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.

  • $\begingroup$ Just to confirm, the atom/group pointed up from Z is supposed to be carbon/methyl? $\endgroup$
    – jerepierre
    Aug 11, 2015 at 15:55
  • $\begingroup$ Alas, that wasn't given in the question. $\endgroup$
    – Bryston
    Aug 11, 2015 at 16:00
  • $\begingroup$ It might not matter, but just checking to see if there was a possible typo. $\endgroup$
    – jerepierre
    Aug 11, 2015 at 16:29
  • $\begingroup$ I don't really understand the question. $\endgroup$
    – Lighthart
    Aug 12, 2015 at 16:46

1 Answer 1


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.

  • $\begingroup$ 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, 2015 at 22:34

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