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Barium Chloride is represented as $\ce{BaCl2}$.

Since chlorine is a diatomic molecule, It should be denoted as $\ce{Cl2}$.

Formulating, we get

\begin{array}{|c:cc|}\hline \small \rm Element & \ce{Ba} & \ce{Cl2}\\ \small \rm Valency & 2 & 1 \\\hline \end{array}

Crisscrossing the valencies of Barium and Chlorine we get $\ce{Ba(Cl2)2}$ — as opposed to the accepted formula of $\ce{BaCl2}$. How is it so?

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    $\begingroup$ In barium chloride, the chlorine is not in the form of a diatomic molecule. Think about how ionic compounds are formed. $\endgroup$ – M.A.R. Jun 19 '16 at 6:17
  • $\begingroup$ @TIPS How are they formed in this case? $\endgroup$ – Good Guy Jun 19 '16 at 6:27
  • $\begingroup$ Valency is a property of an element, not a molecule. $\endgroup$ – Ivan Neretin Jun 19 '16 at 7:55
  • $\begingroup$ @IvanNeretin How do you obtain the formula for Barium Chloride then? $\endgroup$ – Good Guy Jun 19 '16 at 9:12
  • $\begingroup$ Much like you did, except don't mention $\ce{Cl2}$ at all. There is just Cl, its valency is 1, and then there is Ba with valency 2, so... $\endgroup$ – Ivan Neretin Jun 19 '16 at 13:52
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The compound barium chloride is not the same thing as barium and chlorine mixed together.
When they react, a barium atom will give up two electrons to form a action, and a chlorine molecule will pick up two electrons to form a pair of chloride ions: $$\ce{Ba -> Ba^2+ +2e^-}$$ $$\ce{Cl2 +2e^- -> 2Cl^-}$$ When you have both of those things at once, the electrons are "consumed" as fast as they are "produced", so they don't appear at all in the result: $$\ce{Ba +Cl2->Ba^2+ +2Cl^-}$$ which forms an ionic lattice when solid. Since this lattice has overall neutral charge, its ionic charges must balance with integer coefficients.
Funnily enough, these coefficients are $1$ and $2$ for $2+$ and $1-$ respectively, so these are applied to the ions which carry those charges. Thus: $$\ce{Ba1Cl2}$$ or more simply and directly: $$\ce{BaCl2}$$

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  • $\begingroup$ If you think you have the reaction right, you should be able to give the half-equations for it, such that the conservation of atomic numbers and conservation of net charge holds. Your proposed formula would require something to break at least one, probably both of those. $\endgroup$ – Nij Jun 19 '16 at 11:36
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When chlorine is in its free state it is diatomic. But when it reacts with barium it is not in the form of $\ce{Cl2}$. It will be in its ionic state which is $\ce{Cl-}$. The same goes for Barium. Barium is mono-atomic and its ionic state is $\ce{Ba^2+}$. Barium gives one electron to a chlorine atom and another electron to another chlorine atom, as valency of chlorine is 1, so it is $\ce{BaCl2}$.

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