Q. Fe$\backslash ,$3+ electron configuration.

We want the electron configuration of the iron(III) ion, \( \mathrm{Fe^{3+}} \).

Step 1: Start from neutral iron.

Neutral iron has atomic number \(26\), so \( \mathrm{Fe} \) has \(26\) electrons.

The neutral electron configuration (filling in order of increasing energy) is:

\[
\mathrm{Fe:}\ 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^6
\]

Step 2: Determine how many electrons are removed for \( \mathrm{Fe^{3+}} \).

The ion \( \mathrm{Fe^{3+}} \) has a charge of \(+3\), meaning it has lost \(3\) electrons compared to neutral iron.

So the total electron count becomes:

\[
26 – 3 = 23
\]

Step 3: Remove electrons from the highest-energy occupied subshell.

For neutral iron, the outer/valence electrons are in the \(4s\) and \(3d\) subshells, specifically \(4s^2\ 3d^6\).

When forming \( \mathrm{Fe^{3+}} \), electrons are removed first from the \(4s\) subshell.

\[
\mathrm{Fe:}\ 4s^2\ 3d^6
\]

Remove \(3\) electrons total:

– Remove \(2\) electrons from \(4s\) (emptying \(4s\) first)

– Remove the remaining \(1\) electron from \(3d\)

Step 4: Write the electron configuration for \( \mathrm{Fe^{3+}} \).

After removing \(3\) electrons from the neutral configuration:

– \(4s^2\) becomes \(4s^0\) (so \(4s\) is no longer included)

– \(3d^6\) becomes \(3d^5\)

Thus, the electron configuration of \( \mathrm{Fe^{3+}} \) is:

\[
\mathrm{Fe^{3+}:}\ 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5
\]

Final answer:

\[
\boxed{\mathrm{Fe^{3+}:}\ 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5}
\]