
The pair of species with the same bond order is:
A.\[NO,CO\]
B.\[O_2^{2 - },{B_2}\]
C.\[O_2^ - ,N{O^ - }\]
D.None of the above
Answer
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Hint: Bond order is the number of chemical bonds found between a pair of atoms in a given molecule. For example in diatomic nitrogen gas, the bond order is 3 meaning that there are 3 bonds in between the nitrogen atoms. Bond order is a useful method to indicate the stability of a bond. So in this question, we have to find out the species which are having equal numbers of bonds.
Complete answer:
To find the answer to this question we need to find out the number of bonds between the atoms in the given pairs.
To calculate the number of bonds, we need to first write down the molecular orbital configuration of both molecules.
For \[O_2^{2 - },\] we can say that the configuration will be
\[\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2(\pi 2p_x^2 = \pi 2p_y^2)({\pi ^ * }2p_x^2 = {\pi ^*}2p_y^2){\sigma ^*}2p_z^0\]
This means that the bond order of the molecule can be found out by using the following formula:
\[ \Rightarrow {\text{Bond order = }}\dfrac{{{\text{Bonding electrons - Antibonding electrons}}}}{2}\]
This means that the bonding electrons which are in the bonding orbitals and the antibonding electrons which are in the anti-bonding orbitals represented by the asterisk will contribute to the bond order.
Bond order of \[O_2^{2 - },\]
\[ \Rightarrow {\text{Bond order = }}\dfrac{{10 - 8}}{2}\]
\[ \Rightarrow {\text{Bond order}} = 1\]
For \[{B_2}\] we can write the configuration as:
\[\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\]
Bond order of \[{B_2}\] is given as:
\[ \Rightarrow {\text{Bond order = }}\dfrac{{6 - 4}}{2}\]
\[ \Rightarrow {\text{Bond order}} = 1\]
Thus we can say that the bond order of both \[O_2^{2 - } , {B_2}\] are the same
Hence option (B) is the correct answer.
Note:
In questions like these, we need to find out the bond order of each of the pairs and eliminate all the pairs that do not have the same bond order.
An easy way to find the bond order for common molecules is to just count the number of bonds between them in their molecule. Thus hydrogen bond order is 1, oxygen bond order is 2, and nitrogen bond order is 3.
Complete answer:
To find the answer to this question we need to find out the number of bonds between the atoms in the given pairs.
To calculate the number of bonds, we need to first write down the molecular orbital configuration of both molecules.
For \[O_2^{2 - },\] we can say that the configuration will be
\[\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2(\pi 2p_x^2 = \pi 2p_y^2)({\pi ^ * }2p_x^2 = {\pi ^*}2p_y^2){\sigma ^*}2p_z^0\]
This means that the bond order of the molecule can be found out by using the following formula:
\[ \Rightarrow {\text{Bond order = }}\dfrac{{{\text{Bonding electrons - Antibonding electrons}}}}{2}\]
This means that the bonding electrons which are in the bonding orbitals and the antibonding electrons which are in the anti-bonding orbitals represented by the asterisk will contribute to the bond order.
Bond order of \[O_2^{2 - },\]
\[ \Rightarrow {\text{Bond order = }}\dfrac{{10 - 8}}{2}\]
\[ \Rightarrow {\text{Bond order}} = 1\]
For \[{B_2}\] we can write the configuration as:
\[\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\]
Bond order of \[{B_2}\] is given as:
\[ \Rightarrow {\text{Bond order = }}\dfrac{{6 - 4}}{2}\]
\[ \Rightarrow {\text{Bond order}} = 1\]
Thus we can say that the bond order of both \[O_2^{2 - } , {B_2}\] are the same
Hence option (B) is the correct answer.
Note:
In questions like these, we need to find out the bond order of each of the pairs and eliminate all the pairs that do not have the same bond order.
An easy way to find the bond order for common molecules is to just count the number of bonds between them in their molecule. Thus hydrogen bond order is 1, oxygen bond order is 2, and nitrogen bond order is 3.
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