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

Jul 11, 2016

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 ${180}^{\circ}$ 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 ${109.5}^{\circ}$ in three dimensions, but the lone pair 'crunches' the atoms together a little, so the angles become less than ${109.5}^{\circ}$.

This molecule is similar to ${\text{NH}}_{3}$ with one hydrogen replaced with $\text{C"-="N}$. Since $\text{C"-="N}$ is larger than $\text{H}$, the $\text{C"-"N"-"H}$ bond angle should be slightly larger than the $\text{H"-"N"-"H}$ bond angle of ${\text{NH}}_{3}$, but less than ${109.5}^{\circ}$.