Question

write the electronic configuration of a divalent ion of a coinage metal​

Answer 1

Answer:

The energy required to remove one of them is the highest ionization energy of any atom in the periodic table: 24.6 electron volts. The energy required to remove the second electron is 54.4 eV, as would be expected by modeling it after the hydrogen energy levels

Explanation:

The helium ground state consists of two identical 1s electrons. The energy required to remove one of them is the highest ionization energy of any atom in the periodic table: 24.6 electron volts. The energy required to remove the second electron is 54.4 eV, as would be expected by modeling it after the hydrogen energy levels. The He+ ion is just like a hydrogen atom with two units of charge in the nucleus. Since the hydrogenic energy levels depend upon the square of the nuclear charge, the energy of the remaining helium electron should be just 4x(-13.6 eV) = -54.4 eV as observed.

The fact that the second electron is less tightly bound can be interpreted as a shielding effect; the other electron partly shields the second electron from the full charge of the nucleus. Its energy can be used to model the effective shielding as follows.

Another way to view the energy is to say that the repulsion of the electrons contributes a positive potential energy which partially offsets the negative potential energy contributed by the attractive electric force of the nuclear charge. The description of any electron in a multi-electron atom must find a way to characterize the effect of the other electrons on the energy.

Answer 2

The electronic configuration of a divalent ion of a coinage metal (Cu2+) is 1s2,2s2,2p6,3s2,3p6,3d9.

• Gold (Au), Silver(Ag) and Copper (Cu) are refferd toas coinage metals.
• A divalent ion of  a metal is the ion obtained by the removal of two electrons from the outermost orbiatal of the metal.
• The general configuration of a coinage metal consists of an inner noble gas core configuration and an outer ns2,(n-1)d9 configuration.
• So, in a divalent ion two electrons are removed from the ns2 orbital, hence the configuration of the divalent ion is (n-1)d9.