The reduction of \(K_2Cr_2O_7\) by SO2 produces Cr3+ ions, which impart a green colour to the solution.
Sulphur dioxide
Sulphur trioxide
Carbon dioxide
Hydrogen sulphide
When \(K_2Cr_2O_7\) is acidified with dilute \(H_2SO_4 \)and exposed to\( SO_2\), it gets reduced, turning green due to the formation of\( Cr^3+\) ions:
\(Cr_2O ^{2−} _7 + 3SO_2 + 2H ^+ → 2Cr^{3+} + 3SO^{2−}_ 4 + H_2O\).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The elements, in the columns of the periodic table in which d subshells are being occupied are known as d block elements.
These are the elements that have the capability of forming stable cations with incompletely filled d orbitals. Elements like mercury and Zinc are not considered transition metals because they have electronic configurations: (n-1)d10 ns2. These elements have filled d-orbitals in their ground state and, therefore, even in some of their oxidation states.