Atomic Foundations of Matter — NCERT Solutions

We need total positive charge $6+$ and total negative charge $6-$. Check each option:

• (i) $2\,Al^{3+}$ gives $+6$ but $3\,Cl^-$ gives $-3$ → not balanced.
• (ii) $3\,Mg^{2+}$ gives $+6$ but $1\,PO_4^{3-}$ gives $-3$ → not balanced.
• (iii) $2\,Fe^{3+}$ gives $+6$ and $3\,O^{2-}$ gives $-6$ → balanced.
• (iv) $3\,Ca^{2+}$ gives $+6$ and $2\,SO_4^{2-}$ gives $-4$ → not balanced.

So the correct combination is (iii).

(i) False. Correct statement: Elements are made up of atoms, and compounds are made up of molecules or formula units.

(ii) False. Correct statement: The molecule of a compound is made up of two or more atoms of different kinds.

(iii) False. A nitrogen molecule is $N_2$, so one molecule of nitrogen gas contains two nitrogen atoms, not three.

(iv) True. Water is made of two hydrogen atoms covalently bonded with one oxygen atom.

Using the ionic formula-writing method:

• Aluminium nitrate: $Al^{3+}$ and $NO_3^-$ → crossover gives $Al(NO_3)_3$.
• Calcium oxide: $Ca^{2+}$ and $O^{2-}$ → simplest ratio gives $CaO$.
• Ferric oxide: $Fe^{3+}$ and $O^{2-}$ → crossover gives $Fe_2O_3$.

Use criss-cross of charges and simplify where needed:

• $Ca^{2+}$ and $Br^-$ → $CaBr_2$
• $Al^{3+}$ and $CO_3^{2-}$ → crossover gives $Al_2(CO_3)_3$
• $K^+$ and $SO_4^{2-}$ → $K_2SO_4$
• $NH_4^+$ and $Cl^-$ → $NH_4Cl$

A chloride ion, $Cl^-$, has 17 protons and 18 electrons. So the correct diagram must show chlorine with one extra electron and a negative charge. In Fig. 9.18, that is option (iii).

Calculate formula unit mass by adding atomic masses:

(i) Ammonium nitrate, $NH_4NO_3$

Mass = $(1\times 4) + 14 + 14 + (16\times 3)$

$= 4 + 14 + 14 + 48 = 80\,u$

(ii) Phosphoric acid, $H_3PO_4$

Mass = $(1\times 3) + 31 + (16\times 4)$

$= 3 + 31 + 64 = 98\,u$

(iii) Sodium hydrogencarbonate, $NaHCO_3$

Mass = 23 + 1 + 12 + (16\times 3)$

$= 23 + 1 + 12 + 48 = 84\,u$

Form compounds by balancing charges:

• Magnesium and nitrogen: $Mg^{2+}$ and $N^{3-}$ → $Mg_3N_2$
• Lithium and nitrogen: $Li^+$ and $N^{3-}$ → $Li_3N$
• Sodium and sulfur: $Na^+$ and $S^{2-}$ → $Na_2S$
• Aluminium and oxygen: $Al^{3+}$ and $O^{2-}$ → $Al_2O_3$

Complete the table by balancing charges:

• NH$_4^+$:
• with $NO_3^-$ → $NH_4NO_3$
• with $SO_4^{2-}$ → $(NH_4)_2SO_4$
• with $PO_4^{3-}$ → $(NH_4)_3PO_4$

• Li$^+$:
• with $NO_3^-$ → $LiNO_3$
• with $SO_4^{2-}$ → $Li_2SO_4$
• with $PO_4^{3-}$ → $Li_3PO_4$

• Al$^{3+}$:
• with $NO_3^-$ → $Al(NO_3)_3$
• with $SO_4^{2-}$ → $Al_2(SO_4)_3$
• with $PO_4^{3-}$ → $AlPO_4$

• Cu$^{2+}$:
• with $NO_3^-$ → $Cu(NO_3)_2$
• with $SO_4^{2-}$ → $CuSO_4$
• with $PO_4^{3-}$ → $Cu_3(PO_4)_2$

So the completed table entries are:
$NH_4NO_3$, $(NH_4)_2SO_4$, $(NH_4)_3PO_4$;
$LiNO_3$, $Li_2SO_4$, $Li_3PO_4$;
$Al(NO_3)_3$, $Al_2(SO_4)_3$, $AlPO_4$;
$Cu(NO_3)_2$, $CuSO_4$, $Cu_3(PO_4)_2$.

Add the masses of reactants and products:

Reactants:

• Sodium carbonate = 5.3 g
• Acetic acid = 6.0 g

Total mass of reactants = $5.3 + 6.0 = 11.3$ g

Products:

• Carbon dioxide = 2.2 g
• Water = 0.9 g
• Sodium acetate = 8.2 g

Total mass of products = $2.2 + 0.9 + 8.2 = 11.3$ g

Since total mass of reactants = total mass of products, the Law of Conservation of Mass is valid.

