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3 years ago

Use the following MO diagram for Be2, Be2+, and Be2-. Based on this diagram,

2 answers:

6 0

Bond Order = [Σ (bonding e-) - Σ (antibonding e-)]/2
Be2 = 4e = σ1(2e) σ2*(2e) σ3(0) π1(0) π2*(0) σ4*(0) bo = 0 
[Be2]+ = 3e = σ1(2e) σ2*(1e) σ3(0) π1(0) π2*(0) σ4*(0) bo = 0.5 
[Be2]+ would be more likely to exist since it has a bond order of 0.5 whereas Be2 has zero bond order

Fantom [35]3 years ago

4 0

Answer:

Be2+ and Be2- are both more stable than Be2

Explanation:

Beryllium is the first element of Group 2 and. This means that it has 2 electrons in its last layer.

An atom is more stable when its outer shell of electrons is full, as in the case of noble gases.

So, Beryllium would become stable if it had the last complete electronic layer ridding itself of the two electrons of the second layer.

This means that the beryllium oxidation number is +2, because the 2 valence electrons (electrons of the last layer) are gone, leaving only the two in the inner layer. This is represented by Be + 2. And since the inner layer is complete, you can say that Be + 2 is more stable than Be.

A particular case of stability occurs when the layer is half full. This is the case of Be-2. If Be receives 2 electrons, you get Be2-. Now the anion has 2 electrons in the inner layer and 4 valence electrons in the second layer. As the completely filled layer would have 8 electrons; You can see that in this case it has half (8/2 = 4). That is a half-filled shell, so Be2- is also more stable than just Be.

So, based on what was said above, you can say that Be2+ and Be2- are both more stable than Be2

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