# Mathematical form of four hybrid orbitals

The mathematical form of the four $\ce{sp^3}$ hybrid orbitals are given below. \begin{align} \ce{ \tag{a} sp^3 &= 1/2s + 1/2p_x + 1/2p_y + 1/2p_z\\ \tag{b} sp^3 &= 1/2s + 1/2p_x - 1/2p_y - 1/2p_z\\ \tag{c} sp^3 &= 1/2s - 1/2p_x + 1/2p_y - 1/2p_z\\ \tag{d} sp^3 &= 1/2s - 1/2p_x - 1/2p_y - 1/2p_z\\ }\end{align} For each hybrid orbital the number in front of the $\ce{s}$ and three $\ce{p}$ functions, called a coefficient, describes the contribution and relative ratio of each canonical orbital to the hybrid wave function. Add up the coefficients and prove to that these orbitals are $\ce{s^1p^3}$.

Yes, I know that the grammar is terrible on that last sentence, that is how it is written. So, I know that for an $\ce{sp^3}$ orbital it is one part $\ce{s}$ orbital and three parts $\ce{p}$ orbital. However, I have no idea what this question wants me to do. Adding up the coefficients gives me $2$, $0$, $0$, and $-1$, but I have no idea if that is how it wants me to add it up (you could ignore the sign and get $2$ every time, and show that $1/2$ is a quarter of $2$, thus proving that each component contributes a quarter to the hybridized orbital, but this seems too simple and incorrect).

• The sign on the $\text{p}_z$ orbital in (d) is wrong. It should be +, not -. – MaxW Sep 14 '16 at 23:21
• Well, that may be true, but that's how it was given to me. There could be a typo though. – vin2 Sep 14 '16 at 23:26
• Could you provide a source for the question, or was it given to you by your teacher? I would like to remind everyone, that the notation is a contraction: $$\ce{sp^{n} = s^{\frac{1}{n+1}}p^{\frac{n}{n+1}}},$$ omitting the common factor of $\frac{1}{n+1}$. – Martin - マーチン Sep 15 '16 at 4:47
• This was given to me as a practice quiz in my organic chemistry class. – vin2 Sep 15 '16 at 21:28

If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination

$$\psi = \sum_n c_n \phi_n$$

then loosely speaking, the "amount" of $\phi_n$ in $\psi$ is given not by $c_n$, but rather by $\left|c_n\right|^2$. (See, for example, Griffiths, Introduction to Quantum Mechanics, 2nd ed., Section 3.4.)

Therefore, summing the four values of $c_n$ for each AO has no physical meaning.

Instead let's take wavefunction (b) for example. The coefficient in front of the s orbital is $1/2$, so the "amount" of s-orbital character is simply $\left|1/2\right|^2 = 1/4$. The coefficient in front of the $\mathrm{p}_y$ orbital is $-1/2$, so there is $\left|-1/2\right|^2 = 1/4$ $\mathrm{p}_y$-character.

Armed with this knowledge, it's incredibly easy to see that each orbital has exactly $1/4$ s-character and a total of $3/4$ p-character ($1/4$ from each p orbital), because none of the $\pm$ signs matter (the minus signs are removed by the absolute value).

• Thanks for the answer... Unfortunately quantum mechanics isn't a prerequisite for the class that I'm a part of for which this question was used, that's next year, I got this question in sophomore organic chemistry class. Your answer makes sense, but there should be another way to explain this? I dunno, it could just be a bad question. Thanks again though! – vin2 Sep 15 '16 at 4:20
• @vin2 I think this is the only "correct" way of explaining it. You could apply other heuristics to argue that the contribution is 1:3, such as what you described as 1/2 being a quarter of 2, but that would be an incorrect way of reaching the correct answer. – orthocresol Sep 15 '16 at 5:11
• @vin2 The simplest (and ultimately correct) way is to think that individual unhybridized orbitals (s,p,d...) form an ortonormal basis set in multi-dimensional vector space by definition. School teaches to work with such a basis (cartesian coordinates) in 2d and 3d spaces, it should be pretty easy to extrapolate. This approach is surprisingly common in applied mathematics. – permeakra Sep 15 '16 at 6:09