Step 1: Analysis of Statement I
- The metallic radius of N a (neutral sodium atom) is 1.86 ̊A.
- The ionic radius of Na+ is smaller than the neutral atom because Na+ has one less electron, resulting in reduced electron-electron repulsion and greater effective nuclear charge on the remaining electrons. - Thus, Statement I is correct.
Step 2: Analysis of Statement II - While cations (Na+) are always smaller than their corresponding neutral atoms, anions (e.g., Cl-) are larger than their corresponding neutral atoms due to increased electron-electron repulsion in the outer shell.
- Hence, ions are not always smaller than the corresponding elements.
- Thus, Statement II is false.
Step 3: Conclusion - Statement I is correct, but Statement II is false.
Final Answer: (1)
Match the following:
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) 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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :