Question

Write the ground state electron configurations of the following transition metals: ${V^{5 + }},A{u^{3 + }},$ and $F{e^{2 + }}$ ?

Answer 1

Transition metals are the chemical elements that do not have a completely filled d-subshell. Generally, elements undergo loss or gain of electrons and become ions. Ions electronic configuration is different from the electronic configuration of elements. If the positive ion is formed, then the number of electrons lost are equal to the number mentioned.

Complete answer:
Given ions are ${V^{5 + }},A{u^{3 + }},$ and $F{e^{2 + }}$.
Vanadium is an element with atomic number $23$ it undergoes oxidation which is loss of electrons. Here the number of electrons lost were $5$ . ${V^{5 + }}$ has five electrons less than neutral vanadium atom. The ground state electron configuration of ${V^{5 + }}$ is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
Gold is a chemical element with chemical symbol of Au and the atomic number is $79$
Gold undergoes oxidation which is loss of electrons. Here the number of electrons lost were $3$ . $A{u^{ + 3}}$ has three electrons less than a neutral gold atom. The ground state electron configuration of $A{u^{ + 3}}$ is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}4{d^{10}}5{s^2}5{p^6}6{f^{14}}5{d^8}$
Iron is chemical element with chemical symbol of Fe and the atomic number is $26$
Iron undergoes oxidation which is loss of electrons. Here the number of electrons lost were $2$ . 0$F{e^{2 + }}$ has two electrons less than the neutral iron atom. The ground state electron configuration of $F{e^{2 + }}$ is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}$

Note:
The number of electrons lost by the neutral atom can be indicated by the positive charge on the neutral atom. In this case the number of electrons will be less than the neutral atom. If electrons were gained by a neutral atom, then the atom can be represented by a negative charge. In this case the number of electrons will be more than the neutral atom.