- Statement I: Transition metals tend to have more stable higher oxidation states as we move down the group due to the increasing availability of d-orbitals. This is in contrast to p-block elements, where higher oxidation states become less stable down the group.
- Statement II: Copper can indeed liberate hydrogen from weak acids such as hydrochloric acid, which is true for copper in the form of copper(I) chloride.
Thus, both statements are true, and the correct answer is (3).
List I | List II |
---|---|
(A) Cr2O72− | (I) +3 |
(B) MnO4− | (II) +5 |
(C) VO3− | (III) +7 |
(D) FeF63− | (IV) +6 |
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
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: