Chapter 14, Problem 14.117P

(a)

Interpretation Introduction

Interpretation:

Balanced chemical equation, oxidation state of Oxygen, Oxidizing agent and Reducing agent for each of the given reaction has to be given.

Concept-Introduction:

Oxidation reaction: The loss of electrons or the gain of oxygen atoms. And also increase their oxidation number.

Reduction reaction: Gaining electrons or adding hydrogen atoms. And also decrease their oxidation number.

Oxidation number: The total number of electrons in an atom after losing or gaining electrons to make a bond with another atom. It indicates the charge of an ion.

An oxidizing agent is a substance that oxidizes another substance in a chemical reaction and itself gets reduced.

A reducing agent is a substance that reduces another substance in a chemical reaction and itself gets oxidized.

Chemical equation is the representation of a chemical reaction, in which the reactants and products of the reactions are represented left and right side of an arrow respectively by using their respective chemical formulas.

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Explanation of Solution

• Gaseous $F_2$ and water reacts to give ${\text{HF and O}}_{\text{2}}$.

The equation is $F_2\left(g\right)+H_{2}O\left(l\right)\underset{}{\to }\text{ HF}\left(g\right)+O_2\left(g\right)$

The number of each atom on reactant and product side is not equal.  Hence, appropriate coefficient is given to make the number of each atom on both sides equal.  Therefore, balanced chemical equation can be given as,

  $2F_2\left(g\right)+\text{ 2}{H}_{2}O\left(l\right)\underset{}{\to }\text{ 4}\text{HF}\left(g\right)+O_2\left(g\right)$

Oxidation state of Fluorine and Oxygen is given blow,

  $\stackrel{0}{2F_2\left(g\right)}+\text{ 2}{H}_2\stackrel{2-}{O}\left(l\right)\underset{}{\to }\text{ 4}\text{H}\stackrel{1-}{\text{F}}\left(g\right)+{\stackrel{0}{O}}_2\left(g\right)$

The oxidation state of $O$ in $H_{2}O$ is $-2$  whereas in $O_2$ is $0$.

Oxidation number of $O$ is increased from $-2to0$.  Therefore, $F_2$ is the oxidizing agent.

Oxidation number of $F$ is decreases from $0to-1$ when reacting with $H_{2}O$.  Hence, $H_{2}O$ is the reducing agent.

• $F_2$ and $NaOH$ reacts to give ${\text{F}}^{-}{\text{, H}}_2{\text{O and OF}}_{\text{2}}.$.

The equation is $F_2\left(g\right)+NaOH\left(aq\right)\underset{}{\to }\text{ NaF}\left(aq\right)+H_{2}O\left(l\right){\text{ + OF}}_2\left(g\right)$

The number of each atom on reactant and product side is not equal.  Hence, appropriate coefficient is given to make the number of each atom on both sides equal.  Therefore, balanced chemical equation can be given as,

  $2F_2\left(g\right)+\text{ 2}NaOH\left(aq\right)\underset{}{\to }\text{ 2NaF}\left(aq\right)+H_{2}O\left(l\right){\text{ + OF}}_2\left(g\right)$

Oxidation state of Fluorine and Oxygen is given blow,

  $2{\stackrel{o}{F}}_2\left(g\right)+\text{ 2}Na\stackrel{2-}{O}H\left(aq\right)\underset{}{\to }\text{ 2Na}\stackrel{1-}{\text{F}}\left(aq\right)+H_2\stackrel{2-}{O}\left(l\right)\text{ + }\stackrel{2+}{\text{O}}{\stackrel{1-}{\text{F}}}_2\left(g\right)$

The oxidation state of $O$ in $H_{2}O$ and $NaOH$ is $-2$  whereas in $O{F}_2$ is $+2$.

Oxidation number of $O$ is increased from $-2to\text{ +2}$.  Therefore, $F_2$ is the oxidizing agent.

Oxidation number of $F$ is decreases from $0to-1$ when reacting with $NaOH$.  Hence, $NaOH$ is the reducing agent.

• $O{F}_2$ and excess $O{H}^{-}$ reacts to give ${\text{F}}^{-}{\text{, H}}_2{\text{O and O}}_{\text{2}}.$.

