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

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1 Answer
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^@# 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^@#.