Determinng the Shape for BeH₂. The Lewis structure is determined by trial and error.

First thing to do is check the group numbers and count valence electrons

Beryllium is in group 2A. It has 2 valence electrons.

Hydrogen is in group 1A. It has 1 valence electron.

The total valence electrons is 2 + 2 = 4 There is an even number of valence electrons. This means the octet rule could be met. If the count was odd( 5 ,7,9 or 11) the octet rule would be violated.

Draw the symbols for the Be as the central atom and hydrogen atoms on the "outside".

The molecule has a linear shape.

The HCN molecule can be analyzed the following way.

First thing to do is check the group numbers and count valence electrons

Carbon is in group 4A. It has 4 valence electrons.

Hydrogen is in group 1A. It has 1 valence electrons.

Nitrogen is in group 5A. It has 5 valence electrons.

The total valence electrons is 4 + 1 + 5 = 10 . There is an even number of valence electrons. This means the octet rule should be met.

Divide ten by 2 and get the number of electron pairs total.

Draw the symbols for the C as the central atom with hydrogen and nitrogen atoms on the "outside".

Place electron pairs between the atom symbols. This uses only four electrons, two pairs.

Place electron pairs around the atom symbols. This uses all the electrons, but there is no octet on carbon.

Move unshared pairs form the nitrogen to create multiple bonds between the carbon and nitrogen.

The bond angle is 180 degrees.

This is the same type structure seen in CN^1- and N₂. These all have ten valence electrons. All of these have a linear shape.

MAJOR IDEA:

Combinations of atoms with the same number of atoms and electrons will have the same type of Lewis structures. Carbon monoxide and nitrogen both have ten velence electrons and two atoms. They both have a triple bond. What would you expect for CN^1- cyanide ion with 10 valence electrons?