ${\sigma ^*}{2}{s^2}$ ${\pi}$${2}$${p_x^2}$ ${\pi}$${2}$${p_y^2}$ ${\sigma}$${2}$${p_z^2}$ ${\pi ^*}$$2$${p_x^1}$
In the above configuration we can see that there is only one electron in the $({\pi ^*}2{p_x})$ orbital so it will cancel out one electron from the $(\pi 2{p_y})$ orbital. Then the 2 electron in the $(\sigma 2{p_z})$ orbital will form a sigma bond, the 2 electrons in $(\pi 2{p_x})$ will form 1 pi bond and the 1 electron in the $(\pi 2{p_y})$ will also form half of a pi bond. So, in total this molecule forms 1 sigma and 1.5 pi bonds.
-For ${O_2}$: total number of electrons = 16
Configuration: ${\sigma }$${1}$${s^2}$ ${\sigma ^*}$${1}$${s^2}$ ${\sigma}$${2}$${s^2}$ ${\sigma ^*}{2}{s^2}$ ${\pi}$${2}$${p_x^2}$ ${\pi}$${2}$${p_y^2}$ ${\sigma}$${2}$${p_z^2}$ ${\pi ^*}$$2$${p_x^1}$ ${\pi ^*}$$2$${p_y^1}$
In the above configuration we can see that there are two electrons in the $({\pi ^*}2{p_x})$ and $({\pi ^*}2{p_y})$ orbitals so they will cancel out the two electrons from the $(\pi 2{p_y})$ orbital. Then the 2 electrons in the $(\sigma 2{p_z})$ orbital will form a sigma bond and the 2 electrons in $(\pi 2{p_x})$ will form 1 pi bond. So, in total this molecule forms 1 sigma and 1 pi bond.
-From the above discussion we can now conclude that the only molecule in the options which forms half sigma and 2 pi bonds along with bond order 2.5 is $N_2^ + $.

Hence the correct option is: (A) $N_2^ + $.

Note: The electrons should always be filled according to the increasing energy level of the molecular orbitals as shown below:
$\sigma 1s,{\sigma ^*}1s,\sigma 2s,{\sigma ^*}2s,\sigma 2{p_z},\pi 2{p_x} = \pi 2{p_y},{\pi ^*}2{p_x} = {\pi ^*}2{p_y},{\sigma ^*}2{p_z}$ and so on.
Also, the MOT uses the concept of bond order to explain the existence of a molecule on the basis of bond order, but this method can neither be feasible nor appropriate to explain about the molecular existence of polyatomic molecules. Also, MOT does not say anything about the geometry and shape of the molecule. So, this theory also has some drawbacks.