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Chapter 8: Problem 6

Write formulas for the following compounds: (a) Mercury(II) chloride (b) Nickel(II) sulphate (c) Tin(IV) oxide (d) Thallium(I) sulphate (e) Iron(III) sulphate (f) Chromium(III) oxide

Short Answer

Expert verified
(a) HgCl_2, (b) NiSO_4, (c) SnO_2, (d) Tl_2SO_4, (e) Fe_2(SO_4)_3, (f) Cr_2O_3.

Step by step solution

01

Understanding Oxidation States

The Roman numerals in chemical names denote the oxidation states of the metal. For example, in Mercury(II) chloride, 'II' means mercury has an oxidation state of +2. Similarly, for Nickel(II) sulphate, nickel has an oxidation state of +2.
02

Determining Anion Formula and Charge

Identify the anion involved and its typical charge. Chloride (Cl) has a charge of -1, sulphate (SO_4) has a charge of -2, and oxide (O) has a charge of -2.
03

Balancing Charges for Mercury(II) Chloride

Mercury (Hg) has a +2 charge, and chloride (Cl) has a -1 charge. For the compound to be neutral, two chloride ions are needed to balance one mercury ion. Hence, the formula is HgCl_2.
04

Balancing Charges for Nickel(II) Sulphate

Nickel (Ni) has a +2 charge, and sulphate (SO_4) has a -2 charge. Only one sulphate ion is needed to balance one nickel ion. Hence, the formula is NiSO_4.
05

Balancing Charges for Tin(IV) Oxide

Tin (Sn) has a +4 charge, and oxide (O) has a -2 charge. Two oxide ions are needed to balance one tin ion. Hence, the formula is SnO_2.
06

Balancing Charges for Thallium(I) Sulphate

Thallium (Tl) has a +1 charge, and sulphate (SO_4) has a -2 charge. Two thallium ions are needed to balance one sulphate ion. Hence, the formula is Tl_2SO_4.
07

Balancing Charges for Iron(III) Sulphate

Iron (Fe) has a +3 charge, and sulphate (SO_4) is -2. Three sulphate ions will balance two iron ions. Hence, the formula is Fe_2(SO_4)_3.
08

Balancing Charges for Chromium(III) Oxide

Chromium (Cr) has a +3 charge, and oxide (O) has a -2 charge. Two chromium ions and three oxide ions are needed to balance each other. Hence, the formula is Cr_2O_3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Oxidation States
Oxidation states, often represented by Roman numerals in chemical compound names, are crucial for determining how elements combine to form compounds. These numbers indicate the charge of an atom if all bonds were ionic. For metals, oxidation states help us understand how many electrons the atom has lost to reach a stable state in a compound.
- In Mercury(II) chloride, the Roman numeral 'II' tells us that mercury has lost two electrons, giving it an oxidation state of +2.
- Similarly, in compounds like Nickel(II) sulphate, the nickel atom has an oxidation state of +2.
Knowing the oxidation state is essential in predicting how atoms will interact with each other, as it determines the number of anions needed to balance the charges.
Balancing Charges in Ionic Compounds
The principle of charge balance states that in a neutral compound, the sum of positive charges must equal the sum of negative charges. This balance is crucial for the formation of stable ionic compounds.
For example, here's how we find the correct formula using Mercury(II) chloride:
- Mercury ion (Hg) has a +2 charge, and each chloride ion (Cl) has a -1 charge. To balance, two chloride ions are needed for every mercury ion, resulting in the formula HgCl₂.
Other examples include:
  • Nickel(II) sulphate (NiSO₄), where nickel's +2 charge is balanced perfectly by sulphate's -2 charge.

  • Tin(IV) oxide necessitates two oxide ions (each with -2 charge) for every tin ion (+4 charge), resulting in SnO₂.

    • Keep in mind that balancing charges ensures that compounds are electrically neutral, which is pivotal for their stability in nature.
Formation of Ionic Compounds
Ionic compounds are formed when metals transfer electrons to nonmetals, creating ions that attract each other due to opposite charges. This process involves:
  • Metals losing electrons to become positively charged cations.

  • Nonmetals gaining electrons to become negatively charged anions.

    • The resulting electrostatic attraction between cations and anions leads to the formation of a crystal lattice structure, characteristic of ionic compounds.
    - For instance, in Chromium(III) oxide (Cr₂O₃), chromium ions (each with a +3 charge) and oxide ions (each with a -2 charge) combine to form a stable, electrically neutral compound.
    - Similarly, Iron(III) sulphate (Fe₂(SO₄)₃) results from the interaction of iron ions (+3) with sulphate ions (-2), in which two iron ions pair with three sulphate ions.
    Understanding these interactions aids in predicting the structure and formula of ionic compounds accurately.

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Most popular questions from this chapter

Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions: (a) \(2 \mathrm{AgBr}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HBr}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})\) (b) \(\begin{aligned} \mathrm{HCHO}(\mathrm{l})+2\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+4 \mathrm{NH}_{3} &(\mathrm{aq}) \\\\+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{aligned}\) (c) \(\mathrm{HCHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (d) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (e) \(\mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with \(10.00 \mathrm{~g}\). of ammonia and \(20.00 \mathrm{~g}\) of oxygen ?

Fluorine reacts with ice and results in the change: \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g})+\mathrm{HOF}(\mathrm{g})\) Justify that this reaction is a redox reaction.

Justify that the following reactions are redox reactions: (a) \(\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})\) (c) \(4 \mathrm{BCl}_{3}(\mathrm{~g})+\) 3LiAlH \(_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})\) (d) \(2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})\) (e) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Given the standard electrode potentials, \(\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}\) \(\mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}\) \(\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V} \cdot \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}\) arrange these metals in their increasing order of reducing power.
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