Chapter 9: Revise, Reflect, Refine — Master Solutions
This master solution document covers all 15 end-of-chapter exercises ("Revise, Reflect, Refine") from the NCERT Class 9 Science textbook. Every solution provides the conceptual reason, electronic structure, chemical equation, and complete mathematical steps required for maximum score in CBSE examinations.
Question 1: A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.
(i) How many electrons does A tend to give or take to become stable?
(ii) What kind of ion would it form?
(iii) How many electrons does B tend to give or take to become stable?
(iv) What kind of ion would it form?
(v) If A and B were to combine, what kind of bond would be formed?
(vi) What would be the formula for the compound thus formed?
Detailed Solution:
Let us first deduce the electronic configuration and identity of elements A and B:
• Element A: Has 1 electron in its 3rd shell ($M$-shell). Filling preceding shells completely gives configuration $K=2, L=8, M=1$ (Total electrons $= 11$, which is Sodium, $Na$).
• Element B: Has 6 electrons in its 2nd shell ($L$-shell). Filling preceding shell gives configuration $K=2, L=6$ (Total electrons $= 8$, which is Oxygen, $O$).
(i) Tendency of A: Element A has 1 valence electron. It tends to give (lose) 1 electron from its $M$-shell to achieve a stable octet ($2,8$) resembling the nearest noble gas, Neon ($Ne$).
(ii) Type of ion formed by A: By losing 1 negatively charged electron, it forms a monovalent positive ion (cation) represented as $\mathbf{A^+}$ (e.g. $Na^+$):
$$A \longrightarrow A^+ + e^-$$
(iii) Tendency of B: Element B has 6 valence electrons. It needs 2 electrons to complete its octet ($2,8$) in the $L$-shell. Hence, it tends to take (gain) 2 electrons.
(iv) Type of ion formed by B: By gaining 2 negatively charged electrons, it forms a divalent negative ion (anion) represented as $\mathbf{B^{2-}}$ (e.g. Oxide ion, $O^{2-}$):
$$B + 2e^- \longrightarrow B^{2-}$$
(v) Type of bond formed: The bond is formed by the complete transfer of electrons from electropositive metal A to electronegative non-metal B. Strong electrostatic forces bind the oppositely charged ions together. Thus, an ionic bond (electrovalent bond) is formed.
(vi) Chemical formula of the compound:
Applying the Criss-Cross Method:
Question 2: An element X has six electrons in its outer shell and forms a diatomic molecule.
(i) Why would that be so?
(ii) What kind of bond would it form?
(iii) Draw the structure of the molecule it would form.
(iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Detailed Solution:
(i) Reason for forming a diatomic molecule: Element X has 6 valence electrons and requires 2 more electrons to attain a stable octet of 8 electrons. When two atoms of X approach each other, neither is willing to completely lose electrons because both have high electronegativity. Therefore, each atom contributes 2 electrons, sharing a total of two electron pairs (4 electrons). This mutual sharing allows both atoms to simultaneously complete their octets, forming a stable diatomic molecule $X_2$ (identical to Oxygen, $O_2$).
(ii) Type of bond formed in $X_2$: Since the bond results from the mutual sharing of two pairs of electrons between non-metal atoms, it forms a double covalent bond ($X=X$).
(iii) Structure of the $X_2$ molecule:
Each atom retains 2 lone pairs (4 non-bonding electrons) and participates in 2 shared pairs:
$$\mathbf{:X = X: \quad \text{or} \quad : \ddot{X} = \ddot{X} :}$$
$$\text{Lewis representation: Two central shared pairs } (X :: X) \text{ and two unshared pairs on each } X.$$
(iv) Structure of compound formed between X and Y:
• Element Y has 2 electrons in its second shell ($K=2, L=2 \implies \text{Beryllium, } Be$, Atomic number 4; or if in outer shell, like an alkaline earth metal). It readily loses its 2 valence electrons to achieve stability ($Y \to Y^{2+} + 2e^-$).
• Element X has 6 valence electrons and readily gains those 2 electrons ($X + 2e^- \to X^{2-}$).
• Complete electron transfer occurs, producing an ionic compound with formula $YX$ (e.g. $BeO$).
$$\text{Structure: } [Y]^{2+} [:\ddot{X}:]^{2-}$$
Question 3: You want to design a new ionic compound, where the total positive charge is $6+$ and the total negative charge is $6-$. Which of the following combinations gives the correct number of ions?
