The compound \([ \text{Ni}(\text{H}_2\text{O})_6 ]^{2+}\) appears green due to d-d transitions in the visible spectrum.
\([ \text{Ni}(\text{CN})_4 ]^{2-}\) is diamagnetic and does not exhibit any d-d transitions, rendering it colourless.
Thus, the correct answer is option 2.
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32