Question #179ac

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Question #179ac

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Answer 1 · 2014-02-01T20:48:06

Iodine would have a standard electron configuration of

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{5}$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^5$

The noble gas in the row above iodine is krypton.

We can replace $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6}$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6$ with the symbol [Kr] and rewrite the noble gas configuration of iodine as

$[K r] 5 s^{2} 4 d^{10} 5 p^{5}$ $[Kr] 5s^2 4d^10 5p^5$

I hope this was helpful.
SMARTER TEACHER

Answer 2 · 2014-11-21T23:12:25

The full electron structure of iodine is :

$! s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{5}$ $!s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(5)$

The noble gas Krypton has the electron configuration:

$1 s^{2} 2^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$ $1s^(2)2^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)$

We refer to this as the "Noble Gas Core" which we can write as :

$[K r]$ $[Kr]$

So now we can write the electron configuration as :

$[K r] 4 d^{10} 5 s^{2} 5 p^{5}$ $[Kr] 4d^(10)5s^(2)5p^(5)$

You will note that iodine has 7 valence electrons in the outermost n = 5 energy level, $4 d$ $4d$ being lower in energy than $5 s$ $5s$ . These electrons define the chemistry of iodine which can use some, or all of them in chemical reactions with oxidation states ranging from -1 to +7.

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Question #179ac

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