Start by drawing the Lewis structure of $\displaystyle \text{NO}^-_2$. $\displaystyle \text{NO}^-_2$ has $\displaystyle 18$ valence electrons.

Since the nitrogen has $\displaystyle 2$ oxygen atoms bonded to it and a lone pair, the steric number must be $\displaystyle 3$. From that, we know that it must have sp² hybridization.

Look at each of the carbon atoms in the chain. Each carbon atom has $\displaystyle 3$ other atoms bonded to it. This means that the steric number of each carbon is $\displaystyle 3$. Thus, the only type of hybridization found in this molecule is sp².

Start by counting the number of atoms the nitrogen is bonded to. It has $\displaystyle 3$ bonded atoms. The nitrogen also has a lone pair. This means the steric number of the nitrogen is $\displaystyle 4$. Thus, it must have $\displaystyle sp^3$ hybridization.

Notice that there are $\displaystyle 4$ groups bonded to the sulfur. The groups are the double bond to the oxygen, the lone pair, and the bonds to the two separate carbons.

Since there are $\displaystyle 4$ groups on the sulfur, it must have $\displaystyle sp^3$ hybridization.

Start by drawing the Lewis structure.

Notice that selenium has $\displaystyle 5$ groups around it: 4 bromines and 1 lone pair.

Molecules with $\displaystyle 5$ electron groups will exhibit $\displaystyle sp^3d$ hybridization.

For this question, we're asked to identify which kind of hydridization, if any, a carbonyl carbon of a ketone group will have.

First, let's recall that a ketone group is a functional group in which a carbon atom contains a double bond to an oxygen atom. In addition, the carbon also has a single bond to each of two other atoms, neither of which are hydrogen.

Let's also recall that a double bond consists of one sigma bond and one pi bond. Each of the two single bonds that carbon has to the other atoms will also be sigma bonds, because these are both single bonds. Thus, in total, this carbon atom will have three sigma bonds and one pi bond.

In order to have a pi bond, there must be p orbital overlap. Thus, one out of the three p orbitals from the carbon will be used towards contributing to the pi bond with the oxygen. Consequently, we have one s orbital and two p orbitals left over. These three orbitals can hybridize to become three $\displaystyle sp^{2}$ orbitals. Each of these $\displaystyle sp^{2}$ orbitals will take place in the three sigma bonds that the carbon atom has.

Count the number of groups that the sulfur has attached to it. It is attached to $\displaystyle \text{NH}_2$, two oxygen atoms, an done carbon. Since there are four groups attached to the sulfur, it must have $\displaystyle sp^3$ hybridization.

Start by recalling that nitrous acid is $\displaystyle \text{HNO}_2$.

Next, draw its Lewis Structure:

Notice that the nitrogen is bonded to two oxygen atoms. The nitrogen atom also has a lone pair. Since there are $\displaystyle 3$ groups attached to the nitrogen, it must have $\displaystyle sp^2$ hybridization.

In ethylene, the two central Carbon atoms are bonded together with a double covalent bond. In order to form a double bond, Carbon always takes on $\displaystyle sp^{2}$ hybridization. The double bond consists of a sigma bond caused by the overlapping of a pair of $\displaystyle sp^{2}$ hybridized orbitals and a pi bond caused by the overlapping of neighboring unhybridized $\displaystyle p$ orbitals. Further, $\displaystyle sp^{2}$  hybridization always results in hybrid orbitals that are oriented in a trigonal planar manner, with a separation angle of 120° for maximum electron separation between the three sigma-bonding $\displaystyle sp^{2}$ orbitals. Thus, all other answers are incorrect.

In HCN, Carbon is bonded to Nitrogen with a triple covalent bond consisting of one sigma bond and two pi bonds. The sigma bond is formed by overlapping $\displaystyle sp$ hybridized orbitals, with the two remaining unhybridized $\displaystyle p$ orbitals overlapping to form the two pi bonds.