Chapter 3, Problem 3I.13E

Interpretation Introduction

Interpretation:

Lewis structure of orthosilicate anion $\left({\text{SiO}}_4^{4-}\right)$ has to be drawn. Formal charges and oxidation number of atoms in ${\text{SiO}}_4^{4-}$ anion has to be determined. Also, shape of ${\text{SiO}}_4^{4-}$ anion has to be determined.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

• The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
• Total valence electrons are estimated.
• single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
• Add charge on overall structure in case of polyatomic cation or anion.

Hybridization is calculated by the hybrid orbitals and to calculate hybrid orbitals we need to know the steric number that is given by,

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{number of atoms bonded to the central atom}\right)+\\ \left(\text{number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The relation of steric number with geometry or shape is depicted in the table below.

$\begin{array}{cccc}\text{Steric Number}& \text{Number of lone pairs}& \text{Molecular Geometry}& \text{Shape}\\ 2& 0& \text{linear}& \text{linear}\\ 3& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}\text{trigonal planar}\\ \text{trigonal planar}\end{array}& \begin{array}{c}\text{trigonal planar}\\ \text{bent}\end{array}\\ 4& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{tetrahedral}\\ \text{tetrahedral}\\ \text{tetrahedral}\end{array}& \begin{array}{c}\text{tetrahedral}\\ \text{trigonal pyramidal}\\ \text{bent}\end{array}\\ 5& \begin{array}{c}0\\ 1\\ 2\\ 3\end{array}& \begin{array}{c}\text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\\ \text{trigonal bipyramidal}\end{array}& \begin{array}{c}\text{trigonal bipyramidal}\\ \text{see-saw}\\ \text{T-shaped}\\ \text{linear}\end{array}\\ 6& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{octahedral}\\ \text{octahedral}\\ \text{octahedral}\end{array}& \begin{array}{c}\text{octahedral}\\ \text{square pyramidal}\\ \text{square planar}\end{array}\end{array}$

Explanation of Solution

The molecule ${\text{SiO}}_4^{4-}$ consists of one $\text{Si}$ and four $\text{O}$ atoms. Since $\text{S}$ is least electronegative and is drawn as central atom.

The symbol for silicon is $\text{Si}$ with atomic number 14. Its electronic configuration is $\left[\text{Ne}\right]{\text{3s}}^2{\text{3p}}^2$. Thus, it possesses 4 valence electrons.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

Four negative charges are added up in the total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{SiO}}_4^{4-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4+6\left(4\right)+4\\ =32\left(16\text{ pairs}\right)\end{array}$

The skeleton structure ${\text{SiO}}_4^{4-}$ has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-8\\ =24\left(\text{12 pairs}\right)\end{array}$

Hence, 24 electrons are allocated as 3 lone pairs on each oxygen atom and total number of electrons about silicon is 8. The Lewis structure of ${\text{SiO}}_4^{4-}$ is as follows:

As from the structure above, there are 4 bond pairs and 0 lone pairs on central atom.

Substitute 4 for number of atoms bonded to central atom and 0 for number of lone pairs on central atom in equation (1) to determine steric number.

$\begin{array}{c}\text{Steric number}=4+0\\ =4\end{array}$

The shape of the molecule is defined by the lone pairs present on the central atom. Anion ${\text{SiO}}_4^{4-}$ has steric number 4 so the electronic arrangement and shape is tetrahedral.

The formal charges on each atom in Lewis structure can be calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (2)

Here,

$V$ denotes valence electrons in free atom.

$L$ denotes electrons present as lone pairs.

$B$ denotes electrons present as bond pairs.

Substitute 4 for $V$, 0 for $L$ and 8 for $B$ in equation (2) to calculate formal charge on central silicon atom.

  $\begin{array}{c}\text{Formal charge}\left(\text{Si}\right)=4-0-\frac{1}{2}\left(8\right)\\ =0\end{array}$

Substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (2) to calculate formal charge on each oxygen atom.

  $\begin{array}{c}\text{Formal charge}\left(\text{O}\right)=6-6-\frac{1}{2}\left(2\right)\\ =-1\end{array}$

Hence formal charge on silicon is 0 and same on each oxygen atom is $-1$.

Expression to determine oxidation number for ${\text{SiO}}_4^{4-}$ is as follows:

  $\left[\left(\text{oxidation number of Si}\right)+4\left(\text{oxidation number of O}\right)\right]=-4$        (3)

Rearrange equation (3) to determine oxidation number of silicon.

  $\text{Oxidation number of Si}=-4-4\left(\text{oxidation number of O}\right)$        (4)

Substitute $-2$ for oxidation number of $\text{O}$ in Equation (4).

  $\begin{array}{c}\text{Oxidation number of Si}=-4-4\left(-2\right)\\ =+4\end{array}$

The oxidation state of silicon is $+4$ and for oxygen is $-2$.