Question:
As per molecular orbital theory, the pair of molecules which do not exist is
As per molecular orbital theory, the pair of molecules which do not exist is
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A molecule is stable only when its bond order is positive. If
\[
\text{Bond Order}=0,
\]
the molecule is not expected to exist according to Molecular Orbital Theory.
Updated On: Jul 18, 2026
- \(Li_2,\ B_2\)
- \(He_2,\ C_2\)
- \(Be_2,\ C_2\)
- \(Be_2,\ Ne_2\)
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The Correct Option is D
Solution and Explanation
Step 1: Criterion for existence of a molecule.
According to Molecular Orbital Theory, a molecule exists only if its bond order is positive.
\[ \text{Bond Order} = \frac{N_b-N_a}{2} \] where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.
If \[ \text{Bond Order}=0, \] the molecule does not exist.
Step 2: Calculate bond order of \(Be_2\).
Electronic configuration of \(Be\): \[ 1s^2\,2s^2 \] For \(Be_2\), the valence MO configuration is \[ (\sigma_{2s})^2(\sigma_{2s}^{*})^2 \] Therefore, \[ N_b=2,\qquad N_a=2 \] Hence, \[ \text{Bond Order} = \frac{2-2}{2} =0 \] Thus, \[ Be_2 \] does not exist.
Step 3: Calculate bond order of \(Ne_2\).
Each neon atom has 10 electrons. Thus, \[ Ne_2 \] contains 20 electrons.
All bonding and antibonding molecular orbitals become completely filled.
Hence, \[ N_b=N_a \] Therefore, \[ \text{Bond Order} = 0 \] Thus, \[ Ne_2 \] also does not exist.
Step 4: Examine the remaining molecules.
\[ Li_2:\ \text{Bond Order}=\frac{2-0}{2}=1 \] \[ B_2:\ \text{Bond Order}=1 \] \[ C_2:\ \text{Bond Order}=2 \] All of these have positive bond order and hence exist.
Step 5: Final conclusion.
The pair of molecules having zero bond order and therefore not existing is \[ \boxed{Be_2,\ Ne_2} \] Hence, option (4) is correct.
According to Molecular Orbital Theory, a molecule exists only if its bond order is positive.
\[ \text{Bond Order} = \frac{N_b-N_a}{2} \] where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.
If \[ \text{Bond Order}=0, \] the molecule does not exist.
Step 2: Calculate bond order of \(Be_2\).
Electronic configuration of \(Be\): \[ 1s^2\,2s^2 \] For \(Be_2\), the valence MO configuration is \[ (\sigma_{2s})^2(\sigma_{2s}^{*})^2 \] Therefore, \[ N_b=2,\qquad N_a=2 \] Hence, \[ \text{Bond Order} = \frac{2-2}{2} =0 \] Thus, \[ Be_2 \] does not exist.
Step 3: Calculate bond order of \(Ne_2\).
Each neon atom has 10 electrons. Thus, \[ Ne_2 \] contains 20 electrons.
All bonding and antibonding molecular orbitals become completely filled.
Hence, \[ N_b=N_a \] Therefore, \[ \text{Bond Order} = 0 \] Thus, \[ Ne_2 \] also does not exist.
Step 4: Examine the remaining molecules.
\[ Li_2:\ \text{Bond Order}=\frac{2-0}{2}=1 \] \[ B_2:\ \text{Bond Order}=1 \] \[ C_2:\ \text{Bond Order}=2 \] All of these have positive bond order and hence exist.
Step 5: Final conclusion.
The pair of molecules having zero bond order and therefore not existing is \[ \boxed{Be_2,\ Ne_2} \] Hence, option (4) is correct.
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