# How is Be(OH)4 Formed?

I recently did a test paper and the question stated that Be(OH)2 is amphoteric and so I was asked to write equations showing it both acting as an acid and a base.

Acid answer: $$\quad\ce{Be(OH)2 + 2H+ -> Be^2+ + 2H2O}\quad$$ I understand this answer

Base answer: $$\quad\ce{Be(OH)2 + 2OH- -> Be(OH)4-}$$

My question is how is $$\ce{Be(OH)4}$$ able to form if the $$\ce{Be^2+}$$ has a $$2+$$ charge and $$\ce{OH-}$$ has a $$1-$$ charge? Why would $$\ce{Be^2+}$$ form 4 bonds and not 3? It doesn't gain a full outer shell from this bonding anyways. Also why wouldn't the end product have a $$2-$$ charge? Is the bonding covalent or ionic? I think it is covalent due to the electronegativity of the beryllium but then where does it get the electrons from to share with the $$\ce O$$ and the $$\ce H$$ it only has 2 in it's outer shell?

• Maybe this one is not in your syllabus, but other coordination complexes surely must be. – Ivan Neretin Oct 11 at 11:09
• Be would form 4 bonds to complete its octet. Why do you think this is not a "full outer shell"? The end product does have a 2- charge, see wiki: en.wikipedia.org/wiki/Beryllium_hydroxide where the extra sodium cations stabilize an overall 2- charge. The electrons come from the oxygen in the hydroxide anion, ie. oxygen contributes 2 electrons. This is called a dative bond (or maybe the new name is a "coordinate covalent bond"). I would guess there is enough covalent character to call it a covalent bond. – Blaise Oct 11 at 11:40
• So Be has 2 electrons in its outer shell. It gains 4 from oxygens and 4 from hydrogens. That would equal 10 but 8 is a full shell. Unless it gains 6 from the oxygens but then what do they get in return as Be only has 2 to give and what does the hydrogen do? Does it not donate anything? – James Oct 11 at 17:01
• Remember beryllium is in a 2+ state. Otherwise Be(OH)$_4$ would have an overall 4- minus charge instead of a 2- charge. To keep things simple, you can think of beryllium as having zero valence electrons, and then each hydroxide donates 2 electrons from oxygen to form a bond. 4 x 2 = 8. No hydrogen atoms are involved with the beryllium. Hydrogen is bonded with oxygen, sharing its 1 valence electron with 1 electron from oxygen. Draw the lewis structure for OH$^-$, then imagine one of the lone pairs on oxygen being moved into a bond with beryllium. – Blaise Oct 11 at 17:19
• I get you! You are really good at explaining things ngl. Thank you so much! So just another quick question. So even if a metal is forming a covalent bond it still turns into an ion? So like Be formed a covalent bond with OH but it still became +2 or is that only in this case? – James Oct 11 at 17:33