Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
To solve this question, we need to analyze each pair of group 13 elements, considering the trends in atomic and ionic radii as one moves down the group in the periodic table.
Thus, the element with a higher ionic radius in the incorrect pair is Gallium, and its atomic number is \(31\).
1. Incorrect pair: - $\mathrm{Al}<\mathrm{Ga}$
2. Ionic radius comparison: - $\mathrm{Al}^{3+}<\mathrm{Ga}^{3+}$
- The atomic number of $\mathrm{Ga}$ is 31.
Therefore, the correct answer is (1) 31.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.