Beryllium Dichloride
BeCl
2
Valence e = 2 + (2x7) = 16
RHED = 2
(2 individual BeCl bonds)
Electronic geometry is
LINEAR.
No lone pairs on Be so
molecular geometry is
LINEAR
Be
Cl
Cl
180º
δ
+
δ

δ

POLAR covalent bonds;
NON POLAR molecule
3p
[He]
2s
Be
Cl
[Ne]
3s
Need 2
unpaired
e for Be..
How
about
this?
NO Use
this..
See over..
Cl already has
one unpaired e
in a p orbital,
ready to bond
2p
2p
2p
[He]
2s
Be
[He]
sp
Be
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Using a 2s and a 2p orbital for Be and the 3p orbital for each Cl,
we would get:
3p
3p
Cl
Cl
Be
1s
2s
2p
Not very efficient, and very uneven.
Instead, we HYBRIDIZE the single 2s orbital and one of the 2p orbitals, (this
is basically mixing two equations to get two new ones). The resulting sp
hybridized orbitals are degenerate (of the same energy) and look like this:
The two sp orbitals are arranged in space 180º apart, and so we can now
build our symmetrical BeCl
2
molecule:
3p
Cl
3p
Cl
sp
sp
Be
1s