In full detail explain why an iodine atom becomes an iodide ion. be sure to address the following terms in your explanation: valence electrons, octet rule, and noble gases.

Iodine atoms have atomic number 53. Being the atomic number the number of protons in the nucleus of the atom, each atom of iodine posseses 53 protons. Since, the charges of electrons and protons are equal, all neutral atoms have the same number of electrons than protons. So, the neutral atoms of iodine have 53 electrons.

When you depict the electron configuration of iodine, following Aufbau rules, you get that the 53 eletrons are represented by: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁵ or, equivalently, in abbreviated form, [Kr]5s²5p⁵.

The seven electrons 5s²5p⁵ are the outermost electrons and they are the valence electrons, i.e. the electrons que usually take active part in the chemical bonds that iodine atoms form.

The octet rule states that the atoms lose, gain or share electrons to acquire a configuration of 8 valence electrons, which the most stable electron configuration such as the nobles gases'.

So, having iodine atoms 7 valence eletrons, following the above mentioned octet rule, they will easily gain one electron, becoming iodide ions (I⁻, with one negative charge due to the excess of electrons over protons), to get a full valence shell with 8 electrons, which is the configuration of the next noble gas in the periodic table.

In this way, iodide ion (I⁻) has 54 electrons, has 8 electrons in its outermost electron shell, and its electron configuration is the same of the noble gas Xe: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.