Question

The electronic configuration of the free manganese and copper atoms are [Ar]3d54s2 and [Ar]3d104s1, respectively. What are the configurations of the free ions Mn4+ and
Cu3+ ions ?

Answer 1

mn4+ = [Ar]3d14s2

cu3+=[Ar] 3d7 4s1

Answer 2

Answer: The configuration of the manganese (IV) ion will be:

$[Mn^{4+}]=[Ar]3d^34s^0$

The configuration of the copper(III) ion will be:

$[Cu^{3+}]=[Ar]3d^{8}4s^0$

Explanation:

The electronic configuration of the free manganese atom :

$[Mn]=[Ar]3d^54s^2$

When the manganese atom forms the manganese ion with +4 positive charge it looses 4 electrons from its valence shell. So, the configuration of the manganese (IV) ion will be:

$[Mn^{4+}]=[Ar]3d^34s^0$

The electronic configuration of the free copper atom:

$[Cu]=[Ar]3d^{10}4s^1$

When the manganese atom forms the manganese ion with +4 positive charge it looses 3 electrons from its valence shell. So, the configuration of the copper(III) ion will be:

$[Cu^{3+}]=[Ar]3d^{8}4s^0$