Use molecular orbital theory to explain why the $Be_{2}$ molecule does not exist.
Use molecular orbital theory to explain why the $Be_{2}$ molecule does not exist.
The electronic configuration of Beryllium is $1 s^{2}\, 2 s^{2}$
The molecular orbital electronic configuration for $Be _{2}$ molecule can be written as:
$\sigma _{1s}^2\,\,\sigma _{1s}^{ + 2}\,\,\sigma _{2s}^2\,\,\sigma _{2s}^{ + 2}$
Hence, the bond order for $Be _{2}$ is $\frac{1}{2}\left(N_{b}-N_{a}\right)$
Where
$N_{b}=$ Number of electrons in bonding orbitals
$N_{a}=$ Number of electrons in anti-bonding orbitals
$\therefore $ Bond order of $Be_{2}$ $=\frac{1}{2}(4-4)=0$
A negative or zero bond order means that the molecule is unstable. Hence, $Be _{2}$ molecule does not exist.
Similar Questions
According to Molecular Orbital Theory,
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
According to Molecular Orbital Theory,
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
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