# length of bond, bond angle and p character in bond

My reference data: http://www.rsc.org/suppdata/ra/c4/c4ra06781a/c4ra06781a1.pdf

Consider these five chemicals:

• A: $\ce{CH4}$ (methane)
• B: $\ce{CH3Cl}$ (chloromethane)
• C: $\ce{CH2Cl2}$ (dichloromethane)
• D: $\ce{CHCl3}$ (trichloromethane/chloroform)
• E: $\ce{CCl4}$ (tetrachloromethane/carbon tetrachloride)

From what I know, an orbital is longer and has higher energy as its p character increases.

From A to E, both $\ce{C-H}$ bond and $\ce{C-Cl}$ bonds get shorter, and $\ce{Cl-C-Cl}$ bond angle in D is smaller than that in C.

Nothing makes sense, first one implies more p character in $\ce{C-Cl}$ bond i.e. D < C < B and second order implies the opposite. I assumed equal x for every C's orbital taking part in $\ce{C-Cl}$ bond by symmetry.

The angle between two orbitals with $sp^x$ hybridisation is $\cos^{-1} \left( - \frac 1 x \right)$.

Also, from A to E, the electronegativity of central atom C increases, so C-Cl bond should become longer as the difference between EN decreases.

So my question is the p character and hybridization logic only working for 2nd-period elements exclusively?

Maybe relevant: It is a widespread opinion that bond energy in sigma bond p-p>s-p>s-s but sigma bond energies of $\mathrm{sp>sp^2>sp^3}$. The rationalization for the first order is more overlapping is possible due to directional and long nature of p orbital and rationalization for 2nd order is that energy of p is more than s, so the order of energies of molecular orbitals is more for more p character.

EDIT: In compounds containing $\ce{H,\ C,\ N,\ O,\ F}$ only I can predict bond lengths and bond angles almost every time using relative properties of hybrid orbitals and their relation with p character in a respective bond otherwise known as Bent's rule. But, once $\ce{Cl}$ enters the picture, there is no simple trend in any scenario, bond angle could be rationalized due to steric reasons but the cos and x relation should be valid if it follows simple hybridisation which should allow me to compare bond lengths by comparing lengths of participating orbitals (more p in carbon's hybrid orbital towards $\ce{Cl}$, longer orbital, and $\ce{Cl}$ has 3p in every case), but it seems it doesn't follow simple hybridisation. As an example, as bond angle increases, bond length must decrease, which is not the case for $\ce{CCl3}$, $\ce{CH2Cl2}$ etc

• The electronegativity of any element is constant. – DHMO Jan 21 '17 at 8:04
• Maybe you need to note that the active orbital in chlorine is 3p. Comparing 3p with 1s isn't exactly fair. – DHMO Jan 21 '17 at 8:11
• Well, I did not compare 3p with 3s for bond lengths I compared C-Cl bonds in different compounds with different atoms attached to C. Also, Electronegativity of (CHCl2)-X>(CH2Cl)-X, more Cl attached to C make the C slightly positive and increase "electronegativity". I suppose I'm gonna have to scrap every hybridisation concept for every element other than 2nd period, attaching any number of Cl breaks away from every trend. – Mrigank Jan 21 '17 at 8:38
• Your question is a bit confusing, maybe you could clear it up, write down the numbers you mean and why you think they are wrong and so on. What do you mean by first and second order? If I understand it correctly, you want to argue with this hybridization concept on why there are marginal differences in bond lengths? I'm starting to think they should stop teaching that anywhere. If you want to stay in the primitive world of hybridization, you should look at calculations that only consider s and p-orbitals. The basis sets used in your reference have d, f and g-orbitals in them. – AMT Jan 21 '17 at 9:17
• You sure ask a lot of hybridization questions. What exactly is it that you want to do? Is this for a publication? Or just for your personal amusement? You might be ready to leave those concepts behind that as you noticed really only work for the handful of examples they were made for and learn some actual quantum chemistry – AMT Jan 21 '17 at 10:50