Q14.73 P - Chemistry: Molecular Nature Of Matter And Change

Question

Indium (In) reacts with HCl to form a diamagnetic solid with the formula $\mathit{I}\mathit{n}\mathit{C}{\mathit{l}}_{\mathbf{2}}$ . 

(a) Write condensed electron configurations for $\mathit{I}\mathit{n}\mathbf{,}\mathbf{ }\mathit{I}{\mathit{n}}^{\mathbf{+}}\mathbf{,}\mathbf{ }\mathit{I}{\mathit{n}}^{\mathbf{2}\mathbf{+}}$and $\mathit{I}{\mathit{n}}^{\mathbf{3}\mathbf{+}}$. 

(b) Which of these species is (are) diamagnetic and which paramagnetic? 

(c) What is the apparent oxidation state of In in $\mathit{I}\mathit{n}\mathit{C}{\mathit{l}}_{\mathbf{2}}$?

(d) Given your answers to parts (b) and (c), explain how $\mathit{I}\mathit{n}\mathit{C}{\mathit{l}}_{\mathbf{2}}$can be diamagnetic.

Step-by-Step Solution

Answer

a) $$\begin{gathered} In=\left[Kr\right]4d^{10}5s^{2}5p^1 \\ I{n}^{+}=\left[Kr\right]4d^{10}5s^2 \\ I{n}^{2+}=\left[Kr\right]4d^{10}5s^1 \\ I{n}^{3+}=\left[Kr\right]4d^{10} \end{gathered}$$

b) In and $I{n}^{2+}$ are paramagnetic while $I{n}^{+}$ and $I{n}^{3+}$ are diamagnetic.

c) +2

d) Indium occurs in +1 and +3 oxidation states in $InC{l}_2$.

Step 1: Condensed electronic configuration

In condensed electronic configuration, we write the neighbouring noble gas symbol within square brackets followed by the orbitals having valence electrons. 

Indium lies in the p-block. Its neighboring noble gas is Krypton and has thirteen valence electrons. When electrons are lost to form cations, indium loses electrons from p orbital first and then s orbital. 

So,

$$\begin{gathered} In=\left[Kr\right]4d^{10}5s^{2}5p^1 \\ I{n}^{+}=\left[Kr\right]4d^{10}5s^2 \\ I{n}^{2+}=\left[Kr\right]4d^{10}5s^1 \\ I{n}^{3+}=\left[Kr\right]4d^{10} \end{gathered}$$

Step 2: Magnetism

In and $I{n}^{2+}$are paramagnetic because both have unpaired electrons at p and s orbitals respectively. $I{n}^{+}$and $I{n}^{3+}$are diamagnetic because all the valence electrons are paired in both. 

Step 3: Apparent oxidation state of In in I n C l 2

Apparent oxidation state of In in $InC{l}_2$ is +2. Indium exists in two different oxidation states, i.e. +1 and +3, in which both are diamagnetic. So $InC{l}_2$ is represented as $I{n}^{I}I{n}^{III}C{l}_2$.