Homework Help

Explain why hydrogen bonds do no operate in H2S, H2Se and H2Te and why the melting...

user profile pic

lilian-0716 | Student | Salutatorian

Posted July 20, 2013 at 10:45 AM via web

dislike 1 like

Explain why hydrogen bonds do no operate in H2S, H2Se and H2Te and why the melting points and heats of fusion of ammonia, water and hydrogen fluoride are much higher than those of methane?

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

1 Answer | Add Yours

user profile pic

mvcdc | Student, Undergraduate | (Level 1) Associate Educator

Posted July 20, 2013 at 1:42 PM (Answer #1)

dislike 1 like

Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative species (F, O, or N). Electronegativity is the tendency of an atom to attract electrons from a chemical bond towards its nucleus. In covalent bonds, the difference in electronegativity (pull of electrons towards an atom) results to polar covalent bonds. When the difference in EN (with H) is very large (as is the case for F, O, or N), hydrogen bonding occurs. Hydrogen will have a partially positive charge, and the electronegative species, a partially negative charge. Since the difference in electronegativities of F, O, or N with H is very large, H almost has a full positive charge (and the electronegative species almost a full negative charge). Then, the 'positive' H is attracted to the 'negative' species -- causing what is called hydrogen bond.

Let's look at the EN of the species involved (first column is name, second the atom's EN, and third, the difference with EN of H):

H     2.20     -

F     3.98     1.78

O     3.44     1.24

N     3.24     0.94

S     2.58     0.38

Se   2.55     0.35

Te   2.10    -0.10

As you can see, the EN difference between S, Se, or Te and H is very small (H is even more electronegative than Te), thus the given compounds are not capable of hydrogen bonding (though are still covalently bonded to H -- and polar).

Hydrogen bonding also explains some properties of certain compounds such as melting points and heats of fusion (energy required to change from solid to liquid).

In the transition from solid to liquid or liquid to gas, a main contributor to the energy requirement are the so-called intermolecular forces. The stronger the intermolecular forces present between molecules of a given compound, the higher more difficult it will be to convert a solid to liquid, or a liquid to gas (since strong IMF's mean that the 'bonds' or interactions between adjacent molecules is very strong), and the higher the energy (or temperature) needed to enact these changes.

If we are to compare the melting points and heat of fusion of ammonia (`NH_3` ), water (`H_2O` ), hydrogen fluoride (`HF` ), and methane (`CH_4` ), we only need to examine the forces present. We've already mentioned above that when H is bonded to either N, O, or F, we have a strong hydrogen bonding. Now, let's look at the polarity of methane.

Methane is tetrahedral -- consisting of 4 hydrogens bonded to a single carbon. While each C-H bond is polar, since all 4 bonds are identical, their polarities cancel out, leading to methane being a non-polar substance. As opposed to the strong dipole-dipole interactions of the partial charges in a hydrogen bond, methane is only capable of van der Waals (or London Dispersion) forces, which are caused by instantaneous distortion of an electron cloud -- these are the weakest kind of force. 

This explains why methane will have the lowest melting point and heat of fusion among the four compounds.

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes