Atomic Foundations of Matter — Important Questions

Answers

• All matter is made up of very tiny particles called atoms, which participate in chemical reactions.
• Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
• Atoms of a given element are identical in mass and chemical properties.
• Atoms of different elements have different masses and chemical properties.
• Atoms combine in the ratio of simple whole numbers to form compounds.
• The relative number and kinds of atoms are constant in a given compound.
  The postulate that explains why matter occupies space is: "All matter is made up of very tiny particles called atoms." Atoms themselves occupy space.
  6. A molecule is an electrically neutral entity consisting of more than one atom that is capable of independent existence and shows all the properties of that substance.
  Example: Helium (He) is an element that exists as atoms. This is because its atoms have a stable electronic configuration (2 electrons in the K-shell), so they do not need to combine with other atoms.
  7. A covalent bond is formed by the sharing of electrons between atoms.
  Formation of an Oxygen molecule ($O_2$):
  An oxygen atom (atomic number 8) has 6 electrons in its valence shell. It needs 2 more electrons to complete its octet. Two oxygen atoms share two electrons each, forming two shared pairs. This results in a double covalent bond.
  Diagram:
  Oxygen atom + Oxygen atom → Oxygen molecule
  (6 valence e⁻) (6 valence e⁻) (Shared pairs)
  O=O
  8. Formation of Hydrogen Chloride molecule (HCl):
  Hydrogen atom (atomic number 1) has 1 electron in its valence shell. It needs 1 more electron to complete its duplet.
  Chlorine atom (atomic number 17) has 7 electrons in its valence shell. It needs 1 more electron to complete its octet.
  One electron from the hydrogen atom is shared with one electron from the chlorine atom, forming a single shared pair of electrons. This forms a single covalent bond.
  Diagram:
  H . + :Cl: → [H : Cl:] → H—Cl
  9. Formation of a water molecule ($H_{2}O$):
  Oxygen atom (atomic number 8) has 6 valence electrons and needs 2 more electrons.
  Hydrogen atom (atomic number 1) has 1 valence electron and needs 1 more electron.
  One oxygen atom shares one electron each with two hydrogen atoms. Each hydrogen atom shares its one electron with the oxygen atom. This forms two single covalent bonds between oxygen and each hydrogen atom.
  Diagram:
  H . + .Ö. + .H → H—Ö—H
  10. An ionic bond is formed by the transfer of electrons between atoms, resulting in the formation of oppositely charged ions that are held together by electrostatic force of attraction.
  Formation of Sodium Chloride (NaCl):
  Sodium atom (atomic number 11) has 1 valence electron. It readily loses this electron to form a stable sodium cation ($N{a}^{+}$).
  Chlorine atom (atomic number 17) has 7 valence electrons. It readily gains one electron to form a stable chloride anion ($C{l}^{-}$).
  The positively charged sodium ion ($N{a}^{+}$) and the negatively charged chloride ion ($C{l}^{-}$) are held together by electrostatic attraction, forming an ionic bond in sodium chloride.
  Diagram:
  Na . + :Cl: → $N{a}^{+}$ + $C{l}^{-}$ → NaCl
  11. Cations are positively charged ions formed when an atom loses one or more electrons.
  Example: Sodium ($N{a}^{+}$), Calcium ($C{a}^{2+}$), Aluminium ($A{l}^{3+}$).
  Anions are negatively charged ions formed when an atom gains one or more electrons.
  Example: Chloride ($C{l}^{-}$), Oxide ($O^{2-}$), Sulfide ($S^{2-}$).
  12. (i) $C{S}_2$: Carbon disulfide
  (ii) $PC{l}_3$: Phosphorus trichloride
  (iii) $N_2{O}_4$: Dinitrogen tetroxide
  13. (i) Aluminium oxide: $A{l}_2{O}_3$
  (ii) Magnesium hydroxide: $Mg(OH{)}_2$
  (iii) Calcium carbonate: $CaC{O}_3$
  14. Element X has 5 valence electrons (configuration 2, 8, 5). It needs 3 electrons to complete its octet. Element Y has 7 valence electrons (configuration 2, 8, 7). It needs 1 electron to complete its octet.
