CH vs. SiH and CF vs. CH using VB arguments
Sanja Pudar
Study Guide Outline
1. Bond Angles in Hydrides
2. Lobe Orbitals
3. Introduction to CH2
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Bond Angles in Hydrides
We looked at the bond angles of the compounds within a vertical group and saw that for smaller atoms
the bond angles are greater than 90◦ , but as the atoms get larger, the bond angle tends to 90 degrees.
VB predicts 90◦ angles for all these bonds (consider for example H2 O, H2 S, H2 Se, H2 Te) but when the
atoms are small, the s orbitals of the two hydrogens on the molecule are close enough that they run
into one another. The reason why H-O-H angle in H2 O is 104.5◦ is that while the p orbitals on oxygen
are orthogonal to each other, when they form bonds to the s orbitals on hydrogen, the s orbitals in the
hydrogens are not naturally orthogonal to each other. As a result, the bond angle for H2 O opens to a value
larger than 90◦ due to the Pauli principle. In the cases of 2nd row atoms and those lower,the hydride
bond involves p orbitals from higher orbital shells (e.g. 3p, 4p, etc.), and since the higher level orbitals
extend further from the nucleus the resulting bonds keep the hydrogens far enough apart that only minimal
overlap occurs.
However, a reasonable question to ask is why the angle for NH2 smaller than that for H2 O. Recall that as
we move right across the periodic table, Zef f increases, and the size of atoms decrease. So atomistically,
N is larger than O, and so the bond angles need to be opened slightly less than for the case with oxygen.
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Lobe Orbitals
Consider BeH: From everything we have learned thus far, there should not be a bound state since we
would assume bringing a hydrogen towards a doubly occupied 2s orbital is analagous to the He − H
example which is unbound. Alternatively, BeH+ should form a bond. Actually, both BeH and BeH+ form
bonds (with energies 46 kcal/mol and 71 kcal/mol, respectively). The reason for this unexpected bond is
that since the 2s and 2p on Be are relatively close in energy, those orbitals can hybridize or “pooch”.
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Pooching only occurs between one s orbital and one empty p orbital of the same n quantum number. If
one of the electrons in the 2s orbital mixes 2p orbital character, it forms what will be referred to as a lobe
orbital. Lobe orbital can allow for more overlap as a bonding orbital since its mixture of s and p character
allows
it to
bondhas
to species
wouldwhich
otherwise
never bond
due to to
orthogonality
the lobe
orbital
more that
character
has better
penetration
the nucleus,conditions.
and extendsAdditionally,
further from
the
orbital
has
moresorbital.
character which has better penetration to the nucleus, and extends further from
thelobe
nucleus
than
a pure
the
nucleus
a pure
orbital.
In order
to than
describe
thesmathematical
basis for pooching consider two electrons in the orbital with the
In
order
to
describe
the
mathematical
basis
for pooching consider two electrons in the 2s orbital with the
overall wavefunction of
overall wavefunction of
¾ ¾ φ2s (1)φ2s (2)(αβ − βα).
This wavefunction is simply the antisymmetric product of two orbitals, labeled and , where
This wavefunction is simply the antisymmetric product of two 2s orbitals, labeled φa and φb , where
¾ and ¾ . Upon mixing some character into both of these orbitals we can
φa = φ2s α(1) and φb = φ2s β(2). Upon mixing some 2p character into both of these orbitals we can
express and now as linear combinations of the quantities of character we are adding (omitting spin
express φa and φb now as linear combinations of the quantities of p character we are adding (omitting spin
and electron number labels):
and electron number labels):
φa= φ2s¾+ λφ2p¾
φb= φ2s¾− λφ2p¾
Thesenew
newφaand
andφbwavefunctions
wavefunctionsare
areshown
shownininFigure
Figure1.1.
These
Figure 1: φa and φb lobe orbital wavefunctions.
Figure 1: and lobe orbital wavefunctions.
We can see from the amplitude of the wavefunctions that each lobe orbital will form bonds on opposite
sides
of the
value for
calculated by optimizing
for the
bestwill
energy
the lobe
We can
seenucleus.
from theThe
amplitude
of λ
theis wavefunctions
that each lobe
orbital
formofbonds
on orbitals,
opposite
and
its
best
value
is
λ
=
0.4.
sides of the nucleus. The value for is calculated by optimizing for the best energy of the lobe orbitals,
+
The
. shows that Be-H bond is not as good as the bond in BeH , which is a
and dissociation
its best valuecurve
is ofBeH
consequence
of thecurve
need of
to BeH
unpairshows
the electrons
in the
pooched
orbitals.
When
the first
there
, which
is a
The dissociation
that Be-H
bond
is not as
good as
the bond
in bond
BeH ·forms,
will
be two spin-paired
electrons
the bond
pluspooched
one additional
on first
the remaining
side
of
consequence
of the need
to unpairforming
the electrons
in the
orbitals.electron
When the
bond forms,
there
will be two spin-paired electrons forming the bond2plus one additional electron on the remaining side of
the hybrid orbital. This additional electron will experience repulsion from the bond electrons and has to
get orthogonal to them. This costs about 1.5 eV in energy, which is deducted from the total energy that
we gain by forming the bond. Hence, the well depth of this bond is shallower than that of a normal bond.
When we form the second bond to the remaining lobe orbital we don’t have to pay this energy penalty
the hybrid orbital. This additional electron will experience repulsion from the bond electrons and has to
get orthogonal to them. This costs about 1.5 eV in energy, which is deducted from the total energy that
we gain by forming the bond. Hence, the well depth of this bond is shallower than that of a normal bond.
When we form the second bond to the remaining lobe orbital we don’t have to pay this energy penalty
again because the orbitals have already been unpaired.
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Introduction to CH2
Before we introduced lobe orbitals, we found out that the ground state of CH2 is singlet. However, it is
well known that the g.s. is a triplet. This can be explained using the lobe orbitals.
It is found that the first hydrogen bonds to a p orbital, resulting in 2 Π state. The second hydrogen bonds
to a lobe orbital, leaving the system in a triplet state. These two hydrogen atoms have no preference over
either p or lobe orbital, thus the two bonds become equal.
During the next class, we’ll talk about where third hydrogen prefers to bind, and we’ll also compare the
formation of CH4 to SiH4 .
Suggested reading:
• Chapter 6: GVB Model of Bonding
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