I was just introduced to the concept of orbital hybridization. I believe I understand the idea behind it, but there is some accompanying terminology that prevents me from achieving a greater understanding.

In the following image:

I see that one of the electrons in the $$2s$$ orbital of carbon is promoted to one of the $$2p$$ orbitals. I start to get confused when hybridization occurs, simply because there are four $$\ce{sp^3}$$ hybrid orbitals, as also shown in this diagram:

I learned that the name $$\ce{sp^3}$$ comes from the combination of one s-orbital and three p-orbitals ($$\ce{s + p + p + p}$$), and this makes sense to me. What doesn't make sense is why each individual hybrid orbital is called an $$\ce{sp^3}$$ orbital. When the $$\ce{2s}$$ orbital, for example, becomes an $$\ce{sp^3}$$ orbital, why is it referred to by its combination with p-orbitals ($$\ce{sp^3}$$) when it was just an s-orbital before?

Could someone please offer a simple explanation that would help clear this up? Why is each individual hybrid orbital called an $$\ce{sp^3}$$ orbital? It seems like all four orbitals together should be called $$\ce{sp^3}$$, because they all make up the combination defined by $$\ce{sp^3}$$ ($$\ce{s + p + p + p}$$).

• You didn't understand the idea behind it. As far number of orbitals is concerned, the picture you've posted is pretty clear, isn't? Oct 3 '19 at 8:42
• @Alchimista I meant I understood the idea behind hybridization. And yes, the picture was clear. It's the basis of my question. I see that there are four orbitals. I wanted to know why. Oct 3 '19 at 12:52
• Think of it as mixing a substance (say a glass of juice with three of water), where in the end the total amount of the substance has to be conserved. The "mixed" four orbitals have to contain what you started out with. Oct 3 '19 at 18:53
• @Kman3 Buck Thorn's example is the right way to think about it. You go from having one s orbital and three p orbitals to four mixed orbitals that are one-quarter s-type and three-quarters p-type. Oct 3 '19 at 18:58
• @BuckThorn Thanks for your help! So, using your example, does each $\ce{sp^3}$ orbital have $\ce{s}$ and $\ce{p}$ characteristics, just like how four glasses of the juice-water mixture would have both water and juice in them? Oct 4 '19 at 0:34

The number of hybridized orbitals is equal to the number of original orbitals. The name of the hybridized orbitals (e.g., $$sp^{3}$$) means just that the ratio of the contributions of $$s$$ to $$p$$ orbitals is $$1:3$$. We do not write $$s^{\frac{1}{4}}p^{\frac{3}{4}}$$.