# Qual è la configurazione elettronica di  "Cr" ^ (2 +) ?

#### Risposta:

[Ar] 3d^4

or

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)

#### Spiegazione:

Chromium and Copper are two special cases when it comes to their electron configurations- having only 1 electron in the 4s orbital, as opposed to the other transition metals in the first row which has a filled 4s orbitale.

The reason for this is because this configuration minimizes electron repulsion. Half filled orbitals for "Cr" in particular is its most stable configuration.

Così il configurazione elettronica for elemental Chromium is

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5).

And the electrons in the 4s orbital is removed first because this orbital lies further from the nucleus, making electrons easier to remove in ionization.

So if we remove 2 electrons to form the Cr^(2+) ion we remove 1 4s electron and 1 3d electron leaving us with:

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(4)

or

[Ar] 3d^4