Chemistry Of P-Block Elementsgroup13 Topic
JEE Main & Advanced Numericals
- Atomic radius of boron: $$V = \dfrac{m}{d}$$
$$V = \dfrac{10.81 \texttt{ g/mol}}{2.34 \texttt{ g/cm}^3}$$
$$V = 4.62 \texttt{ cm}^3/\texttt{mol}$$
$$r = \left(\dfrac{3V}{4\pi N_A}\right)^{\frac{1}{3}}$$
$$r = \left(\dfrac{3 \times 4.62 \texttt{ cm}^3/\texttt{mol}}{4\pi \times 6.022 \times 10^{23} \texttt{ mol}^{-1}}\right)^{\frac{1}{3}}$$
$$r = 0.88 \times 10^{-8} \texttt{ cm}$$
Atomic radius of boron = 88 pm
- Ionization energy of gallium:
$$IE_\text{Ga} = IE_\text{Al} + \Delta IE$$
$$IE_\text{Ga} = 577 \texttt{ kJ/mol} + 165 \texttt{ kJ/mol}$$
Ionization energy of gallium = 742 kJ/mol
- Electronegativity of indium:
$$\chi_\text{In} = \chi_\text{Tl} - \Delta \chi$$
$$\chi_\text{In} = 1.62 - 0.45$$
Electronegativity of indium = 1.17
- Melting point of thallium: The melting points of aluminum, indium, and thallium follow a linear trend. Let’s assume the linear equation is:
$$T_\text{m} = mx + b$$
where (T_\text{m}) is the melting point, (x) is the atomic number, and (m) and (b) are constants. Using the given data, we can create two equations:
$$660 = m (13) + b \tag1$$
$$156 = m(49) + b\tag2$$
Subtracting equation (1) from equation (2), we get:
$$-504 = m(-36)$$
$$m = 14 \texttt{ °C}$$
Substituting (m) back into equation (1), we get:
$$660 = 14 (13) + b$$
$$b = 482 \texttt{ °C}$$
Now we can use the equation to calculate the melting point of thallium (atomic number 81):
$$T_\text{m} = 14 (81) + 482$$
$$T_\text{m} = 1694 \texttt{ °C}$$
Melting point of thallium = 1694 °C
- Boiling point of gallium: The boiling points of aluminum, indium, and gallium follow a linear trend. Let’s assume the linear equation is:
$$T_\text{b} = mx + b$$
where (T_\text{b}) is the boiling point, (x) is the atomic number, and (m) and (b) are constants.
Using the given data, we can create two equations:
$$2467 = m(13) + b\tag1$$ $$2072 = m(49) + b\tag2$$
Subtracting equation (1) from equation (2), we get:
$$-395 = m(-36)$$
$$m = 11 \texttt{ °C}$$
Substituting (m) back into equation (1), we get:
$$2467 = 11(13) + b$$
$$b = 2224 \texttt{ °C}$$
Now we can use the equation to calculate the boiling point of gallium (atomic number 31):
$$T_\text{b} = 11(31) + 2224$$
$$T_\text{b} = 2575 \texttt{ °C}$$
Boiling point of gallium = 2575 °C
- Standard reduction potential of the Al3+/Al couple: Given, (E^\circ_{Ga3+/Ga} = -0.53 \texttt{V}) and (\Delta E^\circ = 0.29 \texttt{V}).
$$E^\circ_{Al3+/Al} = E^\circ_{Ga3+/Ga} - \Delta E^\circ$$
$$E^\circ_{Al3+/Al} = -0.53 \texttt{V} - 0.29 \texttt{V}$$
Standard reduction potential of the Al3+/Al couple = -0.82 V
- Solubility product constant (Ksp) of In(OH)3: Given, (K_\text{sp,Al(OH)3} = 1.9 \times 10^{-31}) and (\Delta \log K_\text{sp} = 1.5 \times 10^{-31}).
$$\log K_\text{sp,In(OH)3} = \log K_\text{sp,Al(OH)3} + \Delta \log K_\text{sp}$$
$$\log K_\text{sp,In(OH)3} = -30.72 + 1.5 \times 10^{-31}$$
$$K_\text{sp,In(OH)3} = 1.9 \times 10^{-31} \times 10^{1.5 \times 10^{-31}}$$
Solubility product constant of In(OH)3 = 1.98 × 10^⁻³¹
CBSE Board Exam Numericals
- Atomic mass of boron: The average atomic mass of boron is: $$M_\text{avg} = 0.199 \times 10.01 \texttt{ g/mol} + 0.801 \times 11.01 \texttt{ g/mol}$$ $$M_\text{avg} = 10.81 \texttt{ g/mol}$$
Atomic mass of boron = 10.81 g/mol
- Mass of aluminum produced:
First, calculate the number of moles of Al2O3:
$$n_\text{Al2O3} = \frac{100 \texttt{ g}}{101.96 \texttt{ g/mol}} = 0.98 \texttt{ mol}$$
Assuming 100% current efficiency, the number of moles of Al produced would be:
$$n_\text{Al} = 2 \times n_\text{Al2O3} = 1.96 \texttt{ mol}$$
However, since the current efficiency is 85%, the actual number of moles of Al produced is: $$n_\text{Al,actual} = 0.85 \times 1.96 \texttt{ mol} = 1.67 \texttt{ mol}$$
Finally, calculate the mass of aluminum produced: $$m_\text{Al} = n_\text{Al,actual} \times M_\text{Al}$$ $$m_\text{Al} = 1.67 \texttt{ mol} \times 26.98 \texttt{ g/mol} = 45.2 \texttt{ g}$$
Mass of aluminum produced = 45.2 g
- Volume of 0.1 M HCl required:
First, calculate the number of moles of In2O3: $$n_\text{In2O3} = \frac{1.0 \texttt{ g}}{277.64 \texttt{ g/mol}} = 0.0036 \texttt{ mol}$$
The balanced chemical equation shows that 6 moles of HCl are required for every mole of In2O3. Therefore, the number of moles of HCl required is:
$$n_\text{HCl} = 6 \times n_\text{In2O3} = 0.022 \texttt{ mol}$$
Finally, calculate the volume of 0.1 M HCl required: $$V_\text{HCl} = \frac{n_\text{HCl}}{M_\text{HCl}} = \frac{0.022 \texttt{ mol}}{0.1 \texttt{ mol/L}} = 0.22 \texttt{ L}$$
Volume of 0.1 M HCl required = 0.22 L
- Percentage of indium in the alloy:
First, calculate the number of moles of InCl3 produced: $$n_\text{InCl3} = \frac{1.34 \texttt{ g}}{221.18 \texttt{ g/mol}} = 0.00606 \texttt{ mol}$$
The balanced chemical equation shows that 1 mole of InCl3 is produced for every mole
