<article_title>Boron</article_title>
<edit_user>Sbharris</edit_user>
<edit_time>Wednesday, August 11, 2010 3:03:49 AM CEST</edit_time>
<edit_comment>/* Chemistry of the element */ Fix per talk. Some work is going to need to be done to explain boron carbides.</edit_comment>
<edit_text>Oxidation of boron depends upon the crystallinity, particle size, purity and temperature. Boron does not react with air at room temperature, but at higher temperatures it burns to form boron trioxide: [[File:Tetraborate-xtal-3D-balls.png|thumb|left|Ball-and-stick model of tetraborate anion, [B&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;(OH)&lt;sub&gt;4&lt;/sub&gt;]&lt;sup&gt;2−&lt;/sup&gt;, as it occurs in crystalline borax, Na&lt;sub&gt;2&lt;/sub&gt;[B&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;(OH)&lt;sub&gt;4&lt;/sub&gt;]·8H&lt;sub&gt;2&lt;/sub&gt;O. Boron atoms are pink, with bridging oxygens in red, and four hydroxyl hydrogens in white. Note two borons are trigonally bonded sp&lt;sup&gt;2&lt;/sup&gt; with no formal charge, while the other two borons are tetrahedrally bonded sp&lt;sup&gt;3&lt;/sup&gt;, each carrying a formal charge of -1 due to oxidation state IV. The mixture of boron <strong><strike>oxidation states (III</strike></strong><strong>coordination numbers</strong> and IV) is characteristic of natural boron minerals.]]</edit_text>
<turn_user>Sbharris<turn_user>
<turn_time>Wednesday, August 11, 2010 9:23:18 AM CEST</turn_time>
<turn_topicname>+4 Oxidation State</turn_topicname>
<turn_topictext>Boron can form compounds whose formal oxidation state is not three, such as B(IV) in boron carbide BC. What does it mean? If so, we may be able to obtain a helium compound very easily because it has only 2 protons and so the 2S electrons will be much less tightly bound!--Anoop.m (talk) 11:14, 28 February 2010 (UTC) The formula for boron carbide is B4C. Your question assumes facts not in evidence. blueSorangeBHarris 22:23, 22 July 2010 (UTC) B(IV) is mentioned several times in the article - I believe this is a mistake. [B(OH)4]−, for example, contains B(III), but the article implies it is B(IV). Ben (talk) 22:07, 10 August 2010 (UTC) The only evidence for B(IV) in the article is BC in the infobox. The recently added unreferenced part on B(IV) has to be fixed or removed - oxidation state IV does not derive from coordination IV. We don't say Cd has oxidation state IV only because it forms zincblende CdS. Charged species don't count here. Materialscientist (talk) 23:45, 10 August 2010 (UTC) Yep, you're both right. Coordination number 4 with oxygen remains oxidation +3 so long as there's a formal -1 charge on the boron (the fourth bond is a coordinate bond). So both B(OH)3 and B(OH)4- are boron (III) = +3. The oxygens are always -2 and the oxidation number of the boron has to make the ion charge come out to what it is. All the borons in tetraborate are +3 also: there we have 9 oxygens (-18), 4 hydrogens (+4), and 2 negative charges, which leaves us +12 for the 4 borons, which gives them +3 each. Wups. I'll fix this. The oxidation number for MgB2 is wrong also. The borons bonding to each other have no oxidation state. The only electrons are lost from the magnesium, one per boron, so the formal oxidation state for B is -1 in that compound. If we assign nitrogen -3 in BN compounds, the boron is always +3 in these various structures also. That leaves me with no +4. Ideas are welcome about what do do with BC compounds. B4C isn't quite stoichiometric anyway, as noted. If it were, and carbon were assigned +4, the boron would come out -1 as in MgB2 again. But I'm not sure it's fair to assign carbon as the electropositive element here, any more than to choose boron as electronegative in all the hydrides, which are analogous to hydrocarbons. What do you think? blueSorangeBHarris 02:52, 11 August 2010 (UTC)I was actually referring to the BC molecule. As to B4C, it is a complex lattice with B12 icosahedra connected by carbon chains. There is an "explanation" here () why it is non-stoichiometric, and I am just reading it. Anyway, the structure of B4C seems uncertain and can hardly indicate B(IV) state. Materialscientist (talk) 04:36, 11 August 2010 (UTC) Yes, the icosahedra are fascinating, and these compounds go up to B(6.5)C stoichiometrically. They should be neat things to make 10B containing composites for spacecraft with maximum B-10-enriched-boron per structural weight. I don't know of any BC molecules except as isolated gases or free radical trapped in argon. BC. would be a free radical, and BC: with the extra electron would have a negative charge. There should be a compound HCB, rather like HCN:, except the electrons in the sigma lobe of HCN: would be missing in HCB, due to boron's smaller charge by 2. In all cases the bond order is about 3, but (again) I'm not sure how to assign C or B as the negative or positive here (they are nearly the same electronegativity), so the oxidation state of the B is a mystery to me. I guess it's either +3 or -3. If you must have your carbon negative, then the B is still +3. And yes, B(+IV) was my mistake. I admit it! Moving on... blueSorangeBHarris 08:01, 11 August 2010 (UTC)</turn_topictext>
<turn_text>Yep, you're both right. Coordination number 4 with oxygen remains oxidation +3 so long as there's a formal -1 charge on the boron (the fourth bond is a coordinate bond). So both B(OH)3 and B(OH)4- are boron (III) = +3. The oxygens are always -2 and the oxidation number of the boron has to make the ion charge come out to what it is. All the borons in tetraborate are +3 also: there we have 9 oxygens (-18), 4 hydrogens (+4), and 2 negative charges, which leaves us +12 for the 4 borons, which gives them +3 each. Wups. I'll fix this. The oxidation number for MgB2 is wrong also. The borons bonding to each other have no oxidation state. The only electrons are lost from the magnesium, one per boron, so the formal oxidation state for B is -1 in that compound. If we assign nitrogen -3 in BN compounds, the boron is always +3 in these various structures also. That leaves me with no +4. Ideas are welcome about what do do with BC compounds. B4C isn't quite stoichiometric anyway, as noted. If it were, and carbon were assigned +4, the boron would come out -1 as in MgB2 again. But I'm not sure it's fair to assign carbon as the electropositive element here, any more than to choose boron as electronegative in all the hydrides, which are analogous to hydrocarbons. What do you think? </turn_text>