Covalent bonds are formed when two nonmetals are bonded together. This covalent bond means that the electrons are shared by the two atoms in order to satisfy each atom's octet. There is very little difference in the electronegativities of the two atoms involved in the bond, so neither atom pulls the electrons closer to its nucleus.

Ionic bonds are formed between a metal and a nonmetal. Due to the dramatic difference between the electronegativities of metals and nonmetals, the electrons are pulled tightly to the nonmetal, and away from the metal nucleus. This results in each atom having a full octet, even though the electrons are not shared.

Carbon and oxygen are both nonmetals, so we would expect only covalent bonds in carbon dioxide.

In order to determine the charge of the transition metal manganese in the molecule, $\displaystyle MnF_3$, we must first determine the net charge of the molecules it is bonded to. The manganese is bonded to three fluoride ions. Fluoride ion carries a negative one charge. since it is a halogen The subscript in front of the fluoride in the molecule tells us that we have three fluoride ions. Each fluoride ion carries a $\displaystyle -1$ charge and because we have 3 of them, there is a total charge of $\displaystyle -3$ from these fluoride ions. Molecules like to exist in their most stable state which gives them an overall charge of zero. Therefore, the manganese atom will carry a charge of $\displaystyle +3$ to counter the $\displaystyle -3$ charge from the three fluoride ions. 

An ionic bond is a bond that occurs between a metal and a nonmetal, which may dissociate into two ions of opposite charges (positive and negative). These bonds are formed by electrostatic forces. We can determine if a compound is ionic from what it is composed of. Ionic compounds can easily be identified because they are generally composed of a metal and non-metal such as $\displaystyle NaCl$. Metals are cations and non-metals are anions. In the options given, the substance that contains an ionic bond is $\displaystyle AgNO_{3}$. The ions involved are $\displaystyle Ag^{+}$ and $\displaystyle NO_{3}^{-}$. 

In order to determine the charge of the $\displaystyle Pb$ ion in the molecule, $\displaystyle PbSO_{4}$, we must first determine the net charge of the molecules it is bonded to. The $\displaystyle Pb$ ion is bonded to a $\displaystyle SO_{4}$ ions.

$\displaystyle SO_{4}$ ion is one a common anion and carries a -2 charge. Molecules like to exist in their most stable state which gives them an overall charge of zero. $\displaystyle PbSO_{4}$ has an overall charge of zero. Therefore, the $\displaystyle Pb$ atom will carry a charge of +2 to counter the -2 charge from the $\displaystyle SO_{4}$ ion.

In order to determine the charge of the barium ion in the molecule, $\displaystyle BaF_{2}$, we must first determine the net charge of the molecules it is bonded to. The barium ion is bonded to two fluoride ions.

Fluoride ion carries a -1 charge. The subscript in front of the fluoride in the molecule tells us that we have two fluoride ions. Each fluoride ion carries a -1 charge and because we have 2 of them, there is a total charge of -2 from these fluoride ions. Molecules like to exist in their most stable state which gives them an overall charge of zero. Therefore, the barium atom will carry a charge of +2 to counter the -2 charge from the two fluoride ions.

The degree of unsaturation is equal to the number of rings plus the number of pi bonds in a molecule. Ephedrine has one ring and three pi bonds, so its degree of unsaturation is four.

To arrive at this answer, one could also use the formula below, where $\displaystyle C$ is the number of carbon atoms, $\displaystyle H$ is the number of hydrogen atoms, $\displaystyle X$ is the number of halogen atoms, and $\displaystyle N$ is the number of nitrogen atoms.

$\displaystyle unsaturation = \frac{2C+2 - H - X + N}{2}$

 For ephedrine, $\displaystyle C = 10$, $\displaystyle H = 15$, $\displaystyle X = 0$, and $\displaystyle N = 1$.

$\displaystyle \frac{2C+2 - H - X + N}{2}=\frac{2(10)+2 - (15) - (0) + (1)}{2}=\frac{8}{2}=4$

2-butyne has the following chemical structure.

$\displaystyle H_{3}C-C\equiv C-CH_{3}$

The end carbons have $\displaystyle sp^3$ hybridization (form single bonds only), while the middle two carbons have $\displaystyle sp$ hybridization (involved in a triple bond).  There are no $\displaystyle sp^2$ hybridized carbons in this molecule.  

Carbon A is $\displaystyle \small sp$ hybridized because this carbon is only bonded to two other atoms. Carbon B is bonded to four atoms, and would therefore be $\displaystyle \small sp^3$ hybridized.

Keep in mind that a carbon involved in a triple bond will always be $\displaystyle \small sp$ hybridized, a carbon involved in a double bond will be $\displaystyle \small sp^2$ hybridized, and a carbon involved only in single bonds will be $\displaystyle \small sp^3$ hybridized.

The incorrect statement is a property of ionic compounds rather than covalent. Recall that electronegativity is a measure of the ability of an atom to draw electrons to itself. Ionic compounds are formed by elements with very different electronegativities, since elements with different electronegativities will tend to form positive and negative ions (that is, they give away or gain electrons easily). In contrast, covalent bonds are formed by elements which are close in electronegativity and could exist as stable free molecules. All other statements are true of covalent compounds.

$\displaystyle \small KCl$ is an ionic compound, while $\displaystyle \small N_2$ is a nonpolar covalent compound. Remember that polarity results from a difference in the electronegativities of the atoms involved in the bond. Too great of a difference will result in an ionic bond; two of the same atoms will have zero difference, resulting in a nonpolar bond.

$\displaystyle \small SO_3$, $\displaystyle \small P_2O_5$, and $\displaystyle \small CCl_4$ contain polar covalent bonds. In the first two, oxygen will carry a slight negative charge, leaving sulfate and phosphorus with slight positive charges. In $\displaystyle \small CCl_4$, chlorine will carry a slight negative charge, leaving the carbon slightly positive. In $\displaystyle \small CCl_4$ and $\displaystyle \small SO_3$, symmetry helps to balance the polar bonds, resulting in an overall nonpolar molecule, even though the individual bonds are nonpolar.