(i) $2Al^{3+}$ and $3Cl^-$
(ii) $3Mg^{2+}$ and $1PO_4^{3-}$
(iii) $2Fe^{3+}$ and $3O^{2-}$
(iv) $3Ca^{2+}$ and $2SO_4^{2-}$
Detailed Solution:
Let us evaluate the net charges for each combination:
Question 4: Choose the correct statement(s) and correct the false statement(s).
(i) Elements are made up of molecules and compounds are made up of atoms.
(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
(iii) One molecule of nitrogen gas contains three nitrogen atoms.
(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Detailed Solution:
(i) False.
Correction: Elements are pure substances made up of only one kind of atom (which may exist as individual atoms like $He$ or homoatomic molecules like $O_2$), whereas compounds are formed when atoms of two or more different elements combine chemically in a fixed proportion by mass.
(ii) False.
Correction: The molecule of a compound is always made up of two or more atoms of different kinds of elements chemically bonded together (e.g. $H_2O$ contains hydrogen and oxygen; $CO_2$ contains carbon and oxygen). Molecules composed of atoms of the same kind are molecules of elements (e.g. $O_2, N_2, P_4$).
(iii) False.
Correction: One molecule of nitrogen gas is diatomic ($N_2$), meaning it contains two nitrogen atoms bonded by a triple covalent bond ($:N \equiv N:$).
(iv) True.
Scientific justification: In a water molecule ($H_2O$), the central oxygen atom shares one pair of valence electrons with each of the two hydrogen atoms, forming two single covalent bonds ($H—O—H$).
Question 5: Write the chemical formulae for the following compounds:
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
Detailed Solution:
Using the systematic Criss-Cross Method:
Question 6: Write the formulae of the compounds formed from the following pairs of ions:
(i) $Ca^{2+}$ and $Br^-$
(ii) $Al^{3+}$ and $CO_3^{2-}$
(iii) $K^+$ and $SO_4^{2-}$
(iv) $NH_4^+$ and $Cl^-$
Detailed Solution:
Applying the rules of charge neutralization and criss-crossing valency numbers:
Question 7: Which of the following, in Fig. 9.18, correctly represents $Cl^-$ ion (Atomic number of chlorine $= 17$)?
Options: Diagrams (i), (ii), (iii), (iv)
Detailed Solution:
Step 1: Determine the number of electrons in $Cl^-$:
• The atomic number of Chlorine ($Cl$) is $Z = 17$, meaning a neutral chlorine atom has 17 protons and 17 electrons with electronic configuration:
$$K = 2, \quad L = 8, \quad M = 7$$
• To form a chloride anion ($Cl^-$), the atom gains 1 electron:
$$Cl\ (2,8,7) + e^- \longrightarrow Cl^-\ (2,8,8)$$
• Total number of electrons in $Cl^- = 17 + 1 = \mathbf{18\text{ electrons}}$.
Step 2: Shell Distribution of 18 Electrons:
• First shell ($K$-shell): 2 electrons.
• Second shell ($L$-shell): 8 electrons.
• Third shell ($M$-shell): 8 electrons.
Conclusion:
Diagram (ii) correctly represents the chloride ion ($Cl^-$) as it depicts three concentric shells containing 2, 8, and 8 electrons respectively, displaying a completely filled outer octet.
Question 8: Determine the formula unit mass of the following substances:
(i) Ammonium nitrate ($NH_4NO_3$), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid ($H_3PO_4$), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate ($NaHCO_3$), used to relieve acidity and helps in digestion.
[Atomic masses: $H=1\text{ u}, C=12\text{ u}, N=14\text{ u}, O=16\text{ u}, Na=23\text{ u}, P=31\text{ u}$]
Detailed Solution:
The formula unit mass is the sum of the atomic masses of all constituent atoms in the chemical formula unit:
(i) Formula unit mass of Ammonium nitrate ($NH_4NO_3$):
Total atoms present $= 2\text{ atoms of } N + 4\text{ atoms of } H + 3\text{ atoms of } O$.
$$\text{Mass} = (2 \times \text{mass of } N) + (4 \times \text{mass of } H) + (3 \times \text{mass of } O)$$
$$\text{Mass} = (2 \times 14\text{ u}) + (4 \times 1\text{ u}) + (3 \times 16\text{ u})$$
$$\text{Mass} = 28\text{ u} + 4\text{ u} + 48\text{ u} = \mathbf{80\text{ u}}$$
(ii) Molecular mass of Phosphoric acid ($H_3PO_4$):
Total atoms present $= 3\text{ atoms of } H + 1\text{ atom of } P + 4\text{ atoms of } O$.
