Chapter 4, Problem 45QP

Give the oxidation numbers for the underlined atoms in the following molecules and ions:

$\text{(a) }\underset{\_}{\text{Cl}}\text{F, (b) }\underset{\_}{\text{I}}{\text{F}}_{\text{7}}\text{, (c) }\underset{\_}{\text{C}}{\text{H}}_{\text{4}}\text{, (d) }{\underset{\_}{\text{C}}}_{\text{2}}{\text{H}}_{\text{2}}\text{, (e) }{\underset{\_}{\text{C}}}_{\text{2}}{\text{H}}_{\text{4}}{\text{, (f) K}}_{\text{2}}\underset{\_}{\text{Cr}}{\text{O}}_{\text{4}}{\text{, (g) K}}_{\text{2}}{\underset{\_}{\text{Cr}}}_{\text{2}}{\text{O}}_{\text{7}}\text{ (h) K}\underset{\_}{\text{Mn}}{\text{O}}_{\text{4}}\text{, (i) NaH}\underset{\_}{\text{C}}{\text{O}}_{\text{3}}\text{, (j) }{\underset{\_}{\text{Li}}}_{\text{2}}\text{, (k) Na}\underset{\_}{\text{I}}{\text{O}}_{\text{3}}\text{, (l) K}{\underset{\_}{\text{O}}}_{\text{2}}\text{, (m) }\underset{\_}{\text{P}}{\text{F}}_{\text{6}}^{\text{-}}\text{, (n) K}\underset{\_}{\text{Au}}{\text{Cl}}_{\text{4}}\text{.}$

Interpretation Introduction

Interpretation:

The oxidation number of the underlined atoms isto be determined.

Concept introduction:

The oxidation number can be assigned using the following rules:

In free state, an element is always in $0$ oxidation state.

For a neutral compound, oxidation numbers of atoms add upto $0$.

In a polyatomic ion, oxidation number of atoms add up to the charge on the ion.

The oxidation number of monoatomic ions is same as its charge.

Oxygen is usually in $-2$ oxidation state except in peroxides and in compounds with halogen.

Hydrogen is usually in $+1$ oxidation state except in metal hydrides.

Group 1 elements are always in $+1$ oxidation state and group 2 elements in $+2$.

Answer to Problem 45QP

Solution:

(a) $+1$

(b) +7

(c) $-4$

(d) $-1$

(e) $-2$

(f) $+6$

(g) $+6$

(h) $+7$

(i) $+4$

(j) $0$

(k) $+5$

(l) $-1/2$

(m) $+5$

(n) $+3$

Explanation of Solution

(a)

Let the oxidation number of Clbe $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of fluorine is $-1$ in all the compounds.

Hence, the oxidation number ofCl can be calculated as follows:

$\begin{array}{c}x+\left(-1\right)=0\\ x=+1\end{array}$

Thus, the oxidation number of Clis $+1$.

(b)

In ${\text{IF}}_7$, let the oxidation number of iodine be $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of fluorine is $-1$ in all the compounds.

Hence, the oxidation number ofiodine can be calculated as follows:

$\begin{array}{c}x+7\left(-1\right)=0\\ x=+7\end{array}$

Thus, the oxidation number of iodine is $+7$.

(c)

Let the oxidation number of carbonbe $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of hydrogen is $+1$.

Hence, the oxidation number ofcarbon can be calculated as follows:

$\begin{array}{c}x+4\left(+1\right)=0\\ x=-4\end{array}$

Thus, the oxidation number of carbonis $-4$.

(d)

Let the oxidation number of carbonbe $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of hydrogen is $+1$.

Hence, the oxidation number ofcarbon can be calculated as follows:

$\begin{array}{c}2x+2\left(+1\right)=0\\ x=-1\end{array}$

Thus, the oxidation number of carbon is $-1$.

(e)

Let the oxidation number of carbonbe $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of hydrogen is $+1$.

Hence, the oxidation number ofcarbon can be calculated as follows:

$\begin{array}{c}2x+4\left(+1\right)=0\\ x=-2\end{array}$

Thus, the oxidation number of carbon is $-2$.

(f)

Let the oxidation number of chromium (Cr)be $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of potassium is $+1$ and that for oxygen is $-2$.

Hence, the oxidation number ofchromium can be calculated as follows:

$\begin{array}{c}2\left(+1\right)+x+4\left(-2\right)=0\\ x=+6\end{array}$

Thus, the oxidation number of chromium is $+6$.

(g)

Let the oxidation number of chromium be $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of potassium is $+1$ and of oxygen is $-2$.

Hence, the oxidation number ofchromium can be calculated as follows:

$\begin{array}{c}2\left(+1\right)+2x+7\left(-2\right)=0\\ x=+6\end{array}$

Thus, the oxidation number of chromium is $+6$.

(h)

Let the oxidation number of manganese be $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of potassium is $+1$ and that for oxygen is $-2$.

Hence, the oxidation number ofmanganese can be calculated as follows:

$\begin{array}{c}1+x+4\left(-2\right)=0\\ x=+7\end{array}$

Thus, the oxidation number of manganese is $+7$.

(i)

Let the oxidation number of carbonbe $x$. The oxidation numbers must add up to zero for any molecule. Oxygen is assigned theoxidation number $-2$, hydrogen $+1$, and sodium is $+1$.

Hence, the oxidation number ofcarbon can be calculated as follows:

$\begin{array}{c}1+1+x+3\left(-2\right)=0\\ x=+4\end{array}$

Thus, the oxidation number of carbon is $+4$.

(j)

The oxidation number of any atom in its elemental form is zero.

Hence, the oxidation number of lithiumis $0$.

(k)

Let the oxidation number of iodinebe $x$. The oxidation numbers must add up to zero for any molecule. Oxygen is assigned an oxidation number of $-2$ and the oxidation number of sodiumis $+1$.

Hence, the oxidation number ofiodine can be calculated as follows:

$\begin{array}{c}1+x+3\left(-2\right)=0\\ x=+5\end{array}$

Thus, the oxidation number of iodine is $+5$.

(l)

Let the oxidation number of oxygenbe $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of Kis $+1$.

Hence, the oxidation number ofoxygen can be calculated as follows:

$\begin{array}{c}1+2x=0\\ x=-1/2\end{array}$

Thus, the oxidation number of oxygen is $-1/2$.

(m)

Let the oxidation number of phosphorus be $x$. For a polyatomic ion, the oxidation numbers must add up to the charge on the ion. The oxidation number of fluorine is $-1$.

Hence, the oxidation number ofphosphorus can be calculated as follows:

$\begin{array}{c}x+6\left(-1\right)=-1\\ x=+5\end{array}$

Thus, the oxidation number of phosphorus is $+5$.

(n)

Let the oxidation number of Aube $x$. The oxidation numbers must add up to zero for any molecule. The oxidation number of potassium is $+1$ and that for Cl is $-1$.

Hence, the oxidation number of Au can be calculated as follows:

$\begin{array}{c}1+x+4\left(-1\right)=0\\ x=+3\end{array}$

Thus, the oxidation number of Au is $+3$.