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.

This Lewis structure is for nitrate, . 

The nitrogen atom is bonded to three other atoms and doesn't have any lone pair electrons, so this atom will be sp² hybridized. The shape of any sp² hybridized molecule, without lone pairs, is trigonal planar. None of the other ions are both sp² hybridized and without lone pairs.

A molecule consisting of a central atom surrounded by two bonding pair electrons and two lone pairs will always be sp³ hybridized. The molecular shape of an sp³ hybridized molecule with two lone pairs is bent/angular. Here is an example of a sp³ hybridized molecule with two lone pairs.

Note that the electronic geometry for this molecule will be tetrahedral, if the lone pairs are included in the geometry.

The basic principle of VSEPR theory is that atoms bonded to a central atom will all push each other as far away from each other as they can be. They will be pushed even harder by lone pairs. To understand what kind of shape is going to be present in a certain molecule, we need to understand how many bonding atoms there are and how many lone pairs there are on the central atom.

The central atom in this case is xenon (Xe), which we can tell primarily because it is the least electronegative. Xenon has 8 valence electrons, and fluorine has 7 valence electrons. Each of the 4 fluorine atoms needs to borrow one electron from xenon to form the covalent bond. This leaves xenon with 4 bonds and 4 leftover valence electrons, or two lone pairs.

The moelcule's steric number is 4 bonds + 2 lone pairs = 6. We want to think of a shape that has 6 equally spaced corners; usually this is called "octahedral." Think of two square-based pyramids being pressed together, base-to-base. This shape will result from any central atom with six substituents.

However, for our molecule, since two "substituents" are lone pairs and not actual atoms, they are not considered a part of the shape of the molecule, though they still affect the other four bonded atoms. To reduce strain and promote molecular symmetry, the two lone pairs are distanced from each other and the four atoms exist in a single plain, resulting in a square planar geometry. 

According to VSEPR, molecules with 4 bonding atoms and 2 lone pairs, will have the square planar shape. 

For this question, we're asked to identify an answer choice that represents a central atom with a trigonal planar shape.

When looking at each compound, the first thing we need to do is determine the central atom. Generally, this atom will have the lowest electronegativity. Then, we need to determine the number of valence electrons the central atom has. We do the same thing for all the atoms attached to the central atom, and then we draw a bond between the central atom and each connecting atom. Next, we need to check to see if the central atom has at least a full octet of eight electrons (although, elements in the third period and below can go beyond this due to their d orbitals). We also check to see if the connecting atoms have a full octet. If not, then we can either draw additional bonds between the central atom and the connecting atom(s). If the central atom cannot contribute anymore bonds, however, we'll need to add additional electrons to fill the octet of each connecting atom, which would then give us a polyatomic ion. With this set of rules in mind, let's take a look at each answer choice and evaluate the shape of its central atom.

Let's take a look at . In this compound, the central atom is nitrogen, which has five valence electrons. To fill its octet, nitrogen needs to gain three more electrons. In this compound, it does this by forming three single bonds with other atoms (in this case, three hydrogen atoms). Also, since each hydrogen atom now contains two electrons (one from itself, and the other from nitrogen), the hydrogen is stable because it will only go up to two electrons. The nitrogen is also stable. It's important to note that the nitrogen in this case also has a lone pair of electrons. As a result, ammonia will have a pyrimidal conformation.

Next, let's look at . In this case, chlorine is the central atom. After forming a single bond with each of three florine atoms, all of the atoms will be stable. The chlorine atom will have three single bonds and two different lone pairs of electrons. As a result, this compound will have a T-shaped geometry.

The central atom for  is sulfur. In order to make all the atoms happy in this compound, sulfur will form a double bond with the oxygen and a single bond to each of the chlorine atoms. As a result, the central sulfur atom will have three total atoms to which it's bound, as well as one lone electron pair. Thus, this compound will have a pyrimidal geometry.

Lastly, let's look at . To stabilize all the atoms, the central carbon atom will have a double bond with the oxygen and two single bonds, one to each methyl group. Consequently, this molecule's central carbon atom is bound to a total of three atoms, and it has no lone electron pairs. As a result, this compound will adopt a trigonal planar geometry.

has six bonds and no lone pairs, so it has an octahedral geometry and its bonds are hybridized.