$$\text{Mass} = (3 \times 1\text{ u}) + (1 \times 31\text{ u}) + (4 \times 16\text{ u})$$
$$\text{Mass} = 3\text{ u} + 31\text{ u} + 64\text{ u} = \mathbf{98\text{ u}}$$
(iii) Formula unit mass of Sodium hydrogencarbonate ($NaHCO_3$):
Total atoms present $= 1\text{ atom of } Na + 1\text{ atom of } H + 1\text{ atom of } C + 3\text{ atoms of } O$.
$$\text{Mass} = (1 \times 23\text{ u}) + (1 \times 1\text{ u}) + (1 \times 12\text{ u}) + (3 \times 16\text{ u})$$
$$\text{Mass} = 23\text{ u} + 1\text{ u} + 12\text{ u} + 48\text{ u} = \mathbf{84\text{ u}}$$
Question 9: Write the formulae for the compounds formed by the reaction of:
(i) Magnesium and nitrogen
(ii) Lithium and nitrogen
(iii) Sodium and sulfur
(iv) Aluminium and oxygen
Detailed Solution:
Writing symbols and valencies, then applying the Criss-Cross rule:
Question 10: Complete Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. $LiNO_3$ is given as an example.
Detailed Solution:
Using the criss-cross method and applying parentheses whenever a polyatomic ion takes a subscript $\ge 2$:
| Cation / Anion | Nitrate ($NO_3^-$) [Valency = 1] |
Sulfate ($SO_4^{2-}$) [Valency = 2] |
Phosphate ($PO_4^{3-}$) [Valency = 3] |
|---|---|---|---|
| Ammonium ($NH_4^+$) [Valency = 1] |
$NH_4NO_3$ Ammonium nitrate |
$(NH_4)_2SO_4$ Ammonium sulfate |
$(NH_4)_3PO_4$ Ammonium phosphate |
| Lithium ($Li^+$) [Valency = 1] |
$LiNO_3$ (Given Example) |
$Li_2SO_4$ Lithium sulfate |
$Li_3PO_4$ Lithium phosphate |
| Aluminium ($Al^{3+}$) [Valency = 3] |
$Al(NO_3)_3$ Aluminium nitrate |
$Al_2(SO_4)_3$ Aluminium sulfate |
$AlPO_4$ Aluminium phosphate (3:3 → 1:1) |
| Copper(II) ($Cu^{2+}$) [Valency = 2] |
$Cu(NO_3)_2$ Copper(II) nitrate |
$CuSO_4$ Copper(II) sulfate (2:2 → 1:1) |
$Cu_3(PO_4)_2$ Copper(II) phosphate |
Question 11: $5.3\text{ g}$ of sodium carbonate and $6.0\text{ g}$ of acetic acid react to produce $2.2\text{ g}$ of carbon dioxide, $0.9\text{ g}$ of water, and $8.2\text{ g}$ of sodium acetate. Verify whether the law of conservation of mass is valid.
Detailed Solution:
According to the Law of Conservation of Mass proposed by Antoine Lavoisier, the total mass of the products formed during a chemical reaction must be strictly equal to the total mass of the reactants consumed:
Chemical Reaction:
$$\text{Sodium carbonate} + \text{Acetic acid} \longrightarrow \text{Carbon dioxide} + \text{Water} + \text{Sodium acetate}$$
1. Calculate Total Mass of Reactants:
• Mass of sodium carbonate $= 5.3\text{ g}$
• Mass of acetic acid $= 6.0\text{ g}$
$$\text{Total Mass of Reactants} = 5.3\text{ g} + 6.0\text{ g} = \mathbf{11.3\text{ g}}$$
2. Calculate Total Mass of Products:
• Mass of carbon dioxide $= 2.2\text{ g}$
• Mass of water $= 0.9\text{ g}$
• Mass of sodium acetate $= 8.2\text{ g}$
$$\text{Total Mass of Products} = 2.2\text{ g} + 0.9\text{ g} + 8.2\text{ g} = \mathbf{11.3\text{ g}}$$
Conclusion:
Since $\text{Total Mass of Reactants } (11.3\text{ g}) = \text{Total Mass of Products } (11.3\text{ g})$, there is no loss or gain of matter during this chemical change. Hence, the Law of Conservation of Mass is completely valid and verified.
Question 12: If a species has 11 protons, 12 neutrons and 10 electrons then:
(i) What is its atomic number and mass number?
