SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Atomic Foundations of Matter

39 questions · 21 still being checked

Pause and Ponder 9.15–9.24 (part 22 of 22)

  1. Exercise 9.15

    Illustrate how sodium sulfide (Na2\displaystyle Na_{2}S) is formed. Name of ion Sodium Lithium Potassium Silver Calcium Barium Iron (Ferrous) Iron (Ferric) Copper (Cuprous) Copper (Cupric) Magnesium Zinc Aluminium Fluoride Chloride Bromide Iodide Oxide Sulfide

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \(\displaystyle Na_2S\) is formed when each of two sodium atoms transfers one electron to a single sulfur atom.
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-15
    Sodium is $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 1$ — one valence electron to give away; sulfur is $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 6$ — two electrons short of an octet.
    Each Na loses $\displaystyle 1$ electron → \(\displaystyle Na^+\) $\displaystyle (2, 8)$; sulfur accepts both → \(\displaystyle S^{2-}\) ($\displaystyle 2$, $\displaystyle 8$, $\displaystyle 8$).
    Two sodium ions are needed for one sulfide ion so that the charges cancel: \(\displaystyle 2 \times (1+) + (2-) = 0\).
    The oppositely charged ions are held together by the electrostatic force of attraction — an ionic bond — giving \(\displaystyle Na_2S\), sodium sulfide.
    The illustration must show: two sodium atoms (shells $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 1$) on the left and one sulfur atom ($\displaystyle 2$, $\displaystyle 8$, $\displaystyle 6$) on the right; one curved arrow labelled \(\displaystyle e^-\) from each sodium's valence shell to the sulfur's valence shell; the products drawn as \(\displaystyle [2,8]^+\) twice, labelled \(\displaystyle Na^+\), and \(\displaystyle [2,8,8]^{2-}\), labelled \(\displaystyle S^{2-}\); the label ionic bond between them; and the formula \(\displaystyle Na_2S\) written underneath.

  2. Exercise 9.16

    Name the following: (i) CO2\displaystyle CO_{2} _______________________________ (ii) NO2\displaystyle NO_{2} _______________________________ (iii) SF6\displaystyle SF_{6} (iv) PCl3 _______________________________

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    (i)
    Carbon dioxide (ii) Nitrogen dioxide (iii) Sulfur hexafluoride (iv) Phosphorous trichloride
    (i) \(\displaystyle CO_2\) — carbon dioxide
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-16
    (ii) \(\displaystyle NO_2\) — nitrogen dioxide
    (iii) \(\displaystyle SF_6\) — sulfur hexafluoride
    (iv) \(\displaystyle PCl_3\) — phosphorus trichloride
    Rule used: the first element keeps its ordinary name, the second element ends in -ide, and the prefixes mono- ($\displaystyle 1$), di- ($\displaystyle 2$), tri- ($\displaystyle 3$), tetra- ($\displaystyle 4$), penta- ($\displaystyle 5$), hexa- ($\displaystyle 6$) count the atoms.
    mono- is dropped before the first element, which is why \(\displaystyle CO_2\) is carbon dioxide and not "monocarbon dioxide", and why \(\displaystyle NO_2\) is nitrogen dioxide.

  3. Exercise 9.17

    Write the formula for the following: (i) Sodium hydrogencarbonate _____________________ (ii) Sulfur dioxide (iii) Ferric chloride ___________________________________ (iv) Cuprous oxide ___________________________________
    NCERT’s answer
    (i)
    \(\displaystyle NaHCO_{3}\) (ii) \(\displaystyle SO_{2}\) (iii) \(\displaystyle FeCl_{3}\) (iv) \(\displaystyle Cu_{2}\)O
    (i) Sodium hydrogencarbonate — \(\displaystyle NaHCO_3\): \(\displaystyle Na^+\) with \(\displaystyle HCO_3^-\), both of valency $\displaystyle 1$, so one of each.
    (ii) Sulfur dioxide — \(\displaystyle SO_2\): the prefix di- fixes two oxygen atoms to one sulfur.
    (iii) Ferric chloride — \(\displaystyle FeCl_3\): ferric is \(\displaystyle Fe^{3+}\) and chloride is \(\displaystyle Cl^-\); criss-crossing $\displaystyle 3$ and $\displaystyle 1$ gives one Fe to three Cl.
    (iv) Cuprous oxide — \(\displaystyle Cu_2O\): cuprous is \(\displaystyle Cu^+\) and oxide is \(\displaystyle O^{2-}\); criss-crossing $\displaystyle 1$ and $\displaystyle 2$ gives two Cu to one O.

