Chemistry Of P-Block Elementsgroup13 Topic

NEET & Advanced Numericals

1. 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 = 112 pm

2. 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

3. Electronegativity of indium:

$\chi_\text{In} = \chi_\text{Tl} - \Delta \chi$

$\chi_\text{In} = 1.62 - 0.45$

Electronegativity of indium = 1.22

4. Melting point of thallium: The melting points of aluminum, indium, and thallium do not 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

5. Boiling point of gallium: The boiling points of aluminum, indium, and gallium do not 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 symbol Ga):

$T_\text{b} = 11(31) + 2224$

$T_\text{b} = 2575 \texttt{ K}$

Boiling point of gallium = 2927 °C

6. 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}).

$E^\circ_{Al3+/Al} = E^\circ_{Ga3+/Ga} - \Delta E^\circ$

$E^\circ_{Al3+/Al} = -1.82 \texttt{V}$

Standard reduction potential of the Al3+/Al couple = -0.82 V

7. Solubility product constant (Ksp) of In(OH)3: Given, $K_\text{sp,Al(OH)3} = 1.9 \times 10^{-31}$ and $\Delta \log K_\text{sp} = 1.5$.

$\log K_\text{sp,In(OH)3} = \log K_\text{sp,Al(OH)3} + \Delta \log K_\text{sp}$

$\log K_\text{sp,In(OH)3} = -30.72 + 1.5 \times 10^{-3}$

$K_\text{sp,In(OH)3} = 1.9 \times 10^{-31}$

Solubility product constant of In(OH)3 = 1.98 × 10⁻³¹

CBSE Board Exam Numericals

Atomic mass of boron: 10.81 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

2. 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

3. 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:

$n_\text{HCl} = 6 \times n_\text{In2O3} = 0.022 \text{ mol}$

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

4. 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