Atomic Structure & Periodic Table

Solution

1. 1

   Identify what is inside and outside the brackets.

   $\text{Mg} \,(\text{NO}_3)_2$

2. 2

   Multiply every atom inside the brackets by the subscript outside.

   $N: 1 \times 2 = 2; \quad O: 3 \times 2 = 6$

3. 3

   Add the magnesium (outside the brackets).

   $Mg: 1$

Solution

1. 1

   Carbon is already balanced (1 each side). Count hydrogen.

   $H: 4 \text{ left}, 2 \text{ right}$

2. 2

   Place a 2 in front of H₂O to balance hydrogen.

   $CH_4 + O_2 \to CO_2 + 2H_2O$

3. 3

   Count oxygen: 2 on left, 2 (in CO₂) + 2 (in 2H₂O) = 4 on right.

   $O: 2 \text{ left}, 4 \text{ right}$

4. 4

   Place a 2 in front of O₂.

   $CH_4 + 2O_2 \to CO_2 + 2H_2O$

Solution

1. 1

   Count hydrogens. 2 on left, 3 on right (1 in NaOH + 2 in H₂).

   $H: 2 \text{ left}, 3 \text{ right}$

2. 2

   Try doubling water: now 2 H₂O gives 4 H on left. We need 4 on right too — put 2 in front of NaOH.

   $Na + 2H_2O \to 2NaOH + H_2$

3. 3

   Re-check Na: 1 on left, 2 on right. Place 2 in front of Na.

   $2Na + 2H_2O \to 2NaOH + H_2$

4. 4

   Re-check oxygen: 2 on each side. Add state symbols.

   $2Na(s) + 2H_2O(l) \to 2NaOH(aq) + H_2(g)$

Solution

1. 1

   Recall the Rᶠ formula.

   $R_f = \frac{\text{distance moved by spot}}{\text{distance moved by solvent}}$

2. 2

   Substitute the values.

   $R_f = \frac{4.2}{7.0}$

3. 3

   Divide.

   $R_f = 0.60$

Solution

1. 1

   Filter the mixture. The sand is insoluble — it stays as the residue. Salt + water passes through as the filtrate.

2. 2

   Take the filtrate (salt solution). Use crystallisation (or evaporation): warm gently in an evaporating basin until water has evaporated and salt crystals are left.

3. 3

   If pure water is also needed, use simple distillation of the original salt solution instead. Steam condenses to give pure water.

Solution

1. 1

   Identify what was expected under the plum-pudding model — α particles should have passed through with only small deflections because positive charge was assumed to be spread thinly.

2. 2

   Large deflections show that positive charge in an atom must be concentrated — strong enough to repel fast positive α particles.

3. 3

   Bouncing back implies a region of high mass — only something massive could halt the α and reverse its direction.

4. 4

   Combined, the results led to the nuclear model: a tiny, dense, positively charged nucleus surrounded by mostly empty space.

Solution

1. 1

   Protons = atomic number = 13.

2. 2

   Neutrons = mass − atomic number = 27 − 13 = 14.

3. 3

   Atom is neutral, so electrons = protons = 13.

Solution

1. 1

   Protons = atomic number = 13 (does not change when ion forms).

2. 2

   Neutrons = 27 − 13 = 14.

3. 3

   Al³⁺ means it has lost 3 electrons. Electrons = 13 − 3 = 10.

Solution

1. 1

   Divide the atom radius by the nucleus radius.

   $\dfrac{1.0 \times 10^{-10}}{1.0 \times 10^{-14}} = 10^{-10-(-14)} = 10^{4}$

2. 2

   Write the result.

   $10^{4} = 10{,}000$

Solution

1. 1

   Each atom has diameter $2 \times 10^{-10}\,\text{m}$.

2. 2

   Convert 1 mm to metres: $1\,\text{mm} = 1 \times 10^{-3}\,\text{m}$.

3. 3

   Number = $1 \times 10^{-3} \div 2 \times 10^{-10} = 5 \times 10^{6}$.

Solution

1. 1

   Multiply each isotope mass by its abundance.

   $(35 \times 75) + (37 \times 25) = 2625 + 925 = 3550$

2. 2

   Divide by 100.

   $A_r = \dfrac{3550}{100} = 35.5$

Solution

1. 1

   Compute (mass × abundance) for each isotope.

   $(24 \times 79) + (25 \times 10) + (26 \times 11) = 1896 + 250 + 286 = 2432$

2. 2

   Divide by 100.

   $A_r = \dfrac{2432}{100} = 24.3$

Solution

1. 1

   Fill shells in order 2, 8, 8: 19 − 2 − 8 − 8 = 1 left for shell 4.

   $K: 2,8,8,1$

2. 2

   Outer shell = 1 electron → group 1. Number of occupied shells = 4 → period 4.

Solution

1. 1

   Outer-shell electrons = 7 → group 7 (halogens).

2. 2

   Number of shells = 3 → period 3.

3. 3

   X = chlorine (Z = 2+8+7 = 17). Group 7 elements gain 1 electron to form X⁻ ions; very reactive non-metals.

Solution

1. 1

   Outer-shell electrons = 6 → group 6.

2. 2

   Number of shells used = 3 → period 3.

3. 3

   Group 6 is on the right side of the staircase → non-metal.

4. 4

   X = sulfur (S).

Solution

1. 1

   Group 2 metals lose 2 outer electrons → 2+ ion: Ca²⁺.

2. 2

   Group 7 non-metals gain 1 electron to fill outer shell → 1− ion: Cl⁻.

Solution

1. 1

   Modification 1: he left gaps in the table where an element of the expected mass had not yet been discovered.

2. 2

   Reason: to keep chemically similar elements in the same group rather than disrupting the pattern.

3. 3

   Modification 2: he reversed the order of some element pairs (e.g. Te and I).

4. 4

   Reason: strict mass-order put them in the wrong group; he kept them with their chemical relatives.

Solution

1. 1

   Li: group 1 → loses 1 electron → Li⁺.

2. 2

   Mg: group 2 → loses 2 electrons → Mg²⁺.

3. 3

   O: group 6 → gains 2 electrons → O²⁻.

4. 4

   Br: group 7 → gains 1 electron → Br⁻.

Solution

1. 1

   Shiny + conducts + high mp → metal.

2. 2

   3+ ion suggests group 3.

3. 3

   Group 3 metal with mp 660 °C → aluminium.