Electron Configuration Of Copper Ii Ion

Please show the electronic configuration of copper (ii) ion in that way.

Electron configuration of copper ii ion. That type of notation is a bit dated, and not what i would teach as the electron configuration. As cu is a transition metal with electron config [ar] 4s1 3d10, the ion of a transition metal loses its 4s electrons. When it loses that 1 electron, it no longer needs the 4s orbital, and therefore its electron configuration becomes 1s^2\2s^2\2p^6\3s^2\3p^6\3d^10.

It is [ar] 3d7 4s2 or extended it is. But for cu2+ you would write 2,8,17. Electron configuration indicates how many electrons an atom or ion has, and how they are distributed on electron orbitals.

Since 4s^2 (not the 3d shell) is the outermost shell, then those electrons are removed. Erbium [xe]6s 2 4f 12. 2) a) write the complete electron configuration for the manganese(ii) ion?

Now sometimes the noble state is written as $\ce{[ar] 3d^10 4s^1}$ or as $\ce{[ar] 4s^2 3d^9}$. According to the rules of filling electron shells, copper should have a configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d9 instead, but it does not. Rhodium [kr]5s 1 4d 8:

If you don't want explanation, jump to the end of answer. Electronic configuration of cu is 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d9 ([ar] 4s2, 3d9), whereas for cu2+ is [ar], 3d9. Using noble gas notation write the electron configuration for the copper(ii)ion.

Therefore, the electron configuration of oxygen is 1s 2 2s 2 2p 4, as shown in the illustration provided below. Copper has an electron configuration of [ar]3d^10\4s^1. The p orbital can hold up to six electrons.