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### Homework 2D #15

Posted: Fri Nov 01, 2019 1:28 pm
If we are given a list of molecules, such as CF4, CCl4, and CBr4, how can we tell which molecule is predicted to have the strongest CX bond, where X is a halogen? (reference: problem 2D15)

### Re: Homework 2D #15

Posted: Fri Nov 01, 2019 2:54 pm
I think that the molecule with the strongest bonds would be the one where the C and X atoms have the greatest difference in electronegativity.

### Re: Homework 2D #15

Posted: Sat Nov 02, 2019 4:41 pm
This is referencing the concept of bond length that Dr. Lavelle explained during Friday's lecture using the example of the strong acids: HF > HCl > HB > HI. Bond strength decreases as atomic radius of increases because the atoms are farther apart and easier to pull apart. Thus, among CF4, CCl4, CBr4, CF4 would be the strongest because F is the smallest atom and can have a shorter bond length than the others.

### Re: Homework 2D #15

Posted: Sun Nov 03, 2019 2:29 pm
In order to determine the order of weakest bonds it is important to identify which are the larger atoms (ionic radius=decrease left to right and increase top to bottom) because the larger atoms have a larger distance therefore longer bond length which is weaker and easier to break.

### Re: Homework 2D #15

Posted: Sun Nov 03, 2019 6:09 pm
In order to answer this question, we need to look at atomic trends. F, Cl, and Br are all in the same group, but different periods. As you move down a group, atomic radius increases. Why is this important? Increasing atomic radius means increasing bond length. Longer bonds typically require less energy to break. So, we can infer that CF4 will have the strongest CX bond because F has the smallest atomic radius.

### Re: Homework 2D #15

Posted: Sun Nov 03, 2019 10:38 pm
This question is referencing the atomic trend of greater radius as atoms get bigger so they have weaker bonds as the electrons are easier to "push out"