Step 1: Understand the trends in atomic radii in the periodic table.
Across a period (from left to right): Atomic radius generally decreases due to an increase in the effective nuclear charge. Electrons are added to the same energy level, while the number of protons in the nucleus increases, leading to a stronger attraction between the nucleus and the electrons, thus pulling them closer.
Down a group (from top to bottom): Atomic radius generally increases because electrons are added to higher energy levels (increasing the principal quantum number \( n \)), resulting in a larger electron cloud. The effect of increased nuclear charge is outweighed by the addition of electron shells and increased shielding by inner electrons.
Step 2: Analyze each option based on the periodic trends.
(1) \( r_{Br}<r_K \) Bromine (Br) is in Period 4, Group 17, and Potassium (K) is in Period 4, Group 1. Across Period 4, atomic radius decreases from left to right.
Therefore, \( r_K>r_{Br} \), which means \( r_{Br}<r_K \) is a correct trend. (2) \( r_{Mg}<r_{Al} \) Magnesium (Mg) is in Period 3, Group 2, and Aluminum (Al) is in Period 3, Group 13. Across Period 3, atomic radius decreases from left to right. Therefore, \( r_{Mg}>r_{Al} \), which means \( r_{Mg}<r_{Al} \) is an incorrect trend. (3) \( r_{Rb}<r_{Cs} \) Rubidium (Rb) is in Period 5, Group 1, and Cesium (Cs) is in Period 6, Group 1. Down Group 1 (alkali metals), atomic radius increases.
Therefore, \( r_{Cs}>r_{Rb} \), which means \( r_{Rb}<r_{Cs} \) is a correct trend. (4) \( r_{At}<r_{Cs} \) Astatine (At) is in Period 6, Group 17, and Cesium (Cs) is in Period 6, Group 1. Across Period 6, atomic radius generally decreases from left to right.
Therefore, \( r_{Cs}>r_{At} \), which means \( r_{At}<r_{Cs} \) is a correct trend.
Step 3: Identify the incorrect trend.
Based on the analysis, the incorrect trend in atomic radii is given in option (2).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).