Question

How would you predict the ideal bond angle(s) around each central atom in this molecule?

Answer 1

You look at the number of electron groups, and consider how to distribute these groups evenly in space.

CENTRAL CARBON

• The carbon has two electron groups, so a `180^@` bond angle is the furthest even separation between the two atoms.

CENTRAL NITROGEN

• The right nitrogen is the interesting one; with four electron groups (the lone pair, the carbon, and the two hydrogens total four), it has a tetrahedral electron geometry and a trigonal pyramidal molecular geometry in accordance with VSEPR Theory.

These geometries have ideal bond angles of `109.5^@` in three dimensions, but the lone pair 'crunches' the atoms together a little, so the angles become less than `109.5^@`.

This molecule is similar to `"NH"_3` with one hydrogen replaced with `"C"-="N"`. Since `"C"-="N"` is larger than `"H"`, the `"C"-"N"-"H"` bond angle should be slightly larger than the `"H"-"N"-"H"` bond angle of `"NH"_3`, but less than `109.5^@`.