Answer
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Hint: Bond length in simple terms can be referred as the average distance between nuclei of two bonded atoms in a molecule, and when we have all the three type of bonds: the single bond, the double bond, and the triple bond, the bond distance is shortest for the triple bonds and the longest for the single bonds the reason lies in the different hybridizations of the molecules.
Complete answer:
There is one more term one must need to know and that’s bond order which is defined as the difference between the number of electrons which are present in the bonding molecular orbitals and number of electrons which are present in the antibonding molecular orbital and the difference is divided by $2$, which can be represented as follows:
$B.O = \dfrac{{no.\,of\,{e^ - }\,(bonding) - no.\,of\,{e^ - }\,(anti - bonding)}}{2}$
The bond order has an inverse relation to the Bond length. Now, first let’s try to solve the question from the Bond length approach,
As we know, that $CO$ has a triple bond between carbon and oxygen, Hence, is the shortest of all three. And $C{O_2}$ consists of double bonds and $C{O_3}^{2 - }$ has double as well as single bonds hence, the order for $C - O$ bond lengths comes out to be as follows:
\[CO{\text{ }} < {\text{ }}C{O_2}{\text{ }} < {\text{ }}C{O_3}^{2 - }\]
So, the correct answer is Option D.
Note:
There is another method among which we can answer that is on the basis of the bond order, as we are familiar with the formula the bond orders for the given chemical species are:
For, \[CO{\text{ }} = {\text{ }}3\], \[C{O_2}{\text{ }} = {\text{ }}2\] and of \[C{O_3}^{2 - }{\text{ }} = {\text{ }}1.33\]
Hence, the order for C-O bond lengths comes out to be as follows:
\[CO{\text{ }} < {\text{ }}C{O_2}{\text{ }} < {\text{ }}C{O_3}^{2 - }\]
Complete answer:
There is one more term one must need to know and that’s bond order which is defined as the difference between the number of electrons which are present in the bonding molecular orbitals and number of electrons which are present in the antibonding molecular orbital and the difference is divided by $2$, which can be represented as follows:
$B.O = \dfrac{{no.\,of\,{e^ - }\,(bonding) - no.\,of\,{e^ - }\,(anti - bonding)}}{2}$
The bond order has an inverse relation to the Bond length. Now, first let’s try to solve the question from the Bond length approach,
As we know, that $CO$ has a triple bond between carbon and oxygen, Hence, is the shortest of all three. And $C{O_2}$ consists of double bonds and $C{O_3}^{2 - }$ has double as well as single bonds hence, the order for $C - O$ bond lengths comes out to be as follows:
\[CO{\text{ }} < {\text{ }}C{O_2}{\text{ }} < {\text{ }}C{O_3}^{2 - }\]
So, the correct answer is Option D.
Note:
There is another method among which we can answer that is on the basis of the bond order, as we are familiar with the formula the bond orders for the given chemical species are:
For, \[CO{\text{ }} = {\text{ }}3\], \[C{O_2}{\text{ }} = {\text{ }}2\] and of \[C{O_3}^{2 - }{\text{ }} = {\text{ }}1.33\]
Hence, the order for C-O bond lengths comes out to be as follows:
\[CO{\text{ }} < {\text{ }}C{O_2}{\text{ }} < {\text{ }}C{O_3}^{2 - }\]
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