Question

explain the formation of au3+​

Answer 1

First of all, let's see the electronic configuration of gold (i.e. Aurum : Au):

$Au : [Xe] \: 4 {f}^{14} \: 5 {d}^{10} \: 6 {s}^{1}$

Now, 3 electrons have to be ejected from a gold atom in order to achieve +3 state:

• So, 1 electron from 6s orbital and 2 electrons from 5d orbital have to be ejected out to achieve the required ionic configuration.

${Au}^{ 3 + } : [Xe] \: 4 {f}^{14} \: 5 {d}^{8} \:$

This is how , $Au^{3+}$ can be obtained.