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How many electrons will be transferred from Mg to SO4 to form MgSO4?

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The sulfate ion has a net charge of -2, meaning it is deficient in 2 electrons in order to have all 5 atoms have a full octet. Each atom contributes only the valence electrons it has. Since Sulfur has 6, and each Oxygen has 6, there are 5 atoms with 6 electrons. This is a total of 30 electrons. When you draw the Lewis Structure of the sulfate ion, you will see that sulfur has a full octet (eight electrons) around it by the four shared bonds with the four oxygen atoms. So, Sulfur would be in the middle like this:
                                                                O ---- S----O
and each dashed line is a shared pair, so it's worth 2 total electrons. That's 8 of the 30. That leaves 22. 
If we put 6 around 3 of the oxygens, we'd have 18 more for a total of 26. That means the final O would have four electrons around it for a total of 30 electrons. This means that this particular O has the four free electrons plus the 2 in the bond for a total of 6. But since it want's eight, it will try to find 2 more and that is why it will steal 2 from another source like Magnesium. Magnesium has 2 valence electrons and thus it would rather give those up than get 6 more because it takes less energy; so when it encounters the sulfate ion, it will donate 2 to the oxygen with only four leaving it +2 and the sulfate ion -2. They will then form the ionic compound MgSO4.  The program won't let me put drawings in, but I hope that helps. 


It might be worth emphasizing that the language of the question is pretty lax -- there just aren't any neutral SO4 species out there to donate electrons to! (Whereas, there are neutral Mg, as the metal). Sunil notes below that the neutral Mg may encounter sulfuric acid (H2SO4), and donate its 2 electrons, 1 each to each of the hydrogens on the acid, to enable it to form H2 gas, as the Mg(2+) ion then associates (if it does) with the resulting SO4(2-) ion. Of course, in solution, these ions would be complexed by water molecules -- they'd only be recovered as a solid salt when the solution is evaporated. If you input vastly more energy, you can cause an ionic solid to vaporize -- but even in the vapor, individual MgSO4 molecules would tend to clump together, the ionic attractive lattice forces are ferocious!
One further consideration: the transfer of electrons, to make ions, is governed by equilibrium considerations driven by energetics. Species like MgSO4 can be considered totally ionic (across the Mg - SO4 pair; polar covalent across the S - O pairs) in practice, but among binary compounds more generally there's a continuum between "ionic" and "covalent" bond behaviors; if I recall, .delta.electronegativity ~1.7 is considered the boundary point.
Magnesium has the atomic number 12. There are 12 electrons in the outermost shell. Mg atom would look like  this Mg: (if we ignore the full shells). This molecule is very reactive, as it wants to reach a full shell, and it can do that by losing those two electrons.
When magnesium is reacted with sulfuric acid H2SO4, it gives these electrons to the two H+ ions, forming H2 gas, converting Mg: to Mg2+. This reaction proceeds because Mg: is more reactive than H2 towards giving electrons (acting as a reducing agent). H2 is content as it has full shells for its shared two electrons, and Mg is happier (more stable) as the +2 charge than as Mg:. 
  An illustration of the molecule sulfuric acid is as below. The oxygen atoms on the S=O bonds have two electron lone pairs, while the oxygen atoms in the S-O-  bonds have three lone pairs. Two hydrogen atoms have donated their electrons to oxygen. Hydrogen has a +1 charge, while oxygen has a -1 charge. The molecule is ionic because here is no sharing of electrons like in the sulfur - oxygen bonds. If you are studying the octet rule, sulfur is one of the exceptions to the rule because of d-orbitals.  
   Hope this helped.
Hey Nat -- you're likely wondering about the SO4 part ... the Mg: +2 is easier to see ... both S and O have 3 outer shell e-pairs
:S:  :O:  you can get S and 3[O]'s to share, but need another e-pair for [SO4] -2
 ..     ..
Mg: +2 provides that 4th e-pair ... Best wishes, sir :)