-1
$\begingroup$

How is the above mentioned compound stable enough to occur naturally? Also, I've read that it's a covalent compound with a

Trigonal Bipyramidal

Structure. The Arsenic atom is $sp^3d$ hybridized. How does that work out? I don't see how it can attain such a hybridization state.

$\endgroup$

closed as off-topic by Mithoron, Todd Minehardt, tschoppi, Tyberius, Jon Custer Jan 29 at 20:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

One way to visualize this is to consider the As to have three $sp^2$ hybrid orbitals (using $p_x$ and $p_y$) that are involved in bonding to the equatorial ligands and two $pd$ hybrid orbitals (using $p_z$ and $d_{z^2}$) involved in bonding to the axial ligands. Overall, the As is $sp^3d$ hybridized in that 5 orbitals are being used, but that doesn't mean that you necessarily have five equivalent $sp^3d$ orbitals. The hybridization can be different because the axial and equatorial bonds are not equivalent.

The $sp^2$ hybridization of the equatorial ligand bonds gives the trigonal planar shape with 120 degree angles that we expect from $sp^2$ hybridization, and the $p_z$ and $d_{z^2}$ orbitals are both oriented along the $z$ axis to bond well with the axial ligands directly above and below the As, perpendicular to the plane of the equatorial ligands.

[Added clarification: Note that this is a deliberate oversimplification to help visualize. In reality, the s orbital will also be involved in bonding to the axial ligands and the $d_{z^2}$ will also interact with the equatorial ligands, so the hybridization will not be as distinct as I described.]

$\endgroup$
  • $\begingroup$ Thank you @Andrew, but I still have a bit of a confusion - If one looks at the electronic config of As - $[Ar] 3d^{10}4s^24p^3$, one can see that the $p$-orbitals are half filled. But how does this allow for the formation of $sp^3d$ orbital here? $\endgroup$ – Apekshik Panigrahi Jan 27 at 5:37
  • $\begingroup$ It might help to imagine first moving a $4s$ electron to the $4d_{z^2}$ orbital so that each of the As orbitals involved in bonding has one electron to start, even though it's not actually in that excited state. This is called "promotion and hybridization". Another way to think of it is to start with $\ce{As^5+}$ (with all of the $4s$, $4p$ and $4d$ orbitals empty) and five $\ce{F-}$ ions that can each put an electron pair into an empty hybrid bonding orbital. Does that make sense? $\endgroup$ – Andrew Jan 27 at 12:18
  • $\begingroup$ It doesn't allow and your mixing of p and d orbitals is obsolete and even deprecated. $\endgroup$ – Mithoron Jan 27 at 18:48
  • $\begingroup$ @mithoron - If this were PF5, I would agree that using d orbitals in bonding is an outdated model. However, for fourth row elements like As, this model is still commonly taught even if the participation of d orbitals is still questionable, and given that the OP asked specifically about sp3d hybridization, it seemed appropriate to work within that model. It provides a direct connection to the observed structure that is much less clear in a more complex model involving ionic interactions. $\endgroup$ – Andrew Jan 27 at 22:52
  • 1
    $\begingroup$ In $\ce{PF5}$ and to a large extent in $\ce{AsF5}$, the $d$ orbital is high enough in energy that involving it in bonding does not give a benefit. The "extra" electrons instead reside completely on F atoms. The simple picture is for the axial F- ligands to both have approximately 1/2-strength bonds to the $p_z$ orbital, with added strength coming from ionic interaction between the negative charge on F and positive charge on As. This is called a "3-center 4-electron bond". The F atomic orbitals are low enough in energy that the equatorial bonds also have a lot of ionic character. $\endgroup$ – Andrew Jan 28 at 11:29

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