For each of the bonds listed below use hybridized orbitals to classify the type of sigma bond that is formed between the two atoms.

a) The sigma bond between boron and hydrogen in BH3

answer: sp^2 - p

b) the sigma bond between S and fluoride in SF4.

answer: sp^3d - p
I understand how to obtain the first part of the answer, but not the second.

Thank you!

You're forgetting that unshared pair of electrons on the S atom. Here is the Lewis dot structure.

https://www.google.com/search?q=lewis+structure+SF4&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

That AB4U makes it trigonal bi-pyramid and that accounts for the dsp3 for SF4

To determine the type of sigma bond between two atoms, you need to understand the hybridization of the central atom and the atomic orbitals involved in bonding.

In the case of boron and hydrogen in BH3 (a), boron is surrounded by three hydrogen atoms. Boron's electron configuration is 1s2 2s2 2p1. To form four hybrid orbitals, one 2s and two 2p orbitals combine and undergo hybridization. The hybridization involved is sp2, which means three hybrid orbitals are formed.

Next, we need to consider the atomic orbital of the hydrogen atom, which is a 1s orbital. The hybrid orbital of boron that overlaps with the hydrogen 1s orbital to form a sigma bond is categorized as "p."

Therefore, the type of sigma bond between boron and hydrogen in BH3 is sp2 - p.

Now, let's move on to the second part of your question, the sigma bond between sulfur (S) and fluoride (F) in SF4 (b).

Sulfur's electron configuration is 1s2 2s2 2p6 3s2 3p4. To form six hybrid orbitals, one 3s and three 3p orbitals combine and undergo hybridization. The hybridization involved is sp3d, leading to six hybrid orbitals.

Here, we need to consider the atomic orbital of fluoride (F), which is a 2p orbital. The hybrid orbital of sulfur that overlaps with the fluorine 2p orbital to form a sigma bond is categorized as "p."

Therefore, the type of sigma bond between sulfur and fluoride in SF4 is sp3d - p.

I hope this explanation helps clarify the second part of your question!