NH₃ · ammonia
Nitrogen bonds three hydrogens and keeps one lone pair. Four electron domains give a tetrahedral arrangement; the lone pair squeezes the bond angle to about 107°.
A must account for exactly the , so total them first: nitrogen is a Group 15 element, so it has 15 - 10 = 5 valence electrons; each hydrogen is a Group 1 element, so it has 1 valence electron, and with 3 of them that is 1 × 3 = 3, giving 8 valence electrons that the finished structure has to use in full, with none added and none left over.
The must bond to every other , so it cannot be hydrogen: with one , hydrogen forms only one bond and is always terminal. That leaves nitrogen as the central atom.
Draw the by connecting N to each with a . A single bond is one shared pair of electrons, so each of these 3 single bonds accounts for 2 electrons: 6 of the 8 are now placed, and 2 remain to be distributed.
Every here is hydrogen, and a hydrogen is complete with just two electrons (a ), which its already supplies. No are added to the terminals, so all 2 remaining electrons are available for N.
The last 2 electrons are placed on N as 1 . Every is now accounted for, and none remain.
Before continuing, check the total electron count surrounding each . N: 6 shared + 2 nonbonding = 8 (a full ); each H: 2 shared + 0 nonbonding = 2 (a duet, all hydrogen needs).
Evaluate the of every , defined as minus minus one half of the . Lone-pair electrons count in full because they belong to one atom, while bonding electrons are divided evenly between the two bonded atoms. Every atom in this structure comes out to zero, so no atom bears an artificial charge and the structure is favored.
Watch out
On paper it's easy to draw this straight and call it done. The on N makes the real trigonal pyramidal, and the shape is what decides .
Determine the using : N is surrounded by 3 bonding domains and 1 , for 4 in total. Because electron domains repel one another, 4 domains arrange themselves as far apart as possible, giving tetrahedral and sp³ .
But are invisible in the final silhouette. Take it out of the picture and the trace a , with the angles squeezed to about ≈107° because lone pairs push harder than bonds do.
Finally, . A shape can't balance its bond pulls, so they stack into a net dipole pointing toward N. This is polar, and this net dipole governs much of its physical behavior.
Worth knowing
That on nitrogen is ammonia's whole personality: it grabs protons, which makes NH3 a classic base. The Haber process turns N2 and H2 into ammonia fertilizer that feeds roughly half the people on Earth.