Question
Use molecular orbital theory to explain why the $\mathrm{Be}_2$ molecule does not exist.
✓
Answer
The electronic configuration of Beryllium is $1 \mathrm{~s}^2 2 \mathrm{~s}^2$ The molecular orbital electronic configuration for $\mathrm{Be}_2$ molecule can be written as:
$\sigma_{1 \mathrm{~s}}^2 \sigma_{1 \mathrm{~s}}^2 \sigma_{2 \mathrm{~s}}^2 \sigma_{2 \mathrm{~s}}^2$
Hence, the bond order for $\mathrm{Be}_2$ is $\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$.
Where
$N_b=$ Number of electrons in bonding orbitals
$N_a=$ Number of electrons in anti-bonding orbitals
$\therefore$ Bond order of $\mathrm{Be}_2=\frac{1}{2}(4-4)=0$
A negative or zero bond order means that the molecule is unstable. Hence, $\mathrm{Be}_2$ molecule does not exist.
$\sigma_{1 \mathrm{~s}}^2 \sigma_{1 \mathrm{~s}}^2 \sigma_{2 \mathrm{~s}}^2 \sigma_{2 \mathrm{~s}}^2$
Hence, the bond order for $\mathrm{Be}_2$ is $\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$.
Where
$N_b=$ Number of electrons in bonding orbitals
$N_a=$ Number of electrons in anti-bonding orbitals
$\therefore$ Bond order of $\mathrm{Be}_2=\frac{1}{2}(4-4)=0$
A negative or zero bond order means that the molecule is unstable. Hence, $\mathrm{Be}_2$ molecule does not exist.
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