Hybridized/Unhybridized orbitals

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

foxi

Full Member
7+ Year Member
Joined
Dec 17, 2015
Messages
35
Reaction score
0
I stumbled upon this problem:

"The s and p orbitals of Pb are poorly energetically matched, and do not noticeably hybridize. In Pb(II) compounds, the two remaining electrons on Pb reside in an unhybridized s orbital, and bonding utilizes only the p-orbitals. In gaseous PbCl2, what is the approximate angle between the Pb—Cl bonds?"

Answer is ~90 degrees. Can anybody explain why this is so? I figured that the Lewis Dot would look something like lead flanked by two chlorides, with a single lone pair, thus approximating a bent Nitrite type molecule with angle of ~120.

Members don't see this ad.
 
Lewis dot structures do not show 3D space very well. Not to mention that wouldn't satisfy the octet rule. Metals bond funny - at least from the perspective of what you're taught in gen chem. But all you need to answer the question is in the question stem. The s orbitals don't hybridize, so bonding only occurs with the p-orbitals. Remember that there are three p-orbitals, which are all mutually-orthogonal to each other: px, py, and pz. So if you only bond with these, then the bonds must also be orthogonal to each other - orthogonal = 90 degrees.
 
Top