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Chapter 2: Q.49E (page 67)

For each of the following pairs of ions, write the symbol for the formula of the compound they will form:

  1. \(C{a^{2 + }},{S^{2 - }}\)
  2. \(NH_4^ + ,SO_4^{2 - }\)
  3. \(A{l^{3 + }},B{r^ - }\)
  4. \(N{a^ + },HPO_4^{2 - }\)
  5. \(M{g^{2 + }},PO_4^{3 - }\)

Short Answer

Expert verified
  1. \(CaS\)
  2. \({\left( {N{H_4}} \right)_2}S{O_4}\)
  3. \(AlB{r_3}\)
  4. \(N{a_2}HP{O_4}\)
  5. \(M{g_3}{\left( {P{O_4}} \right)_2}\)

Step by step solution

01

Step 1.  Finding the way to combine ions

Since the overall charge of a compound is zero, we can combine ions by balancing their charges.

02

Step 2.  Balancing the charges in each part

a. \(C{a^{2 + }},{S^{2 - }}\)\( \to \) +2 charge in 1 Ca atom can balance -2 charge in 1 S atom. Hence, formula becomes\(CaS\).

b. \(NH_4^ + ,SO_4^{2 - }\)\( \to \) Since ammonium ion has only +1 charge, we will be needing 2 molecules of ammonium ion to balance -2 charge in sulphate ion. . Hence the formula of compound becomes\({\left( {N{H_4}} \right)_2}S{O_4}\).

c. \(A{l^{3 + }},B{r^ - }\)\( \to \) 3 atoms of Br will be required to balance +3 charge in aluminium ion. . Hence the formula of compound becomes \(AlB{r_3}\).

d. \(N{a^ + },HPO_4^{2 - }\)\( \to \) -2 charge in 1 molecule of hydrogen phosphate can be neutralized by 2 atoms of sodium. . Hence the formula of compound becomes\(N{a_2}HP{O_4}\).

e. \(M{g^{2 + }},PO_4^{3 - }\)\( \to \) Using cross-multiplying method, we can conclude that charge in 3 atoms of Mg will be balancing charge in 2 molecules of phosphate ion. Hence the formula of compound becomes\(M{g_3}{\left( {P{O_4}} \right)_2}\).

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Most popular questions from this chapter

An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain and what is it’s charge ?

How are electrons and protons similar? How are they different?

Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ.

Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:

\(\begin{aligned}{}\begin{aligned}{{}{}}{\left( a \right){\rm{ }}{}_5^{10}B}\\{\left( b \right){\rm{ }}{}_{80}^{199}Hg}\\{\left( c \right){\rm{ }}{}_{29}^{63}Cu}\\{\left( d \right){\rm{ }}{}_6^{13}C}\end{aligned}\\\left( e \right){\rm{ }}{}_{34}^{77}Se\end{aligned}\)

Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundance (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.

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