To draw the Lewis structure for TeF6, we start by identifying the central atom, which is tellurium (Te). Tellurium is surrounded by six fluorine (F) atoms. Let’s break down the steps and the information:
Lewis Structure
1. **Count the valence electrons**: Tellurium has 6 valence electrons and each fluorine atom has 7. Therefore, the total number of valence electrons is:
- Te: 6
- 6 F: 6 × 7 = 42
- Total: 6 + 42 = 48
2. **Draw the structure**: Place Te in the center and surround it with six F atoms. Since fluorine is highly electronegative, each F will form a single bond with Te. A single bond consists of 2 electrons.
3. **Assign lone pairs**: After creating 6 bonds (12 electrons) with six fluorine atoms, allocate the remaining valence electrons. In this case, all electrons are used in bonding, so Te has no lone pairs.
Information Breakdown
- a. Number of bonding electron pairs: 6 (each F atom is bonded to Te with a single bond, accounting for 6 bonding pairs).
- b. Number of nonbonding electron pairs: 0 (there are no lone pairs on the tellurium atom).
- c. Electron geometry: Octahedral (the arrangement of electrons around the central atom in three-dimensional space).
- d. Molecular geometry: Octahedral (since all bonding sites are occupied by fluorine atoms).
- e. Approximate bond angle: 90 degrees (in an octahedral geometry, all bond angles are 90 degrees).
In summary, the Lewis structure for TeF6 has 6 bonding pairs and 0 nonbonding pairs, with an octahedral electron and molecular geometry, and bond angles of approximately 90 degrees.