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Question is in the title. I understand that the metals towards the left form stronger bonds with other chemicals compared to metals towards the right, and that the ones with weaker bonds break down easier and pass more current. (if that's wrong, please correct me) Why is this the case? Could someone please explain this concept to me more?

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    $\begingroup$ Al -> Si -> P -> S seems problematic for the assertion. $\endgroup$ – Jon Custer May 21 at 13:13
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Actually, this only applies for metals. This trend does not apply for non-metals: non-metals form covalent bonds, which are due to the sharing of electrons - thus, there are no free electrons that can carry charge. Additionally, atoms are held tightly by covalent bonds and thus cannot dissociate into ions that carry electric charge.

With regards to metals, metallic bonding consists of metal ions surrounded by a sea of delocalized electrons - and it is these delocalized electrons that can carry charge. Each atom in a metal has the same number of electrons as the charge on the ion. For example, Na+ has one delocalized electron for one atom, Mg2+ has two, Al3+ has three, and so forth. So as the charge increases, the number of delocalized electrons per atom increases - and so the number of delocalized electrons in the metal overall also increases. This allows it to carry more charge per unit time, leading to a higher rate of conductivity.

Similarly with regards to ionic lattices, a greater charge means there will be more ions formed per molecule that is made molten. So one mol of NaCl will have 2 x 6.022 x 10^23 ions total, while 1 mole of MgCl2 will have 3 x 6.022 x 10^23 ions. So, MgCl2 has more ions to conduct electricity and thus shows greater conductivity.

I'm not sure if it has much to do with the actual 'ease' of breaking bonds. Both NaCl and MgCl2 would have to be in a molten state to be able to conduct electricity anyway, so I don't think it plays a role for ionic compounds. For metallic compounds, again, as they are simply ions surrounded by electrons there isn't any bond-breaking to be done.

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