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It's easy to do atomic electron configuration, but ionic is confusing.

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    $\begingroup$ Welcome to Chemistry.SE. What about ionic configuration is confusing? The more we know about what trouble you are having, the better an answer we can give. For example, if you tell use what you think it is or where you are getting stuck, we can help you with the specific problem you are having. $\endgroup$ – Ben Norris Nov 9 '13 at 19:55
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The electronic configuration for $\ce{Br-}$ is:

$$\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}$$

Because it have one more electron than bromine, which ends its electronic configuration with $\mathrm{4p^5}$.

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I suppose that you know how to find the atomic electron configuration for Br which is $\mathrm{1s^22s^22p^63s^23p^64s^23d^{10}4p^5}$. Now for ions, it is really simple. All you have to do is look at if the ion is a cation or an anion. If it is an anion then add the same amount of electrons as the number of the charge. If it is a cation subtract the number electrons as the number of the charge.

In the case of $\ce{Br-}$ it is an anion of charge minus 1 meaning that it will have one more electron than a normal bromine atom. So really its electron configuration will be the same as krypton which is the atom after bromine. So all you have to do is add an extra electron in the last orbital, the 4p orbital. Therefore its electron configuration will be: $$\mathrm{1s^22s^22p^63s^23p^64s^23d^{10}4p^6}$$

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