| 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 |
To solve this chemistry question, we need to match each compound from List-I with its corresponding color from List-II. Let's analyze each compound:
Based on the above analysis, the correct matching of compounds to their colors is A-III, B-I, C-II, D-IV. This matches with the provided correct answer.
To solve this matching problem, we need to correctly associate each compound in List-I with its corresponding color in List-II. Let us consider each compound one by one:
Based on this analysis, the correct matching of each compound with its respective color is:
The correct answer is: A-III, B-I, C-II, D-IV.
Match List - I with List - II:
List - I:
(A) \([ \text{MnBr}_4]^{2-}\)
(B) \([ \text{FeF}_6]^{3-}\)
(C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\)
(D) \([ \text{Ni(CO)}_4]\)
List - II:
(I) d²sp³ diamagnetic
(II) sp²d² paramagnetic
(III) sp³ diamagnetic
(IV) sp³ paramagnetic
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
