why CH4 is more electronegative than O?

Answer 1

C and H form polar covalent bonds because of the different electronegativities (2.55 for C and 2.20 for H). However, the bonds are very weakly polar since the electronegativities are relatively close to each other. More important, the structure of the CH4 (methane) molecule is symmetric; that is, each H atom is bonded exactly 109.5 degrees from the others. The polarities of the C-H bonds cancel out each other; so overall, methane is not a polar molecule.

There is no bond in a single atom (O), so I'm assuming you mean O2. O2 forms a non-polar covelent bond (the electrons are shared equally) because the electronegativities of the two atoms are (of course) equal.

Answer 2

Uh, I guess it has something to do with the fact that oxygen has properties in its valence structure that wants to attract electronic charge. But as O2 gas, it's paired in a double-bond with an equally greedy, grasping oxygen atom.

In CH4, however, the carbon atom is not as electronegative as oxygen but it's surrounded by 4 hydrogen atoms who readily give up their charge. Hydrogen ionizes to H+ readily. In chemical bonds it will not ionize, but the electrons are more likely to drift on the side of the carbon than the oxygen. Carbon has a richer electron environment because of the bonds so is more electronegative than the O of an O2 molecule.

Answer 3

You've made a small mistake in the question. A compound doesn't have an electronegativity, only an element. C is less electronegative than O because it is a bigger atom, and has fewer protons in its nucleus.