Explain the geometry and bond angles in a NH3 molecule

So if we think about a molecule of ammonia, NH3, there are 3 hydrogens, each with 1 electron (as its atomic number is 1) covalently bonded to 1 nitrogen atom with 5 outer shell electrons on it due to its atomic number of 7, but having 2 electrons in its innermost shell. As hydrogen can only ever have 1 bond to/from it, we can say that nitrogen is the central atom with hydrogens around it. As covalent bonds involve the sharing of electrons, and there are 3 covalent bonds, nitrogen shares 3 of its 5 outer electrons, 1 each with the 3 hydrogens, and so there are 3 bonding pairs. 5-3=2, so that leaves 1 lone pair of electrons around this centrral nitrogen atom. The valence shell electron pair repulsion (VSEPR) theory states that electron pairs around a central atom will repel so they are as far away as possible. As there are 4 electron pairs around the central nitrogen, you can imagine each spreading out to the 4 corners of a tetrahedron, which would suggest that the bond angle is 109.5 degrees. However, as only 3 of the electron pairs are bonding, there isn't a bond at the 'top' to form a tetrahedron, and the actual shape of NH3 is trigonal pyrimidal. VSEPR theory also states that the lone pair of electrons repels bonding pairs more than bonding-bonding pair repulsion, and so you can imagine the 1 lone pair squeezing the bonding pairs slightly closer together in NH3, which reduces the bond angle by about 2 degrees, making the bond angle 107.5 degrees.