
Find the sum of bond order and the number of $\pi$ bonds in ${{C}_{2}}$ molecule on the basis of molecular orbital theory.
Answer
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Hint: The bond order is defined as the difference between bond number and antibond number. The bond number itself is the number of pairs of electrons (bonds) between a couple of atoms. Molecular orbital theory is mainly used to explain the bonding in molecules which Valence Bond theory cannot explain. Resonance means a bond is not single or double but some combination of the two.
Complete answer:
- The Molecular Orbital Theoretical Rules:
First principle: The number of molecular orbitals formed is always equal to the number of atomic orbitals created by the combined atoms.
Second principle: The bonding of molecular orbitals is lower in energy than the parent orbitals, and the interaction of the antibonding orbitals is higher.
Third principle: molecule electrons are distributed to orbitals from the lowest to the successively higher energy.
Fourth principle: Atomic orbitals more easily combine to form molecular orbitals when the atomic orbitals are of similar energy.
${{C}_{2}}$ has 12 electrons.
$\left( \sigma 1s \right)2,\left( \sigma *1s \right)2,\left( \sigma 2s \right)2,\left( \sigma *2s \right)2,\left( \pi \right)4$
So, Bond Order=21(a−b)
Here, an is the number of electrons in molecular bonding orbitals, and b is the number of electrons in molecular anti bonding orbital.
Bond Order=21(8−4)=2
Number of pi bonds in ${{C}_{2}}$ is 2.
Therefore, sum is 2+2=4
Note:
-The sequence of energy levels of molecular orbitals for ${{O}_{2}}$ and ${{F}_{2}}$ is as :$\sigma 1s<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<\sigma 2{{p}_{z}}<(\Pi 2{{p}_{x}}=\Pi 2{{p}_{y}})<({{\Pi }^{*}}2{{p}_{x}}={{\Pi }^{*}}2{{p}_{y}})<{{\sigma }^{*}}2{{p}_{z}}$
-But this sequence is incorrect for the molecules like $L{{i}_{2}}$, $B{{e}_{2}}$, ${{B}_{2}}$, ${{C}_{2}}$ and ${{N}_{2}}$.
- For these molecules the sequence of energy levels of molecular orbitals are as follows:
$\sigma 1s<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<(\Pi 2{{p}_{x}}=\Pi 2{{p}_{y}})<\sigma 2{{p}_{z}}<({{\Pi }^{*}}2{{p}_{x}}={{\Pi }^{*}}2{{p}_{y}})<{{\sigma }^{*}}2{{p}_{z}}$
-Here, Antibonding Molecular orbitals are: ${{\sigma }^{*}}1s$, ${{\sigma }^{*}}2s$, ${{\sigma }^{*}}2{{p}_{z}}$, ${{\Pi }^{*}}2{{p}_{x}}$, ${{\Pi }^{*}}2{{p}_{y}}$
And Bonding Molecular orbitals are: $\sigma 1s$, $\sigma 2s$, $\sigma 2{{p}_{z}}$, $\Pi 2{{p}_{x}}$, $\Pi 2{{p}_{y}}$.
Complete answer:
- The Molecular Orbital Theoretical Rules:
First principle: The number of molecular orbitals formed is always equal to the number of atomic orbitals created by the combined atoms.
Second principle: The bonding of molecular orbitals is lower in energy than the parent orbitals, and the interaction of the antibonding orbitals is higher.
Third principle: molecule electrons are distributed to orbitals from the lowest to the successively higher energy.
Fourth principle: Atomic orbitals more easily combine to form molecular orbitals when the atomic orbitals are of similar energy.
${{C}_{2}}$ has 12 electrons.
$\left( \sigma 1s \right)2,\left( \sigma *1s \right)2,\left( \sigma 2s \right)2,\left( \sigma *2s \right)2,\left( \pi \right)4$
So, Bond Order=21(a−b)
Here, an is the number of electrons in molecular bonding orbitals, and b is the number of electrons in molecular anti bonding orbital.
Bond Order=21(8−4)=2
Number of pi bonds in ${{C}_{2}}$ is 2.
Therefore, sum is 2+2=4
Note:
-The sequence of energy levels of molecular orbitals for ${{O}_{2}}$ and ${{F}_{2}}$ is as :$\sigma 1s<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<\sigma 2{{p}_{z}}<(\Pi 2{{p}_{x}}=\Pi 2{{p}_{y}})<({{\Pi }^{*}}2{{p}_{x}}={{\Pi }^{*}}2{{p}_{y}})<{{\sigma }^{*}}2{{p}_{z}}$
-But this sequence is incorrect for the molecules like $L{{i}_{2}}$, $B{{e}_{2}}$, ${{B}_{2}}$, ${{C}_{2}}$ and ${{N}_{2}}$.
- For these molecules the sequence of energy levels of molecular orbitals are as follows:
$\sigma 1s<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<(\Pi 2{{p}_{x}}=\Pi 2{{p}_{y}})<\sigma 2{{p}_{z}}<({{\Pi }^{*}}2{{p}_{x}}={{\Pi }^{*}}2{{p}_{y}})<{{\sigma }^{*}}2{{p}_{z}}$
-Here, Antibonding Molecular orbitals are: ${{\sigma }^{*}}1s$, ${{\sigma }^{*}}2s$, ${{\sigma }^{*}}2{{p}_{z}}$, ${{\Pi }^{*}}2{{p}_{x}}$, ${{\Pi }^{*}}2{{p}_{y}}$
And Bonding Molecular orbitals are: $\sigma 1s$, $\sigma 2s$, $\sigma 2{{p}_{z}}$, $\Pi 2{{p}_{x}}$, $\Pi 2{{p}_{y}}$.
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