(ii) Is it neutral, a cation or an anion? Explain.
(iii) Write its electronic configuration.
(iv) Name the species.
Detailed Solution:
(i) Atomic Number and Mass Number:
• Atomic Number ($Z$): Equals the number of protons in the nucleus.
$$Z = \text{Number of protons} = \mathbf{11}$$
• Mass Number ($A$): Equals the total number of protons and neutrons (nucleons).
$$A = \text{Protons} + \text{Neutrons} = 11 + 12 = \mathbf{23}$$
(ii) Charge Status (Neutral, Cation, or Anion):
• Total positive charge from protons $= +11$
• Total negative charge from electrons $= -10$
$$\text{Net Charge} = +11 + (-10) = \mathbf{+1}$$
Since the species carries a positive charge of $+1$ (due to having 1 fewer electron than protons), it is a positively charged ion (cation).
(iii) Electronic Configuration:
With 10 electrons to arrange:
$$K\text{-shell} = 2, \quad L\text{-shell} = 8 \quad \implies \mathbf{2, 8}$$
(iv) Name of the species:
The element with atomic number 11 is Sodium ($Na$). Having lost 1 electron, this positively charged ion is the Sodium ion (or Sodium cation, $\mathbf{Na^+}$).
Question 13: Two elements, A and B, have the following configurations —
$\text{A: } 2, 8, 5 \qquad \text{B: } 2, 8, 7$
(i) Which element is more reactive?
(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.
(iii) Predict the formula of the compound they would form.
Detailed Solution:
• Element A ($2,8,5$) has Atomic Number $15$, which is Phosphorus ($P$). It needs 3 electrons to complete its octet.
• Element B ($2,8,7$) has Atomic Number $17$, which is Chlorine ($Cl$). It needs 1 electron to complete its octet.
(i) Which element is more reactive?
Element B (Chlorine) is more reactive.
Reason: Both elements have 3 electron shells. Across a period from left to right, nuclear charge increases while atomic size decreases. Element B ($Z=17$) has a higher effective nuclear charge ($17$ protons pulling on the $M$-shell compared to $15$ protons in A) and smaller atomic radius, making it attract an extra valence electron much more strongly and rapidly to complete its octet.
(ii) Type of bond formed (Ionic or Covalent):
A and B will form a covalent bond.
Explanation: Both A and B are non-metals with high electronegativities and incomplete valence shells (5 and 7 electrons, respectively). Element A cannot easily lose 5 electrons nor can B lose 7 electrons due to the extremely high ionization energy required. Instead, mutual sharing of electrons takes place. Element A shares 1 electron with each of three B atoms, allowing each atom to achieve a stable octet of 8 electrons without complete electron transfer.
(iii) Chemical Formula:
• Element A needs 3 shared electrons (Valency $= 3$).
• Element B needs 1 shared electron (Valency $= 1$).
Criss-crossing valencies gives:
$$\mathbf{\text{Formula} = AB_3 \quad (\text{e.g. } PCl_3, \text{Phosphorus trichloride})}$$
Question 14 (Assertion-Reason):
Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.
Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.
Choose the correct option:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Detailed Solution:
• Evaluating Assertion (A): Copper sulfate ($CuSO_4$) is an ionic compound. In the solid state, it does not conduct electricity because its ions are held rigidly in fixed positions within the crystal lattice. When melted into molten state (or dissolved in water), the electrostatic lattice breaks down and ions become mobile charge carriers capable of conducting electric current. Hence, Assertion (A) is TRUE.
• Evaluating Reason (R): The statement claims that ions are fixed in the molten state and free in the solid state. This is the exact inverse of reality! In the solid state, ions are fixed; in the molten state, ions move freely. Hence, Reason (R) is completely FALSE.
Correct Answer: (iii) A is true, but R is false.
Question 15: The species $^{27}Al$, $^{80}Br^-$ and $^{201}Hg^{2+}$ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Detailed Solution:
For any nuclide or ionic species represented as $^{A}_{Z}X^{\pm q}$:
Summary Table:
| Species | Mass Number ($A$) | Protons ($p$) | Electrons ($e^-$) | Neutrons ($n = A - p$) |
|---|---|---|---|---|
| $\mathbf{^{27}Al}$ | 27 | 13 | 13 | 14 |
| $\mathbf{^{80}Br^-}$ | 80 | 35 | 36 | 45 |
| $\mathbf{^{201}Hg^{2+}}$ | 201 | 80 | 78 | 121 |
VARDAAN COMET