  When X and Y combine, they will likely share electrons to form covalent bonds. Specifically, X needs 3 electrons and Y needs 1 electron. For stability, it's more likely X would gain 3 electrons and Y would gain 1. However, if they were to form a compound by sharing, the problem might be indicating X is from Group 15 (e.g., Nitrogen) and Y is from Group 17 (e.g., Fluorine). Let's assume X is from Group 15 and Y is from Group 17.
  If X and Y combine, they might form ionic bonds if X is a non-metal that can gain electrons and Y is a metal that can lose electrons. However, the configurations suggest X is a non-metal (e.g., Phosphorus) and Y is a halogen (e.g., Chlorine).
  Let's re-examine. If X has 5 valence electrons and forms a diatomic molecule, it implies it is in Group 15 (like Nitrogen). If Y has 7 valence electrons, it is in Group 17 (like a halogen).
  The question states "a certain other element Y has two electrons in its second shell". This means Y has a total of 4 electrons (2 in the first shell, 2 in the second). This is Beryllium (atomic number 4). This is a metal.
  Let's assume the question meant "two electrons in its outer shell". If Y has 2 electrons in its outer shell, it's an alkaline earth metal (e.g., Magnesium, atomic number 12, configuration 2, 8, 2).
  If X is from Group 15 (e.g., P, configuration 2, 8, 5) and Y is a metal from Group 2 (e.g., Mg, configuration 2, 8, 2).
  X needs 3 electrons to form $X^{3-}$. Y needs to lose 2 electrons to form $Y^{2+}$.
  They would form an ionic bond. To balance charges, we need 2 atoms of X and 3 atoms of Y. Formula: $Y_3{X}_2$.
  If the question implies X and Y form covalent bonds, it's tricky with these configurations for a simple binary compound.
  Let's consider the wording carefully: "An element X has six electrons in its outer shell and forms a diatomic molecule." This implies X is like Oxygen (atomic number 8, configuration 2, 6). If X forms $X_2$, it forms covalent bonds.
  "A certain other element Y has two electrons in its second shell." This is Beryllium (atomic number 4, configuration 2, 2). Beryllium is a metal. It forms ionic bonds.
  So, X (like O) and Y (like Be) would form an ionic compound. X forms $X^{2-}$ and Y forms $Y^{2+}$. The formula would be $YX$.
  Bond type: Ionic bond.
  Formula: $YX$.
  15. Ionic compounds have strong electrostatic forces of attraction between oppositely charged ions in a crystal lattice. These ions are held in fixed positions and cannot move freely, hence they do not conduct electricity in the solid state. When ionic compounds are dissolved in water, the ions become free to move and can carry electric charge, thus conducting electricity.
  16. Molecular mass of methane ($C{H}_4$) = (Atomic mass of C × 1) + (Atomic mass of H × 4) = (12 u × 1) + (1 u × 4) = 12 u + 4 u = 16 u.
  17. Formula unit mass of magnesium oxide (MgO) = (Atomic mass of Mg × 1) + (Atomic mass of O × 1) = (24 u × 1) + (16 u × 1) = 24 u + 16 u = 40 u.
  18. (a) The remaining mass (2 g) might have escaped as a gas from the open flask.
  (b) If the experiment was conducted in a closed container, the total mass of the products would be 20 g. This is because according to the Law of Conservation of Mass, the mass of reactants is equal to the mass of products, and no matter is lost or gained.
  (c) The chemical law being tested here is the Law of Conservation of Mass.
  19. (a) Covalent bond.
  (b) Compounds formed by electron sharing (covalent compounds) are generally insoluble in water but soluble in organic solvents like kerosene and petrol.
  (c) No, Priya's statement is not always true. While many ionic compounds conduct electricity when dissolved in water, some ionic compounds might be insoluble. An example of an ionic compound that conducts electricity in aqueous solution is sodium chloride (NaCl).
  20. (a) This observation supports the Law of Constant Proportions (or Law of Definite Proportions).
  (b) The significance of this ratio is that water always contains hydrogen and oxygen in a fixed mass proportion. If 1 g of hydrogen is present, it is always combined with 8 g of oxygen to form 9 g of water.
  (c) If 3 g of hydrogen is present, then the mass of oxygen required is:
  Mass of Oxygen = (8/1) × 3 g = 24 g