Chapter 7, Problem 131AP

Write the ground-state electron configurations of the following ions, which play important roles in biochemical processes in our bodies: $\left(\text{a}\right){\text{ Na}}^{\text{+}}\text{, }\left(\text{b}\right){\text{ Mg}}^{\text{2+}}\text{, }\left(\text{c}\right){\text{ Cl}}^{\text{-}}\text{, }\left(\text{d}\right){\text{ K}}^{\text{+}}\text{, }\left(\text{e}\right){\text{ Ca}}^{\text{2+}}\text{, }\left(\text{f}\right){\text{ Fe}}^{\text{2+}}\text{, }\left(\text{g}\right){\text{ Cu}}^{\text{2+}}\text{, }\left(\text{h}\right){\text{ Zn}}^{\text{2+}}$.

Interpretation Introduction

Interpretation: The ground state electronic configuration of the given ions is to be written.

Concept introduction:

Elements that lose their valence electron(s) are called cations.

Elements that gain electron(s)from another atom are called anions.

Representative elementstypically tend to lose or gain electrons to attain a stable noble gas electronic configuration.

Answer to Problem 131AP

Solution:

a) $[\text{Ne}]$

b) $[\text{Ne}]$

c) $[\text{Ar}]$

d) $[\text{Ar}]$

e) $[\text{Ar}]$

f) $[\text{Ar}]3{\text{d}}^{\text{6}}$

g) $[\text{Ar}]3{\text{d}}^{\text{9}}$

h) $[\text{Ar}]3{\text{d}}^{10}$

Explanation of Solution

a) ${\text{Na}}^{\text{+}}$

Sodium ion $\left({\text{Na}}^{+}\right)$ forms when sodium $\left(\text{[Ne]}3s^1\right)$ lose one electron.

So, the electronic configuration of sodium $\left({\text{Na}}^{+}\right)$ ion will be $\text{1}{s}^2\text{2}{s}^{2}2p^6$ or $\text{[Ne]}$.

Therefore, the electronic configuration of sodium ion is $\text{[Ne]}$.

b) ${\text{Mg}}^{\text{2+}}$

Magnesium ion $\left({\text{Mg}}^{2+}\right)$ forms when magnesium $\left(\text{[Ne]}3s^2\right)$ lose two electrons.

So, the electronic configuration of magnesium ion $\left({\text{Mg}}^{2+}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^6$ or $\text{[Ne]}$.

Therefore, the electronic configuration of magnesium ion is $\text{[Ne]}$.

c) ${\text{Cl}}^{-}$

Chlorine ion $\left({\text{Cl}}^{-}\right)$ forms when chlorine $\left(\text{[Ne]}3s^{2}3p^5\right)$ gain one electron.

So, the electronic configuration of chlorine ion $\left({\text{Cl}}^{-}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^6$ or $\text{[Ar]}$.

Therefore, the electronic configuration of chlorine ion is $\text{[Ar]}$.

d) ${\text{K}}^{+}$

Potassium ion $\left({\text{K}}^{+}\right)$ forms when potassium $\left(\text{[Ar]4}{s}^1\right)$ lose one electron.

So, the electronic configuration of potassium ion $\left({\text{K}}^{+}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^6$ or $\text{[Ar]}$.

Therefore, the electronic configuration of potassium ion is $\text{[Ar]}$.

e) ${\text{Ca}}^{2+}$

Calcium ion $\left({\text{Ca}}^{2+}\right)$ forms when calcium $\left(\text{[Ar]4}{s}^2\right)$ lose two electrons.

So, the electronic configuration of calcium ion $\left({\text{Ca}}^{2+}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^6$ or $\text{[Ar]}$.

Therefore, the electronic configuration of calcium ion is $\text{[Ar]}$.

f) ${\text{Fe}}^{2+}$

Ferrate ion $\left({\text{Fe}}^{2+}\right)$ forms when iron $\left(\text{[Ar]4}{s}^{2}3d^6\right)$ lose two electrons.

So, the electronic configuration of ferrate ion $\left({\text{Fe}}^{2+}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^{6}3d^6$ or $\text{[Ar]3}{d}^6$.

Therefore, the electronic configuration of ferrate ion is $\text{[Ar]3}{d}^6$.

g) ${\text{Cu}}^{\text{2+}}$

Cuprous ion $\left({\text{Cu}}^{\text{2+}}\right)$ forms when copper $\left(\text{[Ar]4}{s}^{2}3d^9\right)$ lose two electrons.

So, the electronic configuration of cuprous ion $\left({\text{Cu}}^{\text{2+}}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^{6}3d^9$ or $\text{[Ar]3}{d}^9$.

Therefore, the electronic configuration of cuprous ion is $\text{[Ar]3}{d}^9$.

h) ${\text{Zn}}^{2+}$

Zinc ion $\left({\text{Zn}}^{\text{2+}}\right)$ forms when zinc $\left(\text{[Ar]4}{s}^{2}3d^{10}\right)$ lose two electrons.

So, the electronic configuration of zinc ion $\left({\text{Zn}}^{\text{2+}}\right)$ will be $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^{6}3d^{10}$ or $\text{[Ar]3}{d}^{10}$.

Therefore, the electronic configuration of zinc ion is $\text{[Ar]3}{d}^{10}$.