I guess that sodium loses an electron so has an oxidation state of +1 and that oxygen has an oxidation state of −2. But what oxidation state does the aluminium atom have in sodium aluminate ($\ce{NaAlO2}$)?
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For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.
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$\begingroup$ Indeed… also, you can confirm this by noting that +3 is the most common oxidation state for aluminium (look at its position in the periodic table) $\endgroup$– F'xAug 29, 2013 at 13:58
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$\begingroup$ But how does it then work with the electrons : does aluminium get electrons from sodium or ..? $\endgroup$– user2117Aug 29, 2013 at 14:00
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$\begingroup$ @LievenB Aluminum is situated in the 3rd main group. So it has 3 valence electrons. Now in $\ce{NaAlO2}$ it has an oxidation state of $+3$ meaning (loosely speaking) that it gives away it's three electrons. Sodium is has an oxidation state of $+1$, so it gives away one electron. The two oxygen atoms, being much more electronegative than the metals take these four electrons and thus have an oxidation state of $-2$ each. $\endgroup$– PhilippAug 29, 2013 at 14:04
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1$\begingroup$ No, $\ce{NaAlO2}$ consists of the anion $\ce{AlO_{2}^{-}}$ and the cation $\ce{Na+}$. Aluminum will be in the center between the two oxygens. Sodium is only some kind of counterion and will have the highest probability density at one of the two oxygens. $\endgroup$– PhilippAug 29, 2013 at 14:30
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1$\begingroup$ Even found a picture of $\ce {NaAlO2}$ here: img1.guidechem.com/chem/e/dict/37/7758-17-0.jpg $\endgroup$– TomcatAug 30, 2013 at 10:37
