
The correct order of the ${\text{O}} - {\text{O}}$ bond length in ${{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{3}}}$ is
A) ${{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$
B) ${{\text{O}}_{\text{3}}} > {{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{2}}}$
C) ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{O}}_{\text{2}}}$
D) ${{\text{O}}_{\text{2}}} > {{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}}$
Answer
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Hint: We know that molecular orbital diagrams are used to determine the bonding in a diatomic molecule. The molecular orbital diagrams are used to predict the magnetic properties of a molecule. Molecular orbital diagrams help in determining the bond order of the molecule.
Complete solution:
We know that molecular orbital theory (MOT) explains the formation of molecules.
According to molecular orbital theory, the atomic orbitals having comparable energy overlap and result in the formation of the same number of molecular orbitals. The molecular orbitals having the same sign combine and give bonding molecular orbitals.
We are given three molecules ${{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{3}}}$:
${{\text{O}}_{\text{2}}}$ molecule is formed by the combination of two oxygen atoms. The two oxygen atoms are linked by a covalent bond.
The electronic configuration of ${{\text{O}}_{\text{2}}}$ is $\sigma 1{s^2}\,{\sigma ^*}1{s^2}\,\sigma 2{s^2}\,{\sigma ^*}2{s^2}\,\sigma 2p_z^2\,\pi 2p_x^2\,\pi 2p_y^2\,{\pi ^*}2p_x^1\,{\pi ^*}2p_y^1$.
Now, we have to calculate the bond order of ${{\text{O}}_{\text{2}}}$ molecule using the formula as follows:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in BMO}}} \right) - \left( {{\text{Number of electrons in ABMO}}} \right)$
The number of electrons in bonding molecular orbitals are 10 and the number of electrons in antibonding molecular orbitals are 6. Thus,
${\text{Bond order}} = \dfrac{1}{2}\left( {10 - 6} \right) = 2$
Thus, the bond order for ${{\text{O}}_{\text{2}}}$ molecule is 2.
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule is formed by the combination of two hydrogen and two oxygen atoms. The two hydrogen and two oxygen atoms are linked by covalent bonds.
Now, we have to calculate the bond order of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule using the formula as follows:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in BMO}}} \right) - \left( {{\text{Number of electrons in ABMO}}} \right)$
The number of electrons in bonding molecular orbitals are 10 and the number of electrons in antibonding molecular orbitals are 8. Thus,
${\text{Bond order}} = \dfrac{1}{2}\left( {10 - 8} \right) = 1$
Thus, the bond order for ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule is 1.
The structure of ${{\text{O}}_{\text{3}}}$ is as follows:
The number of bonds between the atoms in ${{\text{O}}_{\text{3}}}$ is 3. The number of canonical forms of ${{\text{O}}_{\text{3}}}$ is 2. The bond order of ${{\text{O}}_{\text{3}}}$ is,
${\text{Bond order of }}{{\text{O}}_{\text{3}}} = \dfrac{{\text{3}}}{{\text{2}}} = 1.5$
Thus, the bond order of ${{\text{O}}_{\text{3}}}$ is 1.5.
Thus, the increasing order of bond order is,
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} < {{\text{O}}_{\text{3}}} < {{\text{O}}_{\text{2}}}$
The bond length is inversely proportional to the bond order. Thus, the increasing order of bond length is,
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{O}}_{\text{2}}}$
Thus, the correct order of the ${\text{O}} - {\text{O}}$ bond length in ${{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{3}}}$ is ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{O}}_{\text{2}}}$.
Thus, the correct option is (C).
Note:Remember that bond order is the total number of bonds present between two atoms. The bond order depends on the molecular orbital theory. Remember that bond order is inversely proportional to the bond length.
Complete solution:
We know that molecular orbital theory (MOT) explains the formation of molecules.
According to molecular orbital theory, the atomic orbitals having comparable energy overlap and result in the formation of the same number of molecular orbitals. The molecular orbitals having the same sign combine and give bonding molecular orbitals.
We are given three molecules ${{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{3}}}$:
${{\text{O}}_{\text{2}}}$ molecule is formed by the combination of two oxygen atoms. The two oxygen atoms are linked by a covalent bond.
The electronic configuration of ${{\text{O}}_{\text{2}}}$ is $\sigma 1{s^2}\,{\sigma ^*}1{s^2}\,\sigma 2{s^2}\,{\sigma ^*}2{s^2}\,\sigma 2p_z^2\,\pi 2p_x^2\,\pi 2p_y^2\,{\pi ^*}2p_x^1\,{\pi ^*}2p_y^1$.
Now, we have to calculate the bond order of ${{\text{O}}_{\text{2}}}$ molecule using the formula as follows:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in BMO}}} \right) - \left( {{\text{Number of electrons in ABMO}}} \right)$
The number of electrons in bonding molecular orbitals are 10 and the number of electrons in antibonding molecular orbitals are 6. Thus,
${\text{Bond order}} = \dfrac{1}{2}\left( {10 - 6} \right) = 2$
Thus, the bond order for ${{\text{O}}_{\text{2}}}$ molecule is 2.
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule is formed by the combination of two hydrogen and two oxygen atoms. The two hydrogen and two oxygen atoms are linked by covalent bonds.
Now, we have to calculate the bond order of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule using the formula as follows:
${\text{Bond order}} = \dfrac{1}{2}\left( {{\text{Number of electrons in BMO}}} \right) - \left( {{\text{Number of electrons in ABMO}}} \right)$
The number of electrons in bonding molecular orbitals are 10 and the number of electrons in antibonding molecular orbitals are 8. Thus,
${\text{Bond order}} = \dfrac{1}{2}\left( {10 - 8} \right) = 1$
Thus, the bond order for ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ molecule is 1.
The structure of ${{\text{O}}_{\text{3}}}$ is as follows:
The number of bonds between the atoms in ${{\text{O}}_{\text{3}}}$ is 3. The number of canonical forms of ${{\text{O}}_{\text{3}}}$ is 2. The bond order of ${{\text{O}}_{\text{3}}}$ is,
${\text{Bond order of }}{{\text{O}}_{\text{3}}} = \dfrac{{\text{3}}}{{\text{2}}} = 1.5$
Thus, the bond order of ${{\text{O}}_{\text{3}}}$ is 1.5.
Thus, the increasing order of bond order is,
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} < {{\text{O}}_{\text{3}}} < {{\text{O}}_{\text{2}}}$
The bond length is inversely proportional to the bond order. Thus, the increasing order of bond length is,
${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{O}}_{\text{2}}}$
Thus, the correct order of the ${\text{O}} - {\text{O}}$ bond length in ${{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{3}}}$ is ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}} > {{\text{O}}_{\text{3}}} > {{\text{O}}_{\text{2}}}$.
Thus, the correct option is (C).
Note:Remember that bond order is the total number of bonds present between two atoms. The bond order depends on the molecular orbital theory. Remember that bond order is inversely proportional to the bond length.
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