The carbon in the molecule is  hybridized and contains no lone pairs attached to it. This will give the molecule a tetrahedral geometry, standard of a central atom with four single bonds. Bond angles in a tetrahedral molecule are all .

has two bonding regions and no lone pairs around the central atom, so it has a linear geometry, so its bond angle is .

has four bonding regions and no lone pairs around the central atom, so it has a tetrahedral geometry with bond angles of .

has six bonding regions and no lone pairs around the central atom, so it has an octahedral geometry with bond angles of and . For the sake of ranking, we will only consider the smaller angle.

has three bonding regions and one lone pair around the central atom, so it has a trigonal pyramidal geometry with bond angles of .

has three bonding regions and no lone pairs around the central atom, so it has a trigonal planar geometry with bond angles of .

has four bonding regions and no lone pairs, so it has a tetrahedral geometry. has two bonding regions, but it also has two lone pairs, so it has a bent geometry.

According to hybrid orbital bonding theories, the hybridization state will use a number of orbitals equal to the steric number. The steric number is the number of atoms bonding to the central atom, plus any additional lone pairs.

Nitrogen (N) is the central atom because hydrogen atoms are never central. 

Nitrogen has 5 valence electrons, and in ,  it is bonded with 3 hydrogen atoms, which each borrow one of nitrogen's electrons to create a covalent bond. Subtracting those 3 borrowed electrons from its original 5 leaves 2 left, or 1 lone pair.

In total, the nitrogen atom has 3 bonds and 1 lone pair for a steric number of 4. As a result, its hybridization state must use 4 orbitals. The lowest energy way to do that is to use the lowest energy orbitals that are available. In this case, those are the s and p orbitals. So it uses one s and three p orbitals, giving the sp³ hybridization state. 

It is important to note that electronic geometry includes the orientation fo lone pairs, while molecular geometry considers only the geometry of the atoms present in the compound.

Methane, or CH₄, has four hydrogen atoms bound to a central carbon, resulting in a tetrahedral geometry.

BF₃ has three fluorine atoms bound to a central boron, resulting in a trigonal planar geometry.

Be(Cl)₂ has two chlorine atoms bound to a central beryllium, resulting in a linear geometry.

Water, or H₂O, has two hydrogen atoms bound to a central oxygen atom with two lone pairs. The molecular geometry is bent, but the electronic geometry is tetrahedral.

Ammonia, or NH₃, has three hydrogens bound to a central nitrogen atom with one lone pair. The moelcular geometry is trigonal pyramidal, but the electronic geometry is tetrahedral.

Ammonia, water, and metahne all have the same electronic geometries (tetrahedral), giving us our final answer.

SF₄ has 6 electron domains coming off of it- 4 F molecules and 2 lone pairs of e^–. This is an example of see-saw shape.

There are three bonding electron pairs around the  central atom and one non-bonding pair, for a total of 4 electron groups. They arrange spontaneously to be furthest apart according to VSEPR theory, which corresponds to when they form a trigonal pyramidal shape.

Sulfur dioxide has the formula  and takes on a trigonal planar electronic geometry, with the two oxygen atoms and the lone pair in the same plane. The molecular geometry will be bent, resulting in a oxygen-sulfur-oxygen bond angle of 120 degrees.

This is the Lewis structure of . 

Since the tin atom (Sn) is bonded to two atoms and has one lone pair, the molecule is sp² hybridized and has a bent, or angular shape. Bent molecules have bond angles of approximately 120°.

This is the Lewis structure of .

 has three bound atoms and one lone pair, so it is sp³ hybridized and trigonal pyramidal. Note that the electronic geometry (geometry including the lone pair) is tetrahedral, but the molecular geometry (excluding the lone pair) is trigonal pyramidal.