Construct the wave Function for the sp³ hybrid orbitals

The four sp³ hybrid orbitals are as follows:
Ψ ₁ = a ₁ ф _s + b ₁ ф _py +d ₁ ф _pz
Ψ ₂ = a ₂ ф_s + b ₂ ф _py +d ₂ ф _pz

Ψ ₃ = a ₃ ф _s + b ₃ ф _py +d ₃ ф _pz

Ψ ₄ = a ₄ ф _s + b ₄ ф _py +d ₄ ф _pz
Since the s-orbital is equally distributed among the four hybrid orbital, we will get the following:
a ² ₁ = a ² ₂ = a ² ₃ = a ² ₄ = 1/4
Hence,
a ₁ = a ₂ = a ₃ = a ₄ = 1/4 ^1/2 = 1/2
Let us suppose that ф ₁ is developed along the x-axis, that is this hybrid orbital will get contribution from s and p_x orbitals only. Because of this the contribution from ф _py and ф _pz will get vanished.
Hence we can write:
c ₁ = d ₁ = 0
As Ψ₁ is in normal form, we will get:
a ² ₁ + b ² ₁ + c ² ₁ + d ² ₁ = 1
Or we can rewrite it as:
a ² ₁ + b ² ₁ = 1
(Because c ₁ = d ₁ = 0)
Therefore b ₁ = (1 – a ² ₁) ^1/2
= (1 – 1/4) ^1/2
= (3) ^1/2 / 2
The requirement of orthogonality condition for Ψ₁ and Ψ₂, Ψ₁ and Ψ₃, ф₁ and Ψ₄, it will give:
a ₁ a ₂ + b ₁ b ₂ = 0
a ₁ a ₃ + b ₁ b ₃ = 0
a ₁ a ₄ + b ₁ b ₄ = 0

Hence, b ₂ = b ₃ = b ₄
= – a ₁ a ₂ / b ₁
= – a₁ a ₃ / b₁
= – a ₁ a ₄ / b ₁

= – [ (1/2) (1/2) ]/[ (3)^1/2 / 2]

= – 1/ (2 (3) ^1/2)

Suppose that Ψ₂ lies in the XZ plane, the hybrid orbital will have contributions form s , p_x and p_z orbitals only. The contribution of p_y to ф₂ would be equal to zero. Hence,

c₂ = 0
The normalization requirements for Ψ₂ will be:
a ² ₂ + b ² ₂ + c ² ₂ +
d ² ₂ = 1
Or we can rewrite it as:
a² ₂ + b ² ₂ + d ² ₂ = 1
(Because c ₂ = 0)
Because d ² ₂ = 1 – (a ² ₂ + b ² ₂)
= 1 – (1/4 + 1/12) = 2/3
Therefore d₂ = (2/3)^1/2
The requirement of orthogonality condition for and Ψ₂, Ψ₃ and
Ψ ₂, Ψ ₄ will give:
a ₂ a ₃ + b ₂ b₃ + d₂ d₃ = 0
a ₂ a ₄ + b ₂ b₄ + d₂ d₄ = 0
Hence, d₃ = d₄

= [ – a₂ a₃ – b₂ b₃ ]/ d₂

Or
[- a₂ a₄ + b₂ b₄] / d₂

= – [(1/4 + 1/12)]/[ (2/3) ^1/2]

= – 1/ (6) ^1/2

The normalization requirement for Ψ₃ will be given as:
a ² ₃ + b ² ₃ + c ² ₃ +
d ² ₃ = 1
c ² ₃ = 1- (a ² ₃ + b ² ₃ +
d ² ₃)
= 1 – (1/4 + 1/12 + 1/6) = 1/2
Therefore c₃ = + 1/2