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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?


marked as duplicate by Todd Minehardt, Jan, M.A.R., ringo, Wildcat Nov 24 '16 at 9:23

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    $\begingroup$ 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.

  • $\begingroup$ 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

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