The equation is $O{F}_2\left(g\right)+O{H}^{-}\left(aq\right)\underset{}{\to }{\text{ F}}^{-}\left(aq\right)+H_{2}O\left(l\right){\text{ + O}}_2\left(g\right)$

The number of each atom on reactant and product side is not equal.  Hence, appropriate coefficient is given to make the number of each atom on both sides equal.  Therefore, balanced chemical equation can be given as,

  $O{F}_2\left(g\right)+\text{ 2}O{H}^{-}\left(aq\right)\underset{}{\to }{\text{ O}}_2\left(g\right)+H_{2}O\left(l\right){\text{ + 2F}}^{-}\left(aq\right)$

Oxidation state of Fluorine and Oxygen is given blow,

  $\stackrel{2+}{O}{\stackrel{1-}{F}}_2\left(g\right)+\text{ 2}\stackrel{2-}{O}{H}^{-}\left(aq\right)\underset{}{\to }{\stackrel{0}{\text{O}}}_2\left(g\right)+H_2\stackrel{-2}{O}\left(l\right){\text{ + 2F}}^{-}\left(aq\right)$

The oxidation state of $O$ in $H_{2}O$ and $O{H}^{-}$ is $-2$  whereas in $O{F}_2$ is $+2$.

Here, $O{F}_2$ is the oxidizing agent whereas $O{H}^{-}$ is the reducing agent.

(b)

Interpretation Introduction

Interpretation:

Lewis structure of $O{F}_2$ has to be determined also the molecular shape of it has to be predicted.

Concept-Introduction:

Lewis structure

Electron dot structure also known as Lewis dot structure represents the number of valence electrons of an atom or constituent atoms bonded in a molecule.  Each dot corresponds to one electron.

According to VSEPR theory, the geometry is predicted by the minimizing the repulsions between electron-pairs in the bonds and lone-pairs of electrons.  The VSEPR theory is summarized in the given table as,

  $\begin{array}{cccc}\text{Electron-pair}& \text{lone-pair}& \text{Electron-pair}\text{geometry}& \text{Molecular}\text{shape}\\ 2& 0& \text{Linear}& \text{Linear}\\ 3& 0& \text{Trigonal}\text{planar}& \text{Trigonal}\text{planar}\\ 2& 1& \text{Trigonal}\text{planar}& \text{Bent}\\ 4& 0& \text{Tetrahedral}& \text{Tetrahedral}\\ 3& 1& \text{Tetrahedral}& \text{Pyramidal}\\ 2& 2& \text{Tetrahedral}& \text{V}-\text{shape}\\ 5& 0& \text{Trigonal}\text{bipyramidal}& \text{Trigonal}\text{bipyramidal}\\ 4& 1& \text{Trigonal}\text{bipyramidal}& \text{See}-\text{saw}\\ 3& 2& \text{Trigonal}\text{bipyramidal}& \text{T}-\text{shape}\\ 2& 3& \text{Trigonal}\text{bipyramidal}& \text{Linear}\\ 6& 0& \text{Octahedral}& \text{Octahedral}\\ 5& 1& \text{Octahedral}& \text{Square}\text{pyramidal}\\ 4& 2& \text{Octahedral}& \text{Square}\text{planar}\end{array}$

Explanation of Solution

The Lewis electron dot structure for given molecules are determined by first drawing the skeletal structure for the given molecules, then the total number of valence electrons for all atoms present in the molecules are determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

Outer valence electrons of Fluorine and Oxygen are seven and six respectively.

  $\begin{array}{l}O\text{ = 1}\times \text{6 = 6}\\ \text{F = 2}\times \text{7 = 14}\\ Total{\text{ e}}^{-}\text{ = 6 + 14 = 20}\\ \text{2 Bonds = }\left(2\times 2\right)=4\\ {\text{Remaining e}}^{-}\text{ = 20}-4=16\end{array}$

After the distribution of electrons, each terminal fluorine atom gets 3 pairs of electrons (30 electrons) and the central Oxygen atom gets two lone pair of electrons.

The Lewis structure of $O{F}_2$ follows as,

Oxygen has 2 bond pair and 2 lone pairs (4 electron domains).  Therefore, the molecular geometry of $O{F}_2$ is bent.