What is the hybridization of TeBr6?

Probably sp3d2

To determine the hybridization of TeBr6, we need to follow a few steps:

Step 1: Count the total number of valence electrons in the molecule.
Te (Tellurium) is in Group 16 of the periodic table, so it has 6 valence electrons.
Br (Bromine) is in Group 17 of the periodic table, so each bromine atom contributes 7 valence electrons. Since there are 6 bromine atoms in TeBr6, the total number of valence electrons from bromine is 6 * 7 = 42.

Adding the valence electrons from tellurium (6) and the bromine atoms (42), we get a total of 6 + 42 = 48 valence electrons.

Step 2: Identify the central atom, which is the atom bonded to all other atoms in the molecule. In this case, it is Te (Tellurium).

Step 3: Distribute the valence electrons around the central atom to satisfy the octet rule. Place the remaining valence electrons (after bonding) on the central atom.

Each Br-Te bond contributes two electrons, so the number of electrons used for the Te-Br bonds is 2 * 6 = 12.

Subtracting the electrons used for bonding (12) from the total valence electrons (48), we have 48 - 12 = 36 electrons remaining to be distributed around Te.

Step 4: Determine the hybridization.

The hybridization of an atom is determined by the number of regions of electron density around it. Each bond or lone pair counts as one region of electron density.

In TeBr6, we have 6 Te-Br bonds, which count as 6 regions of electron density.

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, when there are 6 regions of electron density, the central atom adopts an octahedral electron geometry.

For an atom in an octahedral electron geometry, the hybridization is d2sp3.

So, the hybridization of TeBr6 is d2sp3.