Which of the following orbitals of a diatomic molecule AB will not have positive overlap?

1) 2S(A) and 2S(B) 2) 2S(A) and 2Px(A) 3) 2Px(A) and 2Px(B) 3) 2S(A) and 2Pz(B)

Dear Bala,

Note that we''re not taking energy differences into account. In cases where the difference in orbital energy is greater than 15eV you get negligible overlap. Now onto the symmetry-allowed overlaps.

1s and 2s are both spherical and overlap to form a sigma bond.

1s and 2p(x) form a sigma bond. You can tell because if you draw the 2p(x) lobe you''ll see it overlap with the corresponding 1s.

The 2p(x) and the 2p(y) do not overlap. There is no possible bond. Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge). The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out. No net overlap.

3p(y)-3p(y) form a pi bond. Once again, draw these out and see it for yourself. Draw the 3p(y) orbitals as rather fat so you can see how they overlap. Since they are not cylindrically symmetric they are not a sigma bond. Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there''d be less overlap. For the sigma bonds above, rotation does not affect the overlap.

2p(x)-2p(x) have positive overlap. Draw them and see why. They point toward one another. Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis. Thus, you can tell that this gives a sigma bond.

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