Question

What is the hybridization of the central atom in SF5Cl?

Helium has 2 naturally occurring isotopes, He-4 (4.003 u) and He-3 (3.016 u). In an unusual sample of helium taken from a natural gas well the abundance of He-4 is 95.78%. What is the average atomic mass of this sample of helium?

Answer 1

The central atom of SF5Cl exhibits sp3d2 hybridization. The average atomic mass of the unusual helium sample from the natural gas well, given its isotopic abundances, is calculated to be 3.959 u.

Step-by-step explanation:

The hybridization of the central atom in SF5Cl can be determined by counting the number of electron domains (bonds and lone pairs) around the central sulfur atom. Sulfur in SF5Cl has five fluorine atoms and one chlorine atom bonded to it, making a total of six electron domains, which suggests a hybridization state of sp3d2. This is similar to the sulfur in SF6, which also exhibits sp3d2 hybridization.

For the average atomic mass of the helium sample, we take into account the natural abundance and atomic mass of each isotope: He-4 with an abundance of 95.78% and an atomic mass of 4.003 u, and He-3 with the remaining abundance and a mass of 3.016 u. Using the formula for weighted average atomic mass, we multiply each isotope's atomic mass by its fractional abundance and sum the results:

Average atomic mass = (4.003 u × 0.9578) + (3.016 u × (1 - 0.9578))
= (3.833 u) + (0.126 u)
= 3.959 u

The average atomic mass of this unusual helium sample is 3.959 u.