Wave Function for sp, sp² and sp³ Hybrid Orbitals

The wave function of hybrid orbitals are obtained by the linear combination of the angular wave functions of the appropriate atomic orbitals.

sp Hybrid Orbital Wave Function

The combination of one s and one 2p orbital, giving two hybrid orbitals may be expressed as-
1. Ψ₁ = a₁Ψ_s + b₁Ψ_2p       ---Equation-1
2. Ψ₂ = a₂Ψ_s + b₂Ψ_2p       ---Equation-2
The values of the linear combination coefficients a₁, b₁, a₂ and b₂ may be determined by the following considerations-
I. Ψ₁ and Ψ₂ are normalized
II. Ψ₁ and Ψ₂ are orthogonal
III. Ψ₁ and Ψ₂ are equivalent
From Condition-1: Ψ₁ and Ψ₂ are normalized

From Condition-2: Ψ₁ and Ψ₂ are orthogonal

Sine the s-orbital is sperically symmetrical and the two hybrid orbitals Ψ₁ and Ψ₂ are equivalent, the share of 's' function is equal in both Ψ₁ and Ψ₂
i.e.

sp² Hybrid Orbital Wave Function

Wave functions for three hybrid orbitals may be written as-
Ψ_sp²(1) = a₁Ψ_2s + b₁Ψ_2px + c₁Ψ_2py      ---Equation-1
Ψ_sp²(2) = a₂Ψ_2s + b₂Ψ_2px + c₂Ψ_2py      ---Equation-2
Ψ_sp²(3) = a₃Ψ_2s + b₃Ψ_2px + c₃Ψ_2py      ---Equation-3
Since, s orbital is equally distributed among the three hubrid orbitals, so, we have-

Assume that Ψ_sp²(1) points towards the x-axis. Hence, this hybrid orbital will have contribution from 2s and 2p_x orbitals only. The contribution of the 2p_y orbital will be zero. So, c₁ = 0.
Since, Ψ_sp²(1) is normalized,

Since, Ψ_sp²(1) and Ψ_sp²(2) are orthogonal and Ψ_sp²(1) and Ψ_sp²(3) are also orthogonal. Hence-
a₁a₂ + b₁b₂ = 0
a₁a₃ + b₁b₃ = 0
Hence,

Again, Since Ψ_sp²(2) is normalized-

Further, Ψ_sp²(3) is also normalized-

Let c₂ have a positive value, then the value of c₃ will have to be negative as Ψ_sp²(2) and Ψ_sp²(3) are orthogonal.
i.e. a₂a₃ + b₂b₃ + c₂c₃ = 0
Thus, If, c₂ = +1/√2 and c₃ = −1/√2
Hence, the three wave functions are-

sp³ Hybrid Orbital Wave Function

The wave function of four sp³ hybrid orbitals may be written as-
Ψ_sp³(1) = a₁Ψ_2s + b₁Ψ_2px + c₁Ψ_2py + d₁Ψ_2pz      ---Equation-1
Ψ_sp³(2) = a₂Ψ_2s + b₂Ψ_2px + c₂Ψ_2py + d₂Ψ_2pz      ---Equation-2
Ψ_sp³(3) = a₁Ψ_2s + b₃Ψ_2px + c₃Ψ_2py + d₃Ψ_2pz      ---Equation-3
Ψ_sp³(4) = a₄Ψ_2s + b₄Ψ_2px + c₄Ψ_2py + d₄Ψ_2pz      ---Equation-4
Since the s orbital is equally distributed among the four hybrid orbitals. So, we have-

Assuming that Ψ_sp³(1) is developed along x-axis, so the contribution from 2p_y and 2p_z orbitals to the hybrid orbitals is zero.
that means, c₁ = d₁ = 0
Since, Ψ_sp³(1) is normalized.
Hence,

The orthogonality conditions for Ψ_sp³(1) and Ψ_sp³(2), Ψ_sp³(1) and Ψ_sp³(3) and Ψ_sp³(1) and Ψ_sp³(4) are -
a₁a₂ + b₁b₂ = 0
or, b₂ = − a₁a₂/b₁
or, b₂ = − 1/2√3
a₁a₃ + b₁b₃ = 0
b₃ = − a₁a₃/b₁
or, b₃ = − 1/2√3
a₁a₄ + b₁b₄ = 0
b₄ = − a₁a₄/b₁
or, b₄ = − 1/2√3
Assuming that Ψ_sp³(2) lies in the xz-plane, the hybrid orbital will have contributions from 2s, 2p_x and 2p_z orbitals and the contribution of 2p_y to Ψ_sp³(2) is zero.
That means,
c₂ = 0
The normalized condition for Ψ_sp³(2) gives-

The requirement of orthogonality conditions for Ψ_sp³(2) and Ψ_sp³(3), and for Ψ_sp³(2) and Ψ_sp³(4) gives-
a₂a₃ + b₂b₃ + d₂d₃ = 0
or, d₃ = −(a₂a₃ + b₂b₃)/d₂
or, d₃ = − 1/√6
and a₂a₄ + b₂b₄ + d₂d₄ = 0
or, d₄ = −(a₂a₄ + b₂b₄)/d₂
or, d₄ = − 1/√6
The normalization required for Ψ_sp³(3) gives-

Similarly, normalization requirement for Ψ_sp³(4) and the orthogonality condition between Ψ_sp³(3) and Ψ_sp³(4) gives-
c₄ = − 1/√4
Hence, the four sp³ hybrid orbital wave functions can be written as-
Ψ_sp³(1) = (1/2)Ψ_2s + (√3/2)Ψ_2p
Ψ_sp³(2) = (1/2)Ψ_2s − (1/2)Ψ_2px + (√2/√3)Ψ_2py
Ψ_sp³(3) = (1/2)Ψ_2s − (1/2√3)Ψ_2px + (1/√2)Ψ_2py − (1/√6)Ψ_2pz
Ψ_sp³(4) = (1/2)Ψ_2s − (1/2√3)Ψ_2px + (1/√2)Ψ_2py − (1/√6)Ψ_2pz

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