• Number of protons = 11, so the atomic number is 11.
• Mass number = protons + neutrons = $11 + 12 = 23$.
• Electrons = 10, which is one less than protons, so the species is not neutral. It has one extra positive charge, so it is a cation.
• Electronic configuration of 10 electrons is 2, 8.
• The species is $Na^+$, called the sodium ion.

Element A has configuration 2, 8, 5, so it has 5 valence electrons. It tends to gain 3 electrons to complete its octet.

Element B has configuration 2, 8, 7, so it has 7 valence electrons. It tends to gain 1 electron to complete its octet.

(i) B is more reactive because it needs only one electron to complete its octet.

(ii) Since both are non-metals and tend to gain/share electrons, they would form a covalent bond by sharing electrons.

(iii) A needs 3 electrons and B needs 1 electron, so one A atom can combine with three B atoms. The formula is $AB_3$.

Copper sulfate is an ionic compound. It conducts electricity in the molten state because ions are free to move, but not in the solid state because ions are fixed in the lattice.

So Assertion (A) is true.

The reason given is incorrect because it reverses the states: ions are fixed in the solid state and free in the molten state. Therefore Reason (R) is false.

Correct option: (iii).

Use:

• electrons = protons adjusted for charge
• neutrons = mass number − protons

$^{27}Al$

• Protons = 13
• Electrons = 13 (neutral atom)
• Neutrons = $27 - 13 = 14$

$^{80}Br^-$

• Protons = 35
• Electrons = $35 + 1 = 36$
• Neutrons = $80 - 35 = 45$

$^{201}Hg^{2+}$

• Protons = 80
• Electrons = $80 - 2 = 78$
• Neutrons = $201 - 80 = 121$

• Assertion (A) is true: $2\,g$ hydrogen combines with $16\,g$ oxygen to form $18\,g$ water.
• Reason (R) is also true: according to Dalton’s atomic theory, atoms combine in a simple whole-number ratio to form compounds.

This explains why the masses combine in a fixed way here. So the correct option is (i).

The relevant examples given in the chapter are:

• Carbon dioxide: one carbon atom combines with two oxygen atoms, so the structure is $O=C=O$. - Hydrogen sulfide: two hydrogen atoms combine with one sulfur atom, so the structure is $H-S-H$. - Ammonia: one nitrogen atom combines with three hydrogen atoms, so the structure shows nitrogen bonded to three hydrogens. These are the standard structures based on valencies from the chapter.

I need the specific statements/blanks to answer accurately from the chapter.

I need the exact list of items to give the names correctly from the chapter.

Using the ion charges: magnesium forms $\mathrm{Mg}^{2+}$ and hydroxide is $\mathrm{OH}^{-}$. Criss-cross the charges to get $\mathrm{Mg(OH)}_2$. The brackets are needed because there are two hydroxide ions.

Use the charges of the ions and criss-cross them:

• $\mathrm{Ca}^{2+}$ and $\mathrm{Br}^{-}$ give $\mathrm{CaBr}_2$
• $\mathrm{Al}^{3+}$ and $\mathrm{CO}_3^{2-}$ give $\mathrm{Al}_2(\mathrm{CO}_3)_3$
• $\mathrm{K}^{+}$ and $\mathrm{SO}_4^{2-}$ give $\mathrm{K}_2\mathrm{SO}_4$
• $\mathrm{NH}_4^{+}$ and $\mathrm{Cl}^{-}$ give $\mathrm{NH}_4\mathrm{Cl}$

Metal M has two electrons in its valence shell, so it can lose two electrons to form a $\mathrm{M}^{2+}$ cation. Since it reacts with oxygen, the compound formed will be a metal oxide with formula $\mathrm{MO}$. Such ionic oxides are generally slightly soluble in water and form an ionic bond.

For $\mathrm{Mg(OH)}_2$:

• Atomic mass of Mg = 24 u
• Atomic mass of O = 16 u
• Atomic mass of H = 1 u

So,
$$\text{Formula unit mass} = 24 + 2\times(16+1)$$
$$= 24 + 2\times 17 = 24 + 34 = 58\,\text{u}$$

• A has one electron in its third shell, so it tends to give away 1 electron to become stable. It forms a cation, $\mathrm{A}^{+}$.
• B has six electrons in its second shell, so it tends to take 2 electrons to complete its octet. It forms an anion, $\mathrm{B}^{2-}$.
• Since A loses electrons and B gains electrons, they form an ionic bond.
• The charges balance in the ratio 2:1, so the formula is $\mathrm{A}_2\mathrm{B}$.

Element X has six valence electrons, so each atom needs two more electrons to complete its octet. Therefore, two X atoms share two pairs of electrons to form a diatomic molecule with a double covalent bond.

If the element is like oxygen, the structure is X = X.

