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I came across a statement that $\ce{PbO}$ is more stable then $\ce{PbF2}$. Now how can i justify that staement because fluorine being more electronegative and should form stronger bond with metals than oxygen.

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closed as too broad by Mithoron, bon, andselisk, Tyberius, airhuff Sep 28 '17 at 18:32

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You put $\ce{PbO}$ on a shelf and it will stay there forever without any change. Ditto for $\ce{PbF2}$. Now what does "more stable" really mean? $\endgroup$ – Ivan Neretin Aug 21 '17 at 15:23
  • $\begingroup$ Stability obiously mean thermodynamic stability. PbO is thermodynamically more stable then PbF2 $\endgroup$ – Yogi Joshi Aug 21 '17 at 15:25
  • $\begingroup$ This was not obvious at all, so you might want to edit your question to that effect. Also, do you have any numerical evidence to support your claim? $\endgroup$ – Ivan Neretin Aug 21 '17 at 15:37
  • $\begingroup$ I read in a book n even i couldnt justify tht statement hence m askng are oxides more stable then fluorides? $\endgroup$ – Yogi Joshi Aug 21 '17 at 15:40
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    $\begingroup$ "$\ce{PbO}$ is thermodynamically more stable then $\ce{PbF2}$" I don't think this statement has any meaning. You need to specify thermodynamically stable relative to what? The constituent elements? $\endgroup$ – Zhe Aug 21 '17 at 15:54
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According to the formula for lattice energy, it is proportional to the product of the charges on the ions, and inversely proportional to the ionic separation (we shall approximate it to be the sum of the ionic radii). In this case, the product of the charges is higher for PbO, and the sum of the ionic radii is similar between the two compounds. This results in the ionic bond in PbO being stronger than in PbF₂. Thus PbO is more thermodynamically stable.

Note: this may not be correct, my friend and I just thought of this explanation when discussing this question.

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