Answer: The ions in the Periodic Table that have electron configuration of nd⁸ are: Cu⁺², Zn⁺², Ga⁺³, Ge⁺⁴, As⁺⁵, Se⁺⁶, Br⁺⁷, Pd⁺², Cd⁺², In⁺³, Sn⁺⁴, Sb⁺⁵, Te⁺⁶, I⁺⁷, Hg⁺², Tl⁺³, Pb⁺⁴, Bi⁺⁵, At⁺⁷.
Explanation:
To find the ions it is important to remember that an ion is an atom with charge, if it has positive charge it is called cation, and if it has negative charge it is called anion.
In the Periodic Table there are only neutral atoms, so we do not find ions on it. Because of that, to find the ions that have an electron configuration of nd⁸ it is necessary look for atoms that have any <em>d sublevel</em> in its neutral atom electronic configuration.
Let's find them:
1st) Only atoms with an n level equal to 3, 4, 5 or 6 will have a d sublevel, that's why we have to look for them through periods 3, 4, 5 or 6 in the Periodic Table.
2nd) To find an ion that has 8 electrons in the <em>d sublevel</em>, we need to start looking for an atom with al least 29 or 30 electrons, because following the Moeller's Diagram (to order correctly the electrons in the sublevels of an atom) at level n=3 (the first one to have an <em>d sublevel</em>) we complete the configuration 1s², 2s², 2p⁶, 3s², 3p⁶, 4s² and then 3d⁹ or 3d¹⁰, so the total amount of electrons up to here is 29 or 30.
3rd) It is important to remember that an atom becomes an ion when the atom wins electrons or loses electrons, and we can consider this by paying attention to the Oxidation number (<u>this number represents the possible charges that an atom could have if it loses electrons</u>) in the Periodic Table.
<u>For example</u>:
- The Cooper atom (Cu) has 29 electrons, its electron configuration is [Ar] 4s¹ 3d¹⁰ and it has two possible oxidation numbers: 1 and 2. If an cooper atom loses 1 electron the electron configuration of its ion Cu⁺¹ will be [Ar] 4s¹ 3d⁹, but if it loses 2 electrons the electron configuration of its ion Cu⁺² will be [Ar] 4s¹ 3d⁸ and here we found the ion of cooper that has an electron configuration of nd⁸ with n=3.
- Another example it the Cadmium atom (Cd), it has 48 electrons, its electron configuration is [Kr] 5s² 4d¹⁰ and it has only one oxidation number: 2. So, if an cadmium atom loses 2 electrons, the electron configuration of its ion Cd⁺² will be [Kr] 5s² 4d⁸ with an electron configuration of nd⁸ and with n=4.
- Here are other ions that have an electron configuration of nd⁸: Zn⁺², Ga⁺³, Ge⁺⁴, As⁺⁵, Se⁺⁶, Br⁺⁷, Pd⁺², In⁺³, Sn⁺⁴, Sb⁺⁵, Te⁺⁶, I⁺⁷, Hg⁺², Tl⁺³, Pb⁺⁴, Bi⁺⁵, At⁺⁷.
