Chapter 3, Problem 3.85P

Interpretation Introduction

(a)

Interpretation:

The expected coordination number for ${\text{Y}}_{\text{2}}{\text{O}}_{\text{3}}$ compound is determined using the ionic radii in given B appendix.

Concept Introduction:

The ${\text{Y}}_{\text{2}}{\text{O}}_{\text{3}}$ is the yttrium oxide which is stable in the atmosphere and is a white substance that is in the solid form. In inorganic compounds and science the yttrium oxide is used as the starting material. Here, in the compound, cation is ${\text{Y}}^{3+}$ and anion is ${\text{O}}^{2-}$ thus, according to the formula there will be 2 cations and 3 anions. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

6 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{Y}}^{+3}}=0.89\stackrel{\text{0}}{\text{A}}\\ r_{{\text{O}}^{-2}}=1.32\stackrel{\text{0}}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{Y}}^{+3}}}{r_{{\text{O}}^{-2}}}=\frac{\text{0}\text{.89}\stackrel{\text{0}}{\text{A}}}{\text{1}\text{.32}\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.67$

From the table the coordination number of ${\text{Y}}_{\text{2}}{\text{O}}_{\text{3}}$ is given as 6

Interpretation Introduction

(b)

Interpretation:

The expected coordination number for UO₂ compound is determined using the ionic radii in given B appendix.

Concept Introduction:

Uranium dioxide also called as Urania is black in color. This element is radioactive and powdered crystalline and it is found in the uraninite mineral. From the formula, UO₂the cation is ${\text{U}}^{4+}$ and anion is ${\text{O}}^{2-}$. The number of cations is 1 and that of anion is 2. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

6 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\mathrm{U}}^{+4}}=0.97\stackrel{0}{A}\\ r_{{\text{O}}^{-2}}=1.3\text{2}\stackrel{\text{0}}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{U}}^{+4}}}{r_{{\text{O}}^{-2}}}=\frac{\text{0}\text{.97}\stackrel{\text{0}}{\text{A}}}{\text{1}\text{.32}\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.73$

From the table the coordination number of UO₂is given as 6

Interpretation Introduction

(c)

Interpretation:

The expected coordination number for BaO compound is determined using the ionic radii in given B appendix.

Concept Introduction:

The compound Barium oxide is non-flammable. It is white in color and the structure of the compound is cube. It is used in the crown glass, cathode ray tubes and catalyst. From the formula there will be 1 cation that is ${\mathrm{Ba}}^{2+}$ and 1 anion that is ${\text{O}}^{2-}$. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix.

Answer to Problem 3.85P

8 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{Ba}}^{+2}}=1.34\stackrel{\text{0}}{\text{A}}\\ r_{{\text{O}}^{-2}}=1.32\stackrel{\text{0}}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{Ba}}^{+2}}}{r_{{\text{O}}^{-2}}}=\frac{\text{1}\text{.34}\stackrel{\text{0}}{\text{A}}}{\text{1}\text{.32}\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.99$

From the table the coordination number of BaO is given as 8

Interpretation Introduction

(d)

Interpretation:

The expected coordination number for ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ compound is determined using the ionic radii in given B appendix.

Concept Introduction:

The compound silicon nitride is a compound that is made of nitrogen and silicon. This chemical compound is stable thermodynamically and it is of commercial importance. According to the formula, there will be 3 ${\text{Si}}^{4+}$ and 4 ${\text{N}}^{3-}$ ions. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

4 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{N}}^{-3}}=0.15\stackrel{\text{0}}{\text{A}}\\ r_{{\text{Si}}^{+4}}=0.42\stackrel{0}{A}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{N^{-3}}}{r_{S{i}^{+4}}}=\frac{0.15\stackrel{0}{A}}{0.42\stackrel{0}{A}}\Rightarrow 0.36$

From the table the coordination number of ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$ is given as 4

Interpretation Introduction

(e)

Interpretation:

The expected coordination number for ${\mathrm{GeO}}_2$ compound is determined using the ionic radii in given B appendix.

