According to molecular orbital theory molecule does not exist

Asked by admin @ 13/08/2021 in Chemistry viewed by 328 People

Use molecular orbital theory to explain why the Be2 molecule does not exist. Write the molecular orbital electronic configuration of Be2 molecule. Also calculate the bond order in Be2 molecule. Is Be2 a paramagnetic molecule? Explain your answer

Answered by admin @ 13/08/2021

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration of beryllium will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 4 electrons present in beryllium.

The number of electrons present in $Be_2$ molecule = 2(4) = 8

The molecular orbital configuration of $Be_2$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^0=(\pi_{2p_y})^0],(\sigma_{2p_z})^0,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The number of unpaired electron in $Be_2$ molecule is, 0. So, this is diamagnetic.

The formula of bonding order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

The bond order of $Be_2$ = $\frac{1}{2}\times (4-4)=0$

The bond order of $Be_2$ is, zero. So, $Be_2$ will not be exist.

According to molecular orbital theory molecule does not exist

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