Home Page Textbooks Nelson Science Perspectives 10 Section 5-7: Names and Formulas of Ionic Compounds

Nelson Science Perspectives 10

1st Edition

Christy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe

ISBN: 9780176355289

Textbook solutions

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Section 5-7: Names and Formulas of Ionic Compounds

Exercise 1

Step 1

1 of 3

In this question, we are going to describe the skill that I think needs practicing, and how we are planning on improving this skill.

Step 2

2 of 3

The skill that is needed practicing is the writing of chemical formulas of ionic compounds, and the reason that the writing of chemical formulas can sometimes be tricky so you need to practice so you can master it.

Step 3

3 of 3

To improve this skill you need to understand the rules of writing the chemical formula, and then keep practicing on writing so you can develop this skill.

Exercise 2

Step 1

1 of 6

CaF$_2$ is calcium fluoride.

Step 2

2 of 6

K$_2$S is potassium sulfide.

Step 3

3 of 6

Al$_2$O$_3$ is aluminum oxide.

Step 4

4 of 6

LiBr is lithium bromide.

Step 5

5 of 6

Ca$_3$P$_2$ is calcium phosphide.

Result

6 of 6

a. Calcium fluoride.

b. Potassium sulfide.

c. Aluminum oxide.

d. Lithium bromide.

e. Calcium phosphide.

Exercise 3

Step 1

1 of 4

The ionic charge on K ion is $+1$ while that on Br ion is $-1$, therefore the compound potassium bromide is KBr.

Step 2

2 of 4

The ionic charge on Ca ion is $+2$ while that on O ion is $-2$, therefore the compound calcium oxide is CaO.

Step 3

3 of 4

The ionic charge on Na ion is $+1$ while that on S ion is $-2$. This implies that 2 sodium atoms will need to lose each of their valence electrons to 1 atom of sulfur to form the compound, therefore the compound sodium sulfide is Na$_2$S.

Result

4 of 4

a. KBr.

b. CaO.

c. Na$_2$S.

Exercise 4

Step 1

1 of 2

The correct formula of tin(IV) oxide is SnO$_2$. This is because 1 atom of tin loses 4 of its valence electrons to 2 atoms of oxygen. Each atom of oxygen gains 2 electrons to form the oxide.

Result

2 of 2

SnO$_2$.

Exercise 5

Step 1

1 of 3

The ionic charge on the bromide ion $-1$. This implies that if its compound contains 1 ion of copper per bromide ion, then their ratio is $1:1$ in the compound and therefore its chemical formula must be CuBr. This implies that the ionic charge on copper in this compound is $+1$ and therefore the name of this compound is copper(I) bromide.

Step 2

2 of 3

The ionic charge on the bromide ion $-1$. This implies that if its compound contains 1 ion of copper for every 2 bromide ions, then their ratio is $1:2$ in the compound and therefore its chemical formula must be CuBr$_2$. This implies that the ionic charge on copper in this compound is $+2$ and therefore the name of this compound is copper(II) bromide.

Result

3 of 3

Copper(I) bromide – CuBr.

Copper(II) bromide – CuBr$_2$.

Exercise 6

Result

1 of 1

An ionic compound is made by the transfer of electrons making the compound an electrically neutral entity and therefore the net charge on it must always be $0$.

Exercise 7

Step 1

1 of 6

The ionic charge on calcium ion is $+2$ while that on chlorine ion is $-1$. This implies that 2 chloride atoms will be needed to gain electrons lost from 1 calcium atom to form the compound, therefore the compound calcium chloride is CaCl$_2$.

Step 2

2 of 6

The ionic charge on aluminum ion is $+3$ while that on bromide ion is $-1$. This implies that 3 bromine atoms will be needed to gain electrons lost from 1 aluminum atom to form the compound, therefore the compound aluminum bromide is AlBr$_3$.

Step 3

3 of 6

The ionic charge on magnesium ion is $+2$ while that on sulfide ion is $-2$, therefore the compound magnesium sulfide is MgS.

Step 4

4 of 6

The ionic charge on lithium ion is $+1$ while that on nitride ion is $-3$. This implies that 3 lithium atoms will need to lose each of their valence electrons to 1 atom of nitrogen to form the compound, therefore the compound lithium nitride is Li$_3$N.

Step 5

5 of 6

The ionic charge on calcium ion is $+2$ while that on nitride ion is $-3$. This implies that 3 calcium atoms will lose $3times2=6$ of their valence electrons to $dfrac{6}{3}=2$ atoms of nitrogen to form the compound, therefore the compound calcium nitride is Ca$_3$N$_2$.

Result

6 of 6

a. CaCl$_2$.

b. AlBr$_3$.

c. Mg$_2$S.

d. Li$_3$N.

e. Ca$_3$N$_2$.

Exercise 8

Step 1

1 of 2

The systematic naming convention is a more convenient naming convention because it tells the chemical composition of the minerals.

Result

2 of 2

Click to see answer.

Exercise 9

Step 1

1 of 16

Name

Formula

Step 2

2 of 16

iron(II) bromide

FeBr$_2$

Step 3

3 of 16

manganese(IV) oxide

MnO$_2$

Step 4

4 of 16

tin(IV) chloride

SnCl$_4$

Step 5

5 of 16

copper(I) sulfide

Cu$_2$S

Step 6

6 of 16

iron(III) nitride

FeN

Step 7

7 of 16

copper(II) oxide

CuO

Step 8

8 of 16

lead chloride

PbCl$_2$

Step 9

9 of 16

iron(III) oxide

Fe$_2$O$_3$

Step 10

10 of 16

tin(II) sulfide

SnS

Step 11

11 of 16

copper(II) phosphide

Cu$_3$P$_2$

Step 12

12 of 16

calcium bromide

CaBr$_2$

Step 13

13 of 16

copper(II) fluoride

CuF$_2$

Step 14

14 of 16

potassium phosphide

K$_3$P

Step 15

15 of 16

copper(I) phosphide

Cu$_3$P

Result

16 of 16

Click to see table

Exercise 10

Result

1 of 1

Two charges of $Fe^{3+}$ and one charge of $Fe^{2+}$ with four oxygens will give the desired formula

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Section 5-7: Names and Formulas of Ionic Compounds

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