Explanation of Lassaigne’s Test: - Lassaigne’s test involves the conversion of elements such as nitrogen, sulphur, phosphorus, and halogens present in an organic compound into their ionic forms by fusion with sodium metal. - The resulting sodium fusion extract (also known as Lassaigne’s extract) can be tested for these elements using specific reagents.
- Detection of Nitrogen: The sodium fusion extract is treated with ferrous sulphate (FeSO$_4$) and acidified with sulphuric acid. The formation of a Prussian blue colour indicates the presence of nitrogen.
Detection of Sulphur: The extract is treated with lead acetate solution. A black precipitate of lead sulphide (PbS) indicates the presence of sulphur.
Detection of Halogens: The extract is treated with silver nitrate (AgNO$_3$) solution after acidification with nitric acid. The formation of a white, yellow, or pale yellow precipitate indicates the presence of chlorine, bromine, or iodine, respectively.
- Detection of Phosphorous: Phosphorous is detected by the formation of a yellow precipitate of ammonium phosphomolybdate when the extract is treated with ammonium molybdate in the presence of nitric acid.
Conclusion: Lassaigne’s test is used for the detection of nitrogen, sulphur, phosphorus, and halogens.
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ____. Use: Atomic mass of N (in amu) = 14
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions is:
List-I | List-II |
---|---|
(P) Passing H2S in the presence of NH4OH | (1) Cu2+ |
(Q) (NH4)2CO3 in the presence of NH4OH | (2) Al3+ |
(R) NH4OH in the presence of NH4Cl | (3) Mn2+ |
(S) Passing H2S in the presence of dilute HCl | (4) Ba2+ (5) Mg2+ |
Match List I with List II:
Choose the correct answer from the options given below:
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: