First off, I'm going to apologize for how much what you're about to read may or may not confuse you.
Going to -try- to keep this relatively simple, since I don't know how much chem ppl have had, and I want to avoid causing too many headaches here hehe. Also keep in mind i'm a student, not a teacher, and I'm not entirely sure if i'm 100% correct on this. Furthermore, if chemistry really isn't your thing, I would advise not reading below this line. Oh, and it's 3am as I type this, so screw grammar and spelling
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The strength of the bond/release of energy depends on the atoms that are bonded within the molecule. Organic chem is mostly carbon-based, So in many cases the bond that is broken is a bond between a carbon and a hydrogen or halogen. or other nonmetal.
As for what nucleus the nucelophile is attracted to, it's not actually attracted to the nucleus itself, so it's a bit of a misnomer in my opinion. Rather, the compound is attracted to positive charges or partial charges in polar molecules. The incoming nucleophile has to be able to "see" the nucleus of the central atom/carbon (or chiral center) on the molecule it's hitting. This is why steric hinderance is an issue, because large/bulky groups tend to "hide" the nucleus from the incoming nucleophile. There was a nice video we had in class on it, but it can only be accessed from the campus labs :/ Tried to make some acii art of it, but the spacing got messed up so all you're getting is text heh
There are other factors involved as well, such as the types of solvent used, how good the release group is, polarity of the target molecule, how strong the nucleophillic attraction is...it's really a lot to think about.
Given CH3R in a tetrahedral geometry: where R is a release group:
1. An incoming nucleophile is attracted to the positive side of the molecule. However, it will attempt to bond with the central carbon.
2. As Nucleophile gets closer, the Hydrogens bonded to the carbon will actually be repelled.
3. Hydrogens are forced into a planar configuration, and there is partial bond making with the Nucleo - Carbon and partial bond breaking with the Carbon - Release group. The R group leaves, and the H's will continue moving until they re-establish a tetrahedral configuration.
4. The R group is moved away, and the N is bonded to the Carbon now. This type of reaction happens from the opposite side of the Release group, so it's called a backside attack. This basically -inverts- the structure, so if there was a Cl and an OH group instead of 2 of those H's, the structure inverts, and instead of being left handed, it becomes the opposite.
The problem with steric factors is this:
Say instead of hydrogens, a carbon is surrounded by 3 CH3 groups, and an R group. The bulky CH3 groups attached make it harder for the Nucleophile to "see" that central carbon, so the reaction will slow, or in extreme cases, just not happen at all.
This is about as basic as I can get for what I was trying to talk about.
I'm probably sounding like gibberish
Some of this stuff is hard for me to choke down, so I can't imagine what it's like for anyone who hasn't completed the college general chem sequence
I tried to keep things kinda simple though. I'm not exactly the best person to ask about this kidna thing since i'm just a student heh
i'm tired now and sleep calls -.-
Also got a little productive with some music and cleaned up something I was working on over the weekend hehe 
. . . I need a drink . . .