Concept Introduction:

  ${\text{GeO}}_{\text{2}}$ Compound is also called as germanium oxide which is the germanium salt. This compound is inorganic and it is main source for commercial. With contact of pure germanium the compound forms layer of passivation. From the formula of the compound, the anion and cation will be ${\text{O}}^{2-}$ and ${\text{Ge}}^{4+}$ respectively. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

4 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{Ge}}^{+4}}=0.53\stackrel{\text{0}}{\text{A}}\\ r_{{\text{O}}^{-2}}=1.32\stackrel{\text{0}}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{Ge}}^{\text{+4}}}}{r_{{\text{O}}^{-2}}}=\frac{\text{0}\text{.53}\stackrel{\text{0}}{\text{A}}}{\text{1}\text{.32}\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.40$

From the table the coordination number of $Ge{O}_2$ is given as 4

Interpretation Introduction

(f)

Interpretation:

The expected coordination number for $\text{MnO}$ compound is determined using the ionic radii in given B appendix.

Concept Introduction:

The $\text{MnO}$ compound is inorganic and green crystals are formed by them. It acts as the food additives and fertilizer on a large scale. From the formula, the cation and anion will be ${\text{Mn}}^{2+}$ and ${\text{O}}^{2-}$.

The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

6 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{Mn}}^{+2}}=0.80\stackrel{\text{0}}{\text{A}}\\ r_{{\text{O}}^{-2}}=1.32\stackrel{\text{0}}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{Mn}}^{+2}}}{r_{{\text{O}}^{-2}}}=\frac{0.80\stackrel{0}{\text{A}}}{1.32\stackrel{0}{\text{A}}}\Rightarrow 0.61$

From the table B in Appendix, the coordination number of $\text{MnO}$ is given as 6

Interpretation Introduction

(g)

Interpretation:

The expected coordination number for MgS compound is determined using the ionic radii in given B appendix.

Concept Introduction:

The MgS compound is inorganic and white in color. It accumulates in the impure form and it is made of crystalline material. In the impure form it is in the brown color and the material becomes non crystalline powder. From the formula, the cation is ${\text{Mg}}^{2+}$ and anion is ${\text{S}}^{-}$. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix

Answer to Problem 3.85P

6 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{Mg}}^{+2}}=0.66\stackrel{0}{\text{A}}\\ r_{{\text{S}}^{-2}}=1.32\stackrel{0}{\text{A}}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{Mg}}^{+2}}}{r_{{\text{S}}^{-2}}}=\frac{0.66\stackrel{\text{0}}{\text{A}}}{1.32\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.50$

From the table the coordination number of MgS is given as 6

Interpretation Introduction

(h)

Interpretation:

The expected coordination number for KBr compound is determined using the ionic radii in given B appendix.

Concept Introduction:

Potassium bromide is the salt of ions and in the aqueous solution has pH 7 where it is completely dissociated. This compound is the bromide ions source. From the formula, the cation is ${\text{K}}^{+}$ and anion is ${\text{Br}}^{-}$. The ratio of radius of cation to anion can helps in determining the coordination number of the compound from table B in Appendix.

Answer to Problem 3.85P

6 is the coordination number.

Explanation of Solution

For the selected elements the radii of atoms and ions from the table are given as,

  $\begin{array}{l}{r}_{{\text{K}}^{+1}}=1.33\stackrel{\text{0}}{\text{A}}\\ r_{{\text{Br}}^{-1}}=1.96\stackrel{0}{A}\end{array}$

The ratio of the radius is given as,

  $\frac{r_{{\text{K}}^{+1}}}{r_{{\text{Br}}^{-1}}}=\frac{\text{1}\text{.33}\stackrel{\text{0}}{\text{A}}}{\text{1}\text{.96}\stackrel{\text{0}}{\text{A}}}\Rightarrow 0.68$

From the table the coordination number of KBr is given as 6