0
$\begingroup$

The following electronic configuration and drawing of the structure are given:

$\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (3d)^{10} (4s)^2 (4p)^4}$

enter image description here

How does this image correspond with the electronic configuration?

My train of thoughts so far:

Since the configuration has 34 electrons, I assume the element Z is actually selenium (atomic number 34). It could also be gallium since Z has no charge and has five covalent bonds to look like krypton. This is mere speculation since I just don't know how to approach the problem.

I am unsure what kind of bonds are present here. The tetrahedral form, even though it consists of a total of five bonds, indicates hybridization, but I am completely unfamiliar with hybridization outside of carbon-like hydbridization (four bonds). I am guessing one double bond leads to sp$^2$ hybridization.

$\endgroup$
  • $\begingroup$ Just to confirm, the atom/group pointed up from Z is supposed to be carbon/methyl? $\endgroup$ – jerepierre Aug 11 '15 at 15:55
  • $\begingroup$ Alas, that wasn't given in the question. $\endgroup$ – Bryston Aug 11 '15 at 16:00
  • $\begingroup$ It might not matter, but just checking to see if there was a possible typo. $\endgroup$ – jerepierre Aug 11 '15 at 16:29
  • $\begingroup$ I don't really understand the question. $\endgroup$ – Lighthart Aug 12 '15 at 16:46
4
$\begingroup$

The compound is $\ce{SeO3}$, and here is my reasoning:

From the electronic configuration given in the question, the central atom is selenium (atomic weight 34).

As you can see from its electronic configuration, the valence shell has 6 electrons and it goes on to form 4 covalent bonds with oxygen, of which 3 are $\sigma$ bonds and 1 is a $\pi$ bond. Therefore, it shares 4 electrons and has a lone pair. Its steric number is 4 and hence its hybridisation is $sp^3$.

With the tetrahedral geometry, the lone pair will occupy the uppermost position (which is shown by a dash) and the three oxygen atoms will occupy the base of the distorted tetrahedron (due to the lone pair and arising from VSEPR theory). Further, there would be a negative charge on two of the oxygen atoms.

The central atom is not gallium since its electronic configuration is $(4s)^2(4p)^1$. And krypton doesn't react with oxygen as far as I know.

$\endgroup$
  • $\begingroup$ SeO3 doesn't have lone pair on selenium, also there are two negative charges so it's an anion not oxide like OP suggests. $\endgroup$ – Mithoron Sep 13 '15 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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