Chapter 8, Problem 8.105QE

Interpretation Introduction

Interpretation:

The masses of iron and ${\text{Cl}}_2$ needed to prepare $7.88\text{ g}$ of the metal halide have to be determined.

Concept Introduction:

The electronic configuration is defined as the distribution of electrons in various atomic orbitals of the atom. The electrons that are present in an outermost orbital are known as valence electrons whereas those present in the orbitals with lower quantum numbers are called core electrons. The general outer electronic configuration of $s\text{ block}$ elements is $n{s}^{1-2}$, that of $p\text{ block}$ elements is $n{s}^{2}n{p}^{1-6}$, that of $d\text{ block}$ elements is $\left(n-1\right)d^{1-10}n{s}^{0-2}$ and that of $f\text{ block}$ elements is $\left(n-2\right)f^{1-14}\left(n-1\right)d^{0-10}n{s}^2$.

Electrons are filled in orbitals in accordance with three rules: Aufbau principle, Hund’s rule, and Pauli’s exclusion principle. Aufbau principle states that electrons are filled in the orbitals from lower to higher energy level as follows:

  $1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s$

Hund’s rule states that initially each orbital is singly occupied and then pairing occurs and Pauli’s exclusion principle states that the spin of two electrons in one orbital is always different.

Answer to Problem 8.105QE

$3.47\text{ g}$ of $\text{Fe}$ and $4.41\text{ g}$ of ${\text{Cl}}_{\text{2}}$ is needed to prepare $7.88\text{ g}$ of metal halide $\left({\text{FeCl}}_{\text{2}}\right)$.

Explanation of Solution

The atomic number of iron is 26 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$. The configuration for its cation is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$. This clearly indicates that iron cation is formed by the removal of two electrons from the neutral atom. So the given cation of iron is ${\text{Fe}}^{2+}$. So the metal halide formed is ${\text{FeCl}}_{\text{2}}$. The reaction between iron and ${\text{Cl}}_2$ occurs as follows:

  $\text{Fe}+{\text{Cl}}_2\to {\text{FeCl}}_{\text{2}}$

According to the balanced chemical equation, one mole of $\text{Fe}$ reacts with one mole of ${\text{Cl}}_2$ to form one mole of ${\text{FeCl}}_{\text{2}}$. So the stoichiometric ratio between $\text{Fe}$ and ${\text{FeCl}}_2$ is $1:1$ and that between ${\text{Cl}}_2$ and ${\text{FeCl}}_{\text{2}}$ is also $1:1$.

$126.75\text{ g}$ of ${\text{FeCl}}_{\text{2}}$ is formed by $55.85\text{ g}$ of $\text{Fe}$. The mass of $\text{Fe}$ needed to form $7.88\text{ g}$ of ${\text{FeCl}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}\text{Mass of Fe}=\left(\frac{55.85\text{ g}}{126.75\text{ g}}\right)\left(7.88\text{ g}\right)\\ =3.472\text{ g}\\ \approx 3.47\text{ g}\end{array}$

$126.75\text{ g}$ of ${\text{FeCl}}_{\text{2}}$ is formed by $70.90\text{ g}$ of ${\text{Cl}}_2$. The mass of ${\text{Cl}}_2$ needed to form $7.88\text{ g}$ of ${\text{FeCl}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Mass of Cl}}_{\text{2}}=\left(\frac{70.90\text{ g}}{126.75\text{ g}}\right)\left(7.88\text{ g}\right)\\ =4.407\text{ g}\\ \approx \text{4}\text{.41 g}\end{array}$