Chapter 2, Problem 64QP

Interpretation Introduction

Interpretation:

The chemical formula for the given compounds isto be written.

Concept introduction:

The naming of a chemical compound is based on the charges of the cations and anions. For naming ionic compounds, the name of the cation is written after the name of ananion.

To form a compound from the given anion and cation, first identify the charges on thecation and the anion, as chargestransfer to the subscript of the other ion, which means the charge on the cation transfers to the subscript of the anion and the charge of the anion goes to the subscript of the cation.

Answer to Problem 64QP

Solution:

(a) $\text{CuCN}$

(b) $\text{Sr}{\left({\text{ClO}}_{\text{2}}\right)}_{\text{2}}$

(c) ${\text{HBrO}}_{\text{4}}$

(d) $\text{HI}$

(e) ${\text{Na}}_{\text{2}}\left({\text{NH}}_{\text{4}}\right){\text{PO}}_{\text{4}}$

(f) ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$

(g) ${\text{IF}}_{\text{7}}$

(h) ${\text{P}}_{\text{4}}{\text{S}}_{\text{10}}$

(i) $\text{HgO}$

(j) ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$

(k) ${\text{SeF}}_{\text{6}}$

Explanation of Solution

a) Copper (I) cyanide

The copper (I) cation has a charge of $+1$ always, and the cyanide anion, ${\text{CN}}^{\text{-}}$, has a charge of $-1$.

So, the compoundcopper (I) cyanide is formed from the cation ${\text{Cu}}^{\text{+}}$ and the anion ${\text{CN}}^{\text{-}}$, the charge of copper (I) transfers to the subscript of cyanide and charge of cyanide goes to the subscript of copper (I) and the ions combine in a ratio of one-to-one. Therefore, the correct formula for compoundcopper (I) cyanideis $\text{CuCN}$.

Hence, the formula for copper (I) cyanide is $\text{CuCN}$.

b) Strontium Chlorite

The strontium cation has charge of $+2$ and chlorite anion has charge of $-1$.

So, the compoundstrontium chlorite is formed from the cation ${\text{Sr}}^{\text{+2}}$ and the anion ${\text{ClO}}_{\text{2}}^{\text{-}}$. The charge of strontium goes to the subscript of chlorite and the charge of chlorite transfers to the subscript of strontium and the correct formula of the compound strontium chlorite is $\text{Sr}{\left({\text{ClO}}_{\text{2}}\right)}_{\text{2}}$.

Hence, the formula for strontium chlorite is $\text{Sr}{\left({\text{ClO}}_{\text{2}}\right)}_{\text{2}}$.

c)Perbromic acid

The compoundperbromic acidis formed from bromium oxide and hydrogen. Therefore, formula ofperbromic acidis ${\text{HBrO}}_{\text{4}}$.

Hence, the formula for perbromic acidis ${\text{HBrO}}_{\text{4}}$.

d) Hydroiodic acid

The hydrogen cation has a charge of $+1$ always, and the iodine anion has a charge of $-1$.

So, the compoundhydroiodic acidis formed from cation ${\text{H}}^{\text{+}}$ and anion ${\text{I}}^{\text{-}}$. The charge of hydrogentransfers to the subscript of iodine and charge of iodine goes to the subscript of hydrogen, which makes the formula of the compoundhydroiodic acidto be $\text{HI}$.

Hence, the formula for hydroiodic acidis $\text{HI}$.

e) Disodium ammonium phosphate

The compounddisodium ammonium phosphate is formed fromsodiumNa+ cation and ammonium phosphate ${\text{(NH4)PO4}}^{\text{2-}}$ anion. There is a disodium ion present, which means there are two sodium atoms and ammonium phosphate present, which makes the formula of the compounddisodium ammonium phosphateto be ${\text{Na}}_{\text{2}}\left({\text{NH}}_{\text{4}}\right){\text{PO}}_{\text{4}}$.

Hence, the formula for disodium ammonium phosphateis ${\text{Na}}_{\text{2}}\left({\text{NH}}_{\text{4}}\right){\text{PO}}_{\text{4}}$.

f) Potassium dihydrogen phosphate

The compoundpotassium dihydrogen phosphateis formed from the potassium ${\text{K}}^{\text{+}}$ cationand the hydrogen phosphate anion. The formula of the compoundpotassium dihydrogen phosphateis ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$.

Hence, the formula for potassium dihydrogen phosphateis ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$.

g) Iodine heptafluoride

The compoundiodine heptafluoride is formed from iodine and fluorine. There is an iodine atom present and heptafluoride means there are seven fluoride atoms present, which makes the formula of the compoundiodine heptafluorideto be ${\text{IF}}_{\text{7}}$.

Hence, the formula for iodine heptafluorideis ${\text{IF}}_{\text{7}}$.

h)Tetra phosphorous decasulfide.

The compoundtetra phosphorous decasulfideis formed from the phosphorus cation and the sulphur anion. There are tetra phosphorus present, which means there are four phosphorus atoms and decasulfide means there are10sulfur atoms are present, which makes the formula of the compoundtetra phosphorous decasulfideto be ${\text{P}}_{\text{4}}{\text{S}}_{\text{10}}$.

Hence, the formula for tetra phosphorous decasulfideis ${\text{P}}_{\text{4}}{\text{S}}_{\text{10}}$.

i) Mercury (II) oxide

The mercury cation has a charge of $+2$ always and the iodide anion has a charge of $-1$.

Mercury (II) oxideis formed from the cation ${\text{Hg}}^{\text{2+}}$ and the anion ${\text{O}}^{\text{2-}}$. The charges of the cation and anion negate each other as it is the same and the charge of mercury transfers to the subscript of oxide and the charge of oxidetransfers to the subscript of mercury, which makes the formula of the compoundmercury (II) oxideto be $\text{HgO}$.

Hence, the formula for mercury (II) oxideis $\text{HgO}$.

j) Mercury (I) iodide

The mercurycation has a charge of $+2$ always, and the iodide anion has a charge of $-1$.

So, mercury (I) iodide is formed from the cation ${\text{Hg}}^{\text{2+}}$ and the anion ${\text{I}}^{\text{2-}}$. The charge of mercury transfers to the subscript of iodide and the charge of iodide transfers to the subscript of mercury, which makes the formula of the compoundmercury (I) iodideto be ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$.

Hence, the formula for mercury (I) iodideis ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$.

k)Selenium hexafluoride

The selenium cation has a charge of $+6$ always, and the fluoride anion ${\text{F}}^{-}$ has a charge of $-1$.

So, the selenium hexafluoride is formed from the cation ${\text{Se}}^{\text{6+}}$ and the anion ${\text{F}}^{-}$. The charge of selenium goes to the subscript of fluoride and the charge of fluoridetransfers to the subscript of selenium, which makes the formula of the compoundselenium hexafluorideto be ${\text{SeF}}_{\text{6}}$.

Hence, the formula for selenium hexafluorideis ${\text{SeF}}_{\text{6}}$.