List-I (Compound) | List-II (Colour) |
---|---|
(A) \(Fe_4[Fe(CN)_6]_3.xH_2O\) | (I) Violet |
(B) \( [Fe(CN)_5NOS]^{4–} \) | (II) Blood Red |
(C) \( [Fe(SCN)]^{2+}\) | (III) Prussian Blue |
(D) \((NH_4)_3PO_4.12MoO_3\) | (IV) Yellow |
Matching Explanation :
Fe$_4$[Fe(CN)$_6$]$_3$ $\cdot$ $x$H$_2$O: Prussian Blue (Colour: III)
[Fe(CN)$_5$NOS]$^{4-}$: Violet (Colour: I)
[Fe(SCN)]$^{2+}$: Blood Red (Colour: II)
(NH$_4$)$_3$PO$_4$ $\cdot$ 12MoO$_3$: Yellow (Colour: IV)
A-III, B-I, C-II, D-IV
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