Question

A d-block element iron ${\text{(Fe)}}$ forms two types of compounds,
\[{\text{FeS}}{{\text{O}}_{\text{4}}}\] and ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$.
A. Which of these compounds contains ferric ${\text{(F}}{{\text{e}}^{{\text{3 + }}}}{\text{)}}$ ion?
B. What is the number of electrons present in the d-subshell of ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ ion? Find out by writing the sub-shell electronic configuration.
[Given: Atomic number of ${\text{Fe}}$ is ${\text{26}}$]

Answer 1

Hint:
A. The two compounds, \[{\text{FeS}}{{\text{O}}_{\text{4}}}\] and ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ are ionic compounds and dissociate into individual ions when dissolved in water. The net charge on an ionic compound is always zero.
On dissociation, ${\text{SO}}_4^{2 - }$ is produced. ${\text{SO}}_4^{2 - }$ is a polyatomic ion having charge $ - 2$.
B. While writing the electronic configuration, the order of filling the electrons is: $1s\,2s\,2p\,3s\,3p\,3d$. The $s$ subshell can hold $2$ electrons, $p$ can hold $6$ electrons and $d$ can hold ${\text{10}}$ electrons.

Complete step by step answer:
A. Step 1:
The dissociation reaction of \[{\text{FeS}}{{\text{O}}_{\text{4}}}\] is as follows:
\[{\text{FeS}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{F}}{{\text{e}}^{x{\text{ + }}}} + {\text{SO}}_{\text{4}}^{2 - }\]
The net charge on the ionic compound is always zero.
Thus, to balance the $ - 2$ charge on the ${\text{SO}}_4^{2 - }$ ion, the \[{\text{F}}{{\text{e}}^{x{\text{ + }}}}\] ion must be present as a cation with $ + 2$ charge.
Thus, \[{\text{FeS}}{{\text{O}}_{\text{4}}}\] contains the iron ion ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ (ferrous).
Step 2:
The dissociation reaction of ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ is as follows:
\[{\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}} \rightleftharpoons 2{\text{F}}{{\text{e}}^{x{\text{ + }}}} + 3{\text{SO}}_{\text{4}}^{2 - }\]
The net charge on the ionic compound is always zero.
There are $3$ ions of ${\text{SO}}_4^{2 - }$ having charge $ - 2$. Thus, a total of $3 \times \left( { - 2} \right) = - 6$ negative charges are present.
Thus, to balance the $ - 6$ charge, the \[{\text{2F}}{{\text{e}}^{x{\text{ + }}}}\] ion must contribute $ + 3$ charge each.
Thus, ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ contains ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ (ferric) ion.
Thus, out of the two types of compounds, \[{\text{FeS}}{{\text{O}}_{\text{4}}}\] and ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$.formed by iron ${\text{(Fe)}}$, ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ contains ferric ${\text{(F}}{{\text{e}}^{{\text{3 + }}}}{\text{)}}$ ion.

B.Step 1:
The atomic number of ${\text{Fe}}$ is ${\text{26}}$. Thus, the number of electrons in ${\text{Fe}}$ ${\text{26}}$.
The $s$ subshell can hold $2$ electrons, $p$ can hold $6$ electrons and $d$ can hold ${\text{10}}$ electrons.
Thus, the electronic configuration of ${\text{Fe}}$ is:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^8}$
${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ ion suggests that $3$ electrons are removed from the outermost ${\text{3}}d$ subshell. Thus, the electronic configuration of ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ is:
$1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^5}$
Step 2:
The electronic configuration of ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ is $1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^5}$.
Thus, the d-subshell $3{d^5}$ contains ${\text{5}}$ electrons.

Note: The net charge on an ionic compound is always zero. The charge on the polyatomic ion ${\text{SO}}_4^{2 - }$ ion is $2 - $. Remember in the compound ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ there are three ${\text{SO}}_4^{2 - }$ ions which comprise a total of $ - 6$ negative charge.

The $3 + $ charge on the ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ ion suggests that $3$ electrons are removed from the outermost subshell. On removal of electrons, the neutral ${\text{Fe}}$ atom gains positive charge.