2
$\begingroup$

My text says it to be 4. I believe it's 2. Is there any compound where covalency of 4 is observed?

$\endgroup$
5
  • $\begingroup$ Guess they mean coordination compounds. $\endgroup$ Mar 21 '17 at 12:12
  • $\begingroup$ Can you give me an example? $\endgroup$
    – AdiC
    Mar 21 '17 at 12:25
  • 3
    $\begingroup$ Basic beryllium acetate will do. $\endgroup$ Mar 21 '17 at 12:34
  • 2
    $\begingroup$ Yeah basic beryllium acetate has four coordinate O and can be thought of as significantly covalent due do the polarisability of Be. There are also numerous three coordinate species containing oxonium ions eg Me3O+ $\endgroup$
    – RobChem
    Mar 21 '17 at 12:47
  • $\begingroup$ $\ce{H4O^{2+}}$ is also an option. $\endgroup$ Apr 19 at 12:07
5
$\begingroup$

We know the maximum covalent is at least 3 with such species as $\ce{H3O+}$. Also the pyrylium ion (https://en.m.wikipedia.org/wiki/Pyrylium_salt) has an oxygen covalency of 3 by forming a pi bond for the third bond.

Can we get 4-coordination? As mentioned by @Rob, yes. @Rob's comment states that basic beryllium acetate appears to have 4-coordinate, covalently bound oxygen.

And it check out. Basic beryllium acetate, $\ce {Be4O(C2H3O2)6}$, has the four-coordinate oxygen in the center, then four beryllium atoms coordinated to this oxygen at the vertices of a tetrahedron, then an acetate group bridging each edge of the tetrahedron. In addition to beryllium, zinc forms a similarly structured basic acetate, which can be obtained by heating normal zinc acetate (https://en.wikipedia.org/wiki/Zinc_acetate).

Might the central oxygen be mostly ionically bonded? Magnesium and heavier alkaline earth metals, which form more strongly ionic bonds with oxygen, are not known to form such a basic acetate structure. This corroborates the hypothesis that covalent bonding of the central, four-coordinate oxygen is involved.

$\endgroup$

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.