There is a problem that states: Find the molecular geometry of each molecule and the hybridization of each atom in the molecule. The two molecules are HNO and HCN. I found HNO to be sp3 hybridized and HCN to be sp hybridized. However, it says that only N and O are hybridized in the first. While all three are hybridized in the second. Why is hydrogen listed as sp hybridized?

  • 1
    $\begingroup$ That is a very good question; I can only assume an error in your book. Note that hybridisation should always be discussed on a per-atom level, and note that both the $\ce{O}$ in $\ce{HNO}$ and the $\ce{N}$ in $\ce{HCN}$ can also be assigned various possible hybridisations. P.S: Welcome to chemistry.stackexchange.com. Take a tour and refer to the help center for any questions about the site. $\endgroup$ – Jan Sep 27 '15 at 0:08

Hybridization refers to the orbitals that each atom is using to make bonds. In $\ce{HNO}$, the $\ce{N}$ and $\ce{O}$ are $sp^2$ hybridized. This means they form a sigma bond with an $sp^2$ orbital from each and the pi bond from the interaction of the non-hybridized $p$ orbital on each. The $\ce{N}$ also has a $sp^2$ orbital overlap with the $1s$ of $\ce{H}$, which forms a sigma bond. Each of the lone pairs on the $\ce{N}$ and $\ce{O}$ is in an $sp^2$ orbital.

In $\ce{HCN}$ the carbon is $sp$ hybridized, as is the $\ce{N}$. $\ce{C}$ and $\ce{N}$ form a sigma bond with $sp$ overlap, and the pi bond from the interaction of both of their non-hybridized $p$ orbitals (to form a triple bond). The lone pair of $\ce{N}$ is in an $sp$ orbital, and the $\ce{C-H}$ bond is from an $sp$ overlap with the $1s$ of $\ce{H}$.

None of these molecules have $sp^3$ hybridization.

| improve this answer | |
  • $\begingroup$ Note that OP explicitly also asked why hydrogen is listed as hybridised. Care to expand on that? $\endgroup$ – Jan Sep 27 '15 at 0:13
  • $\begingroup$ Hydrogen only forms bonds with its 1s orbital. It can't hybridize to form bonds as its electron fills a 1 s orbital which is too different in energy from the 2s and 2p oritals to mix. I believe either OP misinterpreted something regarding this statement or there was an error in the book. $\endgroup$ – Mecury-197 Sep 28 '15 at 3:01
  • $\begingroup$ You should edit your answer to include that ;) $\endgroup$ – Jan Sep 28 '15 at 9:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.