What is the hybridization of the central atom in the sulfur trifluoride (SF3^-) anion?
Answer (1)
To determine the hybridization of the central atom in the sulfur trifluoride ion ( SF 3 − ), let's break down the process step-by-step.
Identify the Central Atom and Calculate the Total Number of Valence Electrons:
Sulfur (S) is the central atom here. Sulfur has 6 valence electrons.
Each fluorine (F) atom contributes 7 valence electrons. Since there are three fluorine atoms, that's a total of 3 × 7 = 21 electrons from fluorine.
The ion has a − 1 charge, meaning there’s one additional electron.
Therefore, the total number of valence electrons is 6 + 21 + 1 = 28 .
Draw the Lewis Structure:
Connect each F atom to the central S atom with a single bond. Each bond represents 2 electrons, so 3 × 2 = 6 electrons are used for bonding.
Distribute the remaining 22 electrons (28 total minus 6 used in bonds) to satisfy the octet rule for the fluorine atoms. Each fluorine atom will need 6 more electrons to complete its octet, taking 3 × 6 = 18 electrons.
The remaining electrons (22 total minus 18 used for fluorine octets) total 4 electrons, which are placed as lone pairs on the sulfur atom.
Determine the Hybridization:
Sulfur has three sigma bonds (one with each fluorine) and two lone pairs. The total count of electron groups around sulfur is 5.
A central atom with 5 electron groups is typically sp 3 d hybridized.
This means, in SF 3 − , sulfur is sp 3 d hybridized. This allows the structure to accommodate three bonded atoms and two lone pairs.
