Question-1(a) :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Given: Iron may be extracted from iron (II) sulfide, $\mathrm{FeS}$.
Discuss: why metals, like iron, can conduct electricity.
Answer/Explanation
Solution:
Metals like iron can conduct electricity because they have a unique atomic structure in which the valence electrons, the outermost electrons in the atom, are delocalized and can move freely throughout the metal lattice. This allows the electrons to carry an electric current from one end of the metal to the other.
In more detail, the atoms in a metal are arranged in a regular, repeating lattice structure. The valence electrons of the metal atoms occupy a band of energy levels, called the conduction band, which overlaps with the next highest energy level, called the valence band. This means that the valence electrons can easily move from one atom to another, and as a result, a flow of electrons, or an electric current, can be established.
Furthermore, the delocalized electrons also interact with positively charged metal ions in the lattice, creating a “sea” of electrons that is free to move throughout the metal. This is known as metallic bonding and contributes to the high electrical conductivity of metals like iron.
Question-1(b) :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Discuss: Justify why sulfur is classified as a non-metal by giving two of its chemical properties.
Answer/Explanation
Solution:
Sulfur is classified as a non-metal because it exhibits several characteristic properties of non-metals, including:
Electronegativity: Non-metals tend to have a higher electronegativity, meaning they have a greater attraction for electrons in chemical bonds. Sulfur has an electronegativity of 2.58, which is higher than that of many metals, including iron (1.83). This makes sulfur more likely to form covalent bonds with other non-metals rather than forming ionic bonds with metals.
Non-metallic character: Non-metals tend to be brittle, have low melting and boiling points, and are poor conductors of heat and electricity. Sulfur is a brittle solid at room temperature and has a melting point of 115 degrees Celsius and a boiling point of 444 degrees Celsius. It is also a poor conductor of heat and electricity, further supporting its classification as a non-metal.
In addition, sulfur is located on the right side of the periodic table, along with other non-metals such as oxygen, nitrogen, and chlorine. It also tends to form oxides and acids, which are typical of non-metals rather than metals.
Question-1[(c) (i)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Given: Iron (II) sulfide, FeS, is ionically bonded.
Describe: the bonding in this type of solid.
Answer/Explanation
Solution:
In ionic bonding, the atoms transfer electrons to each other to form ions with opposite charges that are held together by the electrostatic attraction between them. In the case of iron (II) sulfide, FeS, iron (Fe) loses two electrons to become a positively charged ion ($\mathrm{Fe}^{2+}$) and sulfur (S) gains two electrons to become a negatively charged ion ($\mathrm{S}^{2-}$). The resulting ionic compound is held together by the attraction between the oppositely charged ions.
Question-1[(c) (ii)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Discuss: the full electron configuration of the sulfide ion.
Answer/Explanation
Solution:
The electron configuration of a sulfide ion, $\mathrm{S}^{2-}$, can be determined by adding two electrons to the neutral sulfur atom’s electron configuration, which is 1s2 2s2 2p6 3s2 3p4. Therefore, the full electron configuration of the sulfide ion is:
$\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$
Question-1[(c) (iii)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Discuss: in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.
Answer/Explanation
Solution:
Both sulfide ($\mathrm{S}^{2-}$) and oxide ($\mathrm{O}^{2-}$) ions are anions, meaning they have gained electrons to achieve a stable noble gas configuration. Sulfide has a larger ionic radius compared to oxide because of two main reasons:
Sulfur is a larger atom than oxygen: Sulfur has more electrons and energy levels than oxygen, so its outermost electrons are farther from the nucleus. As a result, when sulfur forms an ion by gaining two electrons, the ion’s overall size is larger than an oxide ion with the same charge.
The electronic configuration of sulfur: Sulfur has six valence electrons, which means that when it gains two electrons to become $\mathrm{S}^{2-}$, it has a stable noble gas configuration like argon. In contrast, oxygen has only four valence electrons, and when it gains two electrons to become $\mathrm{O}^{2-}$, it has to accommodate these extra electrons in the same energy level. This creates more electron-electron repulsion, which results in a smaller ionic radius compared to sulfide.
Overall, the larger size of sulfur and the higher electron-electron repulsion in oxide ions make sulfide ions larger than oxide ions.
Question-1[(c) (iv)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Discuss: why chemists find it convenient to classify bonding into ionic, covalent and metallic.
Answer/Explanation
Solution:
Chemists classify bonding into ionic, covalent, and metallic because it provides a useful framework for understanding the different types of bonding that occur in chemical substances. Each type of bonding has distinctive characteristics, and understanding these characteristics helps chemists to predict the properties of chemical compounds and materials. Here are some reasons why chemists find it convenient to classify bonding into these categories:
Predicting the behavior of substances: Understanding the nature of bonding in a substance can help predict its behavior in different environments, such as in reactions with other substances, in the presence of heat or light, or when exposed to different forms of energy.
Understanding properties: The type of bonding in a substance affects its physical and chemical properties, such as melting and boiling points, electrical conductivity, and solubility in different solvents. By classifying bonding into categories, chemists can easily compare and contrast the properties of different substances with similar bonding types.
Simplification: The three categories of bonding – ionic, covalent, and metallic – are broad and cover most bonding situations. By using these categories, chemists can simplify complex bonding situations into more manageable groups.
Teaching and learning: The classification of bonding into these three categories is commonly taught in introductory chemistry courses, making it an important part of chemical education. The categories provide a useful framework for students to understand the basics of bonding and how it relates to the properties and behavior of substances.
Overall, the classification of bonding into ionic, covalent, and metallic is a useful tool for chemists to understand and predict the behavior of substances and materials, compare and contrast their properties, and teach and learn the basics of chemical bonding.
Question-1[(d) (i)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Given: The first step in the extraction of iron from iron (II) sulfide is to roast it in air to form iron (III) oxide and sulfur dioxide.
Write: the equation for this reaction.
Answer/Explanation
Solution:
The correct balanced chemical equation for the roasting of iron (II) sulfide is:
$$4 \mathrm{FeS}(\mathrm{s})+7 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+4 \mathrm{SO}_2(\mathrm{~g})$$
This reaction shows that four moles of iron (II) sulfide react with seven moles of oxygen gas to form two moles of iron (III) oxide and four moles of sulfur dioxide gas. The coefficients in the balanced equation represent the stoichiometry of the reaction, which means the relative amounts of each reactant and product involved in the reaction.
Question-1[(d) (ii)] :2021-may-Chemistry_paper_2__TZ1_SL
Topic:
Calculate: the change in the oxidation state of sulfur.
Answer/Explanation
Solution:
The oxidation state of sulfur in $\mathrm{FeS}$ is $-2$ because iron (II) has a $+2$ oxidation state and the compound is electrically neutral. The oxidation state of sulfur in $\mathrm{SO_2}$ is $+4$ because oxygen has a $-2$ oxidation state and the compound is electrically neutral.
Therefore, the change in the oxidation state of sulfur is from $-2$ in $\mathrm{FeS}$ to $+4$ which is equivalent to $+6$ in $\mathrm{SO_2}$. This means that sulfur is being oxidized during the reaction.