Complete the following for the compound NF3: A) State its number of valence electrons B) Draw its Lewis structure C) Describe its VSEPR shape

A) Number of Valence Electrons:
Nitrogen (N) is in group 15 of the periodic table and has 5 valence electrons. Each fluorine (F) is in group 17 and has 7 valence electrons. Since there are three fluorine atoms, they contribute a total of 21 valence electrons. Therefore, the total number of valence electrons in NF3 is:

5 (from N) + 3 × 7 (from F) = 5 + 21 = 26 valence electrons.

B) Lewis Structure:
The Lewis structure for NF3 can be drawn by following these steps:

1. Place the nitrogen atom in the center because it is less electronegative than fluorine.
2. Connect each fluorine atom to the nitrogen atom with a single bond. This uses 6 electrons (2 for each bond).
3. Distribute the remaining valence electrons to complete the octets of the fluorine atoms. Each fluorine needs 8 electrons around it, and since they each already have 2 from the bond, you will add 6 more electron pairs (3 lone pairs) around each fluorine.

The final structure looks like this:

N:       F:
  |       |||
  F: — N — F:
  |       |||
  •      • •   (lone pairs shown as dots)

C) VSEPR Shape:
According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the presence of three bonding pairs (N-F bonds) and one lone pair of electrons on the nitrogen creates a particular molecular geometry. The lone pair occupies more space and pushes the three fluorine atoms closer together.

The resulting shape is trigonal pyramidal, which is similar to a tetrahedral arrangement, but with one vertex (the lone pair) missing. The bond angles are slightly less than 109.5 degrees due to the repulsion from the lone pair, which takes up more space than the bonded pairs.