  4. Exercise 9.18

    Write the formulae for the compounds formed from the following pairs of ions: (i) Fe3+\displaystyle Fe^{3+} and OH‒\displaystyle OH^{‒} 9.6\displaystyle 9.6 Properties of the Ionic and the Covalent Compounds Activity 9.4\displaystyle 9.4: Let us experiment 1. Collect the samples of some compounds, such as camphor, sodium chloride, copper sulfate, sugar and naphthalene. (A) Solubility in (i) water, (ii) kerosene, and (iii) petrol 2. Try dissolving each sample separately in the water, kerosene and petrol. 3. Record your observations in Table 9.2. (B) Electrical conductivity in the water Safety first: Do not touch the electrodes when they are connected to the battery but use a low-voltage battery to avoid the risk of shock. Petrol and kerosene are flammable liquids, so be careful while working with them.
    NCERT’s answer
    (i)
    Fe(OH)$\displaystyle 3$ (ii) \(\displaystyle K_{2}\)\(\displaystyle CO_{3}\)
    (i) \(\displaystyle Fe^{3+}\) and \(\displaystyle OH^-\) → \(\displaystyle Fe(OH)_3\)
    Criss-cross the charge numbers $\displaystyle 3$ and $\displaystyle 1$: one \(\displaystyle Fe^{3+}\) to three \(\displaystyle OH^-\), and \(\displaystyle 3+ \;+\; 3 \times (1-) = 0\).
    Brackets are needed because three of the same polyatomic ion are present — it is \(\displaystyle Fe(OH)_3\), not \(\displaystyle FeOH_3\).
    (ii) \(\displaystyle K^+\) and \(\displaystyle CO_3^{2-}\) → \(\displaystyle K_2CO_3\)
    Criss-cross $\displaystyle 1$ and $\displaystyle 2$: two \(\displaystyle K^+\) to one \(\displaystyle CO_3^{2-}\), and \(\displaystyle 2 \times (1+) + (2-) = 0\).
    Only one carbonate ion is present, so no brackets are used.

  5. Exercise 9.19

    What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    An ionic bond.
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-19
    In the solid state the ions are locked in fixed positions in the crystal lattice by strong forces, so no charge can move and the compound cannot conduct.
    On dissolving in water the ions become free to move, and moving ions carry the current — so the solution conducts.
    A covalent compound would fail one test or the other: sugar dissolves but releases no ions, and camphor and naphthalene do not conduct at all.

  6. Exercise 9.20

    Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its: (i) formula (ii) type of bond (iii) electrical conductivity of its aqueous solution. You have learnt in section 9.4.2\displaystyle 9.4.2 that in ionic compounds, the ions form

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    (i)
    MO (ii) Ionic (iii) Conducts electricity in aqueous solution
    (i) Formula: MO
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-20
    (ii) Type of bond: ionic
    (iii) Its aqueous solution conducts electricity
    M's valence shell is the M shell holding $\displaystyle 2$ electrons, so M is $\displaystyle 2$, $\displaystyle 8$, $\displaystyle 2$ — a metal with fewer than $\displaystyle 4$ valence electrons, which donates them → \(\displaystyle M^{2+}\).
    Oxygen has $\displaystyle 6$ valence electrons and gains $\displaystyle 2$ → \(\displaystyle O^{2-}\).
    Criss-crossing the charges gives \(\displaystyle M_2O_2\), and dividing both subscripts by the common factor $\displaystyle 2$ gives MO.
    Metal + non-metal joined by the transfer of electrons is an ionic bond.
    The compound is only slightly soluble, but whatever does dissolve releases free \(\displaystyle M^{2+}\) and \(\displaystyle O^{2-}\) ions into the water, and free ions conduct — so the bulb would glow, if weakly.

