Generally speaking (outside of water) the more polar the bonds the harder it is to break the bonds (inside water, the logic is opposite to this and the polar bonds are easier to break than the non-polar bonds due to the ability of polar bonds to interact with water, which is where a lot of the confusion about bond strength arises).
In covalent compounds, the more the partial ionic character the stronger the bonds. For example, H
2 and Cl
2 are non polar covalent compounds . When they react (again NO water in this comparison, this is in the gas phase) to form 2HCl, the product HCl is about 80% covalent in character and 20% ionic in character. So this extra percent ionic character means the bonds in the product are stronger than the reactants by about 44 kcal/mol for the above reaction.
On the other hand, if you react H
2 and I
2 to form 2HI, the HI bond has very little partical ionic character (since the electronegativity of hydrogen and iodine are so close to one another) and as a result the products are only slighlty more polar than the reactants and far less energy is liberated in forming the products (4 kcal/mol for the above reaction). Since more energy is liberated forming HCl, you can see that the bonds in HCl are stronger than those in HI (again, outside of water) so HI < HCl.
You don't have to take my word for it, here is the explanation from the great Linus Pauling (I took some numbers from this lecture in my post, but its better to watch the real thing:)
http://osulibrary.oregonstate.edu/specialcollections/coll/pauling/bond/video/1957v.1-17.htmlNow if you are comparing the bond energies between compounds of a similar bond type, typically smaller "harder" molecules have stronger
primary bonds with one another than larger "soft" molecules. For example, I
2 has a bond energy of 151 kj/mol while Cl
2 has a bond energy of 242 kj/mol. These are both non polar covalent compounds but the chlorine bond is closer to the nucleus of each of the two chlorine atoms than the iodine bond is and this makes the chlorine bond tighter and harder to break. In general this holds. Fluorine just happens to be a weirdo because it is so darn small. In making the bond for F
2, the electrons are forced into so small a place that they tend to repel each other weakening the bond. You see this behavior quantitatively by looking at the electron affinity for the various halogens (see:
http://www.chemguide.co.uk/atoms/properties/eas.html) where Fluorine clearly falls out of line with the other three halogens (this weirdness carries over into the covalent bond strengths).
In general though, I
2<Br
2< Cl
2 when it comes to
primary bonding. With
secondary bonding though the order of bond strength is again opposite! And I
2 forms a solid at standard conditions, Br
2 a liquid, and Cl
2 a gas due to secondary bonding interactions such as the london forces.
As to the final question, generally bonds between two s orbitals are stronger than the bond between two p orbitals due to the "degree of overlap" between the two types of orbitals (see your text for examples of this, it is hard to draw out online but the p bonds have farther to "reach" to touch each other in comparison to the two s orbitals). Taken that fact, if you were asked which bonds were stronger between an sp2 bond and an sp3 bond you would answer sp2. The reason is that the sp2 bond has 33% s character whereas the sp3 bond has 25% s character. The more s character, the more overlap, and the stronger the bond. You apply this third concept mostly in organic chemistry when predicting bond strengths for carbon compounds.