Please compose a dissociation-in-water equation for the compound $\ceBeI2$ and $\ceLiI$. Make sure to add the claims of matter after each compound.

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Currently, because that $\ceBeI2$ I have actually the equation $$\ceBeI2 (s) -> Be^2+ (s) + I2^2- (g).$$ I have actually yet to attempt the second one.


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Your ionic charges room not correct because that iodine. Spring at your attempt.

As $\ceBe$ is in group 2, the ionic charge for beryllium ion is fine, yet iodine is in group 17, so its ion is $\ceI-$. Once the ion dissociate, they end up being aqueous or (aq) as the state the matter.

Then the ionic charges have to balance, thus:

$$\ceBeI2 (s) -> Be^2+ (aq) + 2I- (aq)$$

To balance the ionic dues in this example, you need 2 $\ceI-$ (iodine ions) to balance the $\ceBe^2+$ (beryllium ion).

A similar example (and further explanations) are detailed on the UC Davis ChemWiki page distinctive Features of Aqueous services (including an instance of the dissolved of $\ceMgCl2$ - another compound with team 2 and 17 elements).

So,

determine the group, for this reason ionic charge of each dissociated ionbalance these chargesstate that the dissociated ions are aqueous

Now, use the process to identify the dissociation that $\ceLiI$


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edited january 15 "17 in ~ 12:03
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martin - マーチン♦
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$\begingroup$
By definition, a dissociation-in-water reaction outcomes in aqueous, elevation ions, no negatively charged diatomic molecules favor the $\ceI_2^2-$ in the original question (not to point out that diatomic molecules are typically not negatively charged).

The answer should an outcome in solely single-atom ions, together so:

$$\ceBeI2 (s) -> Be^2+ (aq) + 2I- (aq)$$

And using this very same logic for $\ceLiI$:

$$\ceLiI (s) -> Li+ (aq) + I- (aq)$$


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edited january 15 "17 at 12:02
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answered january 15 "17 in ~ 5:26
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