The goal of applying Valence Bond concept to water is to define the bonding in (H_2O) and also account for its structure (i.e., proper bond angle and also two lone bag predicted native VSEPR theory).
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The floor state electronic configuration of atom oxygen atom is (1s^2,2s^2,2p_x^2,2p_y^1 , 2p_z^1) and also of food the floor state electronic configuration of atom hydrogen atom is (1s^1), i.e., a spherical atomic orbital v no preferential orientation. If just the unfilled (2p_y) and also (2p_z) atomic orbitals the the oxygen were used as bonding orbitals, then 2 bonds would certainly be predicted. This bonding wavefunctions would certainly be mixture of only (|2p_y angle) and also (|2p_z angle) orbitals ~ above oxygen and the (|1s angle) orbitals on the hydrogens ((H_1) and (H_2)):
< | chi_1 angle = a_1 |1s angle_H_1 + b_1 |2p_y angle_O labelwrong1>
< | chi_2 angle = a_2 |1s angle_H_2 + b_2 |2p_z angle_O labelwrong2>
However, with a H-O-H bond angle for these bonds would be intended to be 90° because (2p_y) and also (2p_z) space oriented 90° through respect to every other. Note that ( | chi_1 angle ) and ( | chi_2 angle ) room two-center bonding orbitals usual to Valence bond theory.
Using the oxygen atom orbitals straight is obviously not a good model because that describing bonding in water, due to the fact that we understand from experiment the the bond angle because that water is 104.45° (Figure 10.2.2), i m sorry is additionally in covenant with VSEPR theory. Due to the fact that the (2s) orbit is spherical, mixing part (2s) character right into the (2p_z) orbitals can adjust the bond edge as debated previously through creating brand-new hybrid orbitals.
Historically, Valence Bond theory was used to describe bend angle in tiny molecules. That course, it was only qualitatively exactly in doing this, as the following instance shows. Let united state construct the Valence link wavefunctions for the 2 bonding bag in (H_2O) by mix the (|2s angle), (|2p_x angle), (|2p_y angle), and also (|2p_z angle) into four new (sp^3) hybrid orbitals:
<eginalign*chi_1 (r) &= dfrac12 left
Hence, the 3 (2p) orbitals of the oxygen atom combined with the (2s) orbitals of the oxygen to form four (sp^3) hybrid orbitals (Figure 10.2.3).
The bond edge for 4 groups that electrons about a central atom is 109.5 degrees. However, for water the experimental bond edge is 104.45°. The VSPER picture (general chathamtownfc.netistry) for this is that the smaller sized angle deserve to be defined by the visibility of the 2 lone-pairs of electron on the oxygen atom. Due to the fact that they take it up an ext volume of an are compared to a bonding pair of electron the repulsions between lone pairs and bonding pairs is expected to be better causing the H-O-H bond edge to be smaller sized than the appropriate 109.5°.
< |chi_i angle = N ( p + gamma s) labelH1>
where (N) is a normalization constant and (gamma) is the loved one contribution the s character to the hybrid orbital. Because that a pure (sp^3) hybrid, (gamma) would certainly be 0.25 and for a pure (sp) hybrid, (gamma) would be 1. The inquiry is exactly how to identify (gamma) to gain a better picture the the hybridization of water. Starting with the normalization criteria for wavefunctions:
< langle chi_i | chi_i angle =1>
and substituting Equation ( efH1) right into to get
< langle N ( ns + gamma s) | N ( p + gamma s) angle =1>
which in integral notation is
< int N^2 ( ns + gamma s)^2 d au =1 >
where (d au) represents all space. This is then broadened to
< N^2 cancelto1int p^2; d au + N^2 2 gamma cancelto0 int sp; d au + N^2 gamma^2 cancelto1 int s^2 ; d au =1labelH3 >
These terms leveling either due to orthogonality or normality of the constitute atomic orbitals. Equation ( efH3) simplifies to
< N^2 + N^2 gamma^2 =1 >
and hence the normalization element can it is in expressed in terms of (gamma)
< N = dfrac1sqrt1+gamma^2>
and the generic normalized (sp^x) hybrid orbit (Equation ( efH1)) is
< |chi_i angle = dfrac1sqrt1+gamma^2 ( p + gamma s) labelH4>
The s and also p personalities to a hybrid orbital are now easy to obtain by squaring ( |chi_i angle )The size of p-character is < left(dfrac1sqrt1+gamma^2 ight)^2 = dfrac11+gamma^2 labelp> as (gamma ightarrow 0), then the ns character that the hybrid goes to 100% The size of s-character is < left(dfrac1sqrt1+gamma^2 gamma ^2 ight)^2 = dfracgamma^21+gamma^2 labels> as (gamma ightarrow 1), then the s character of the hybrid goes to 50%
As pointed out above, the geometric arrangement will not it is in formed because a molecule is hybridized in a certain way, it is the other means around. Exactly how do we select the correct value of (gamma) for the hybrid orbitals? The mix coefficient (gamma) is plainly related to the bond edge θ. Using some simple trigonometric relationships, it deserve to be proven that:
Equation ( efangle) is an important equation as it connected experimentally identified structure come the nature of the bonding and also specifically, the composition of the atom orbitals that develop the hybrid orbitals supplied in the bonding.
What is the s-character in the hybrid orbitals for (CO_2).
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We recognize from an easy VSEPR theory that the geometry of (CO_2) is a linear triatomic molecule.