GCSE question Lead (II) Oxide

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    I've been asked to work out the chemical formula of Lead (II) Oxide using the periodic table.

    I know that the (II) after Lead means that there is a positive 2+ charge for Lead. Therefore there must also be a negative 2- charge for the Oxygen ion.

    Lead is in Group 4 of the table so therefore has 4 outer electrons. Oxygen is in Group 6 so has 6 outer electrons.

    What I cannot grasp is that if Lead loses 2 of those electrons to Oxygen then it still has 2 electrons left in it's outer shell. If the formula is PbO as the answer says then what happens to the last two electrons in Lead? That means it is unstable. It cannot give more than 2 to Oxygen as Oxygen can only take 2 to make an outer shell.

    I'm confused ????
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    (Original post by Darwinion)
    I've been asked to work out the chemical formula of Lead (II) Oxide using the periodic table.

    I know that the (II) after Lead means that there is a positive 2+ charge for Lead. Therefore there must also be a negative 2- charge for the Oxygen ion.

    Lead is in Group 4 of the table so therefore has 4 outer electrons. Oxygen is in Group 6 so has 6 outer electrons.

    What I cannot grasp is that if Lead loses 2 of those electrons to Oxygen then it still has 2 electrons left in it's outer shell. If the formula is PbO as the answer says then what happens to the last two electrons in Lead? That means it is unstable. It cannot give more than 2 to Oxygen as Oxygen can only take 2 to make an outer shell.

    I'm confused ????
    I'm assuming that you're at GCSE level. The electron model that you're taught at GCSE is actually a pretty major simplification of reality so it's not too long before you start finding situations where it doesn't really work (like you have, here). Thinking about atoms having electron shells of just 2,8,18 etc. works okay for lighter elements but in this case, it doesn't really work.

    You're over-complicating it. As you've correctly written, the fact that it's Lead (II) means the Lead will form a Pb2+ ion. Oxygen (almost) always forms O2- ions. So you need to think about what combination of those two ions forms a neutral combination and of course, it's 1:1. So it's PbO. You may still be annoyed about the fact that you can't properly reconcile this with the ionic bonding model that you've been taught but unfortunately this is something that you'll just need to accept until you study more Chemistry
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    (Original post by Plagioclase)
    I'm assuming that you're at GCSE level. The electron model that you're taught at GCSE is actually a pretty major simplification of reality so it's not too long before you start finding situations where it doesn't really work (like you have, here). Thinking about atoms having electron shells of just 2,8,18 etc. works okay for lighter elements but in this case, it doesn't really work.

    You're over-complicating it. As you've correctly written, the fact that it's Lead (II) means the Lead will form a Pb2+ ion. Oxygen (almost) always forms O2- ions. So you need to think about what combination of those two ions forms a neutral combination and of course, it's 1:1. So it's PbO. You may still be annoyed about the fact that you can't properly reconcile this with the ionic bonding model that you've been taught but unfortunately this is something that you'll just need to accept until you study more Chemistry
    OK many thanks for that! Yes I can see the logic of the 1:1 ratio for the ionic charges as you put it. Was just those remaining electrons.

    I am at GCSE level and I had a list of compounds to work out in this exercise. This was the first one and it threw me totally.

    Thanks again
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    Ending up with a full outer shell isn't what atoms are necessarily trying to do when they react.

    Using Al as my example:

    It takes a bit of energy to steal an e- off an Al atom and for example give it to a Cl atom. When the Cl- ion forms*, some energy is released BUT loads of energy is then released then the ionic bond forms between the Al+ and Cl- ions.

    It takes quite a bit more energy to steal a second e- off the Al+ ion, but the two lots of energy from Cl- + e- >Cl- and the ionic bond are bigger.

    It takes a huge amount of energy to steal the third e-, but again, the release of energy are bigger.

    The fourth e- takes sooo much energy to steal that the energy release isn't big enough to pay for the theft. So only three are stolen. It just happens that this leaves Al3+ with a full outermost shell. That's because the fourth e- would need stealing from a lower energy shell compared to the 1st-3rd e-.

    In the case of lead, the first and second are relatively easy to steal and even if the robber doesn't form strong ionic bonds, the energy needed to do the stealing can easily be paid back.

    But, if the robber is able to form very strong ionic bonds, it can pay back the energy needed. The reason for it is essentially the same, the 1st two e- are from one (high) energy sub-shell and the next two are from a lower energy sub-shell.

    I trust my usage of the term sub-shell doesn't alarm. It isn't something you should hear in a GCSE class, but is a simple idea that within a shell, there are various subshells, e.g. within the third shell, there are three subshells, with differing energy levels, which is why you get 2.8.8.2 and why the fourth shell starts to fill before the third shell is full (it can actually hold 18 e-).

    *there are other things going on like the breaking of the Cl-Cl bond, but I've ignored them for clarity.
 
 
 
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