Check the total charges:

• (i) $2\mathrm{Al}^{3+}$ gives $+6$, but $3\mathrm{Cl}^-$ gives only $-3$.
• (ii) $3\mathrm{Mg}^{2+}$ gives $+6$, but one $\mathrm{PO}_4^{3-}$ gives $-3$.
• (iii) $2\mathrm{Fe}^{3+}$ gives $+6$, and $3\mathrm{O}^{2-}$ gives $-6$.
• (iv) $3\mathrm{Ca}^{2+}$ gives $+6$, and $2\mathrm{SO}_4^{2-}$ gives $-4$.

So the combination that gives total positive charge $6+$ and total negative charge $6-$ is (iii). However, the printed options in the question list one correct combination by ion count for charge balance; the correct neutral compound is $\mathrm{Fe}_2\mathrm{O}_3$ from option (iii).

1. False. Correct statement: Elements are made up of atoms and compounds are made up of molecules or formula units.
2. False. Correct statement: The molecule of a compound is made up of atoms of different elements.
3. False. Correct statement: One molecule of nitrogen gas contains two nitrogen atoms.
4. True. Water is made of two hydrogen atoms covalently bonded with one oxygen atom.

Use ionic charges:

• Aluminium nitrate: $\mathrm{Al}^{3+}$ and $\mathrm{NO}_3^{-}$ give $\mathrm{Al}(\mathrm{NO}_3)_3$
• Calcium oxide: $\mathrm{Ca}^{2+}$ and $\mathrm{O}^{2-}$ give $\mathrm{CaO}$
• Ferric oxide: $\mathrm{Fe}^{3+}$ and $\mathrm{O}^{2-}$ give $\mathrm{Fe}_2\mathrm{O}_3$

Combine the ions so the total positive and negative charges balance:

• $\mathrm{Ca}^{2+}$ and $\mathrm{Br}^{-}$ → $\mathrm{CaBr}_2$
• $\mathrm{Al}^{3+}$ and $\mathrm{CO}_3^{2-}$ → $\mathrm{Al}_2(\mathrm{CO}_3)_3$
• $\mathrm{K}^{+}$ and $\mathrm{SO}_4^{2-}$ → $\mathrm{K}_2\mathrm{SO}_4$
• $\mathrm{NH}_4^{+}$ and $\mathrm{Cl}^{-}$ → $\mathrm{NH}_4\mathrm{Cl}$

Calculate each formula unit mass by adding atomic masses:

(i) Ammonium nitrate, $\mathrm{NH}_4\mathrm{NO}_3$
$$(14 + 4\times 1) + (14 + 3\times 16) = 18 + 62 = 80\,\text{u}$$

(ii) Phosphoric acid, $\mathrm{H}_3\mathrm{PO}_4$
$$3\times 1 + 31 + 4\times 16 = 3 + 31 + 64 = 98\,\text{u}$$

(iii) Sodium hydrogencarbonate, $\mathrm{NaHCO}_3$
$$23 + 1 + 12 + 3\times 16 = 23 + 1 + 12 + 48 = 84\,\text{u}$$

Write the formulae of the ionic compounds formed:

• Magnesium and nitrogen: $\mathrm{Mg}^{2+}$ and $\mathrm{N}^{3-}$ combine to give $\mathrm{Mg}_3\mathrm{N}_2$
• Lithium and nitrogen: $\mathrm{Li}^{+}$ and $\mathrm{N}^{3-}$ combine to give $\mathrm{Li}_3\mathrm{N}$
• Sodium and sulfur: $\mathrm{Na}^{+}$ and $\mathrm{S}^{2-}$ combine to give $\mathrm{Na}_2\mathrm{S}$
• Aluminium and oxygen: $\mathrm{Al}^{3+}$ and $\mathrm{O}^{2-}$ combine to give $\mathrm{Al}_2\mathrm{O}_3$

• Number of protons = 11, so the atomic number = 11.
• Number of neutrons = 12, so mass number = 11 + 12 = 23.
• Number of electrons = 10, so it has one fewer electron than protons and is a cation with charge $+1$.
• Electronic configuration: 2, 8.
• The species is sodium ion, $\mathrm{Na}^{+}$.

• A: 2, 8, 5 has 5 valence electrons, so it tends to gain 3 electrons to become stable.
• B: 2, 8, 7 has 7 valence electrons, so it tends to gain 1 electron to become stable.

(i) B is more reactive, because it needs only one electron to complete its octet.

(ii) They will form an ionic bond. A tends to lose/share electrons to attain stability, and B tends to gain one electron. Since the chapter emphasizes ionic bonding by electron transfer when one atom loses and another gains, this pair would combine through transfer only if a suitable metal were involved; but between two non-metals like these, they would generally form a covalent bond by sharing electrons.

(iii) Using valencies 3 and 1, the formula would be $\mathrm{AB}_3$ if treated by valency criss-crossing.

Copper sulfate is solid ionic compound: ions are fixed in the lattice, so it does not conduct in solid state. In the molten state, ions become free to move, so it does conduct electricity.

Thus, Assertion (A) is true and Reason (R) is false because the reason says the opposite about mobility of ions in solid and molten states.