# Hybridized orbital scheme for H2O

by newton2008
Tags: hybridized, orbital, scheme
 P: 134 Orbital outlay on O: $$1s^2 2s^2 2p^6$$ Unhybradized bonding would have a sigma bond between the s orbital of hydrogen and a parallel p orbital, the next bond would also be between an s orbital of hydrogen and another, parallel yet available p orbital (lets say pz for one py for the other). This would result in the lone pairs being in the 2s and 2px orbitals, and give a funny othogonal geomety. Hybradization of the orbitals to $$sp^3$$ would darastically decrease the repulsion the previous model produces. so the hydrogens would bond to the two avaliable sp3 orbitals. As for the molecular orbital diagram, the lowest energy bonding interaction is a sigma interaction between hydrogen and the afore mentioned hybridization. So you will have two low energy bonding interactions, and two higher energy lone pairs that are not involved in bonding. This is my guess anyway, when I do molecular orbital diagrams recently it is usually after doing a whole load of group theory, and we generally dont take lone pairs into consideration in lectures (yet).
 P: 111 Here is a thought about hybrid orbitals to get you started. You could construct hybrid orbitals h1 and h2 from an s-orbital, s, and two p-type orbitals p1 and p1. Choose p1 to be px and p2 to be $$p_2 = p_x \cos{\theta} + p_y \sin{\theta}$$, where $$\theta = 104.5^o$$ is the bond angle and the px and py are orthonormal. The hybrid orbitals are then $$h_1 = N(s + a p_1)$$ $$h_2 = N(s + a p_2)$$. The requirement that h1 and h2 be orthogonal leads to the condition $$a^2 = -1/\cos{\theta}$$. N is obtained by normalising. If desired, a third hybrid could be constructed with a third p-type orbital, p3, to accommodate a lone pair, leaving the other lone pair in the pz orbital. Or a set of four orbitals could be constructed by including pz in the linear combination defining the hybrids. Perhaps this will give you a start in the right direction.