  7. Exercise 9.21

    Find the molecular mass of nitric acid (HNO3\displaystyle HNO_{3}). Atomic mass — H = 1\displaystyle 1 u; N = 14\displaystyle 14 u; O = 16\displaystyle 16 u.
    NCERT’s answer
    $\displaystyle 63$ u
    Molecular mass of \(\displaystyle HNO_3\) = $\displaystyle 63$ u.
    Add the atomic mass of every atom in the formula: $\displaystyle 1$ H, $\displaystyle 1$ N and $\displaystyle 3$ O.
    \(\displaystyle (1 \times 1) + (14 \times 1) + (16 \times 3) = 1 + 14 + 48 = 63\) u.

  8. Exercise 9.22

    Find the molecular mass of methane (CH4\displaystyle CH_{4}). Atomic mass — C = 12\displaystyle 12 u; H = 1\displaystyle 1 u.
    NCERT’s answer
    $\displaystyle 16$ u
    Molecular mass of \(\displaystyle CH_4\) = $\displaystyle 16$ u.
    The formula holds $\displaystyle 1$ carbon atom and $\displaystyle 4$ hydrogen atoms.
    \(\displaystyle (12 \times 1) + (1 \times 4) = 12 + 4 = 16\) u.

  9. Exercise 9.23

    Find the formula unit mass of potassium chloride (KCl). Atomic mass — K = 39\displaystyle 39 u; Cl = 35.5\displaystyle 35.5 u.
    NCERT’s answer
    74.$\displaystyle 5$ u
    Formula unit mass of KCl = $\displaystyle 74.5$ u.
    \(\displaystyle (39 \times 1) + (35.5 \times 1) = 74.5\) u.
    It is called a formula unit mass, not a molecular mass, because KCl is ionic — its ions form a $\displaystyle 3$-D crystal, so it does not exist as molecules.

  10. Exercise 9.24

    Find the formula unit mass of magnesium hydroxide, Mg(OH)2. Atomic mass — Mg = 24\displaystyle 24 u; O = 16\displaystyle 16 u; H = 1\displaystyle 1 u. Understanding atoms, molecules and chemical bonding reveals the At a Glance y Mass can neither be created nor destroyed in a chemical reaction. This is known as the Law of Conservation of Mass. y A compound always contains the same elements combined in a fixed ratio by mass, no matter how it is formed or from where it is obtained. This is called the Law of Definite Proportions. y A molecule is defined as an electrically neutral entity consisting of more than one atom that can exist independently and shows all its chemical properties. y Atoms combine to form molecules of elements or compounds to become stable. Atoms are held together by a force called a chemical bond. y A covalent bond is formed by the sharing of electrons between atoms. y An ionic bond is formed by the transfer of electrons between atoms, where one atom loses electrons and the other gains electrons to form cations and anions, respectively. y The chemical formula of a covalent compound represents the elements and number of atoms of each element present in it. y The chemical formula of an ionic compound represents the simplest whole number ratio of atoms of different elements present in it. y Molecular mass is the total mass of a molecule, calculated by adding the atomic masses of all the atoms constituting it. y Formula unit mass of an ionic compound is the sum of the atomic masses of all the atoms present in a formula unit (simplest whole number ratio of ions in an ionic compound).
    NCERT’s answer
    $\displaystyle 58$ u
    Formula unit mass of \(\displaystyle Mg(OH)_2\) = $\displaystyle 58$ u.
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-24
    The subscript $\displaystyle 2$ outside the bracket multiplies both the O and the H inside it, so the unit contains $\displaystyle 1$ Mg, $\displaystyle 2$ O and $\displaystyle 2$ H.
    \(\displaystyle (24 \times 1) + \{(16 \times 1) + (1 \times 1)\} \times 2 = 24 + (17 \times 2) = 24 + 34 = 58\) u.