# How do I identify lone pairs and bond pairs in beryllium(II) chloride? [closed]

How do I identify the lone and bonding pairs in $$\ce{BeCl2}$$?

In a simple sense, bond pairs are a pair of electrons (one from the central atom and one with the atom which is bonding) and participate in the bonding of the atom. Whereas lone pairs are the pairs of electron on an atom that do not participate in the bonding of two atoms.

To identify lone pairs in a molecule, figure out the number of valence electrons of the atom and subtract the number of electrons that have participated in the bonding. However, remember that the lone pairs are pairs and therefore if you ever find just one free electron that does not participate, it would mean that the compound has a charge.

Now for BeCl2

Usually, we only show the bond and lone pair of the central atom but if you want, just for your information, each chlorine atom in the reaction has got 3 lone pairs (the red dots.)

Be has atomic number 4 so its electronic configuration is $$1s^22s^2$$. Cl has atomic number 17 so it electronic configuration is $$1s^22s^22p^63s^23p^5$$.

It is given that two Cl atoms bond with one Be atom so that means one $$2s$$ electron of Be gets excited and transitions to the empty $$2p$$ orbital that Be has. So now $$2s$$ and $$2p$$ of Be atom have one lone electron each. These two orbitals hybridize and form two $$sp$$ orbitals. These $$sp$$ orbitals have one lone electron each and these orbitals participate in bonding with the $$p$$ orbital of a Cl atom which has only one lone electron. On bonding, the valency of the Cl atoms is satisfied.

There are only two bond pair of electrons in this case so according to VSEPR theory, these electron pairs have a tendency to be at an angle of 180 degrees to each other. By this, we can conclude that $$\ce{BeCl2}$$ has a linear shape.

On a side note, the valencies of Be are not fully satisfied as it does not achieve an octet. This is the reason why $$\ce{BeCl2}$$ acts as a Lewis acid because it has a tendency to accept more electrons in its remaining two $$2p$$ orbitals to complete its octet.