# Question 179ac

Feb 1, 2014

Iodine would have a standard electron configuration of

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

The noble gas in the row above iodine is krypton.

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

$\left[K r\right] 5 {s}^{2} 4 {d}^{10} 5 {p}^{5}$

SMARTER TEACHER

Nov 21, 2014

The full electron structure of iodine is :

!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:

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

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

$\left[K r\right]$

So now we can write the electron configuration as :

$\left[K r\right] 4 {d}^{10} 5 {s}^{2} 5 {p}^{5}$

You will note that iodine has 7 valence electrons in the outermost n = 5 energy level, $4 d$ being lower in energy than $5 s$. 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.