Complete the electron dot structure below to show how beryllium flouride (BeF2) is formed.
F F
Be+ -> Be
F F
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To complete the electron dot structure for beryllium fluoride (BeF2), we need to determine the number of valence electrons present for each atom.
Beryllium (Be) is an element in Group 2, so it has two valence electrons.
Fluorine (F) is an element in Group 17, so it has seven valence electrons.
Step 1: Place the beryllium atom in the center and surround it with the fluorine atoms.
F F
Be
Step 2: Place two valence electrons on the beryllium atom, represented as dots.
F F
Be • •
Step 3: Place one valence electron on each fluorine atom, represented as a dot.
F • F •
Be • •
Step 4: Connect the remaining valence electrons as pairs around the fluorine atoms until they have a full octet. In this case, it means each fluorine atom will have eight electrons. Use dashes to represent the shared pairs.
F • F •
Be • • F •
Step 5: Check if all atoms have a full octet. In this case, beryllium only has four electrons around it, which is less than an octet.
To complete the octet for beryllium, we need to use a different approach. We will form a double bond between beryllium and one of the fluorine atoms. This means that the two atoms will share two pairs of electrons instead of one.
F •
Be = F
Step 6: Distribute the remaining valence electrons.
F • F •
Be = F
Now, the beryllium atom has a full octet with eight electrons around it, and each fluorine atom also has a full octet with eight electrons. Therefore, the electron dot structure for beryllium fluoride (BeF2) is:
F • F •
Be = F