Home
Class 11
CHEMISTRY
(a). If both (A) and (R) are correct and...

(a). If both (A) and (R) are correct and (R) is the correct explanation for (A).
(b). If both (A) and (R) are correct and (R) is not the correct explanation for (A).
(c). If (A) is correct and (R) is incorrect.
(d). If (A) is incorrect and (R) is correct.(e). If both (A) and (R) are incorrect.
Q. Assertion (A): Hydroxylamine `(NH_2OH)` contains N, and hence gives prussian blue colour in lassaigne's test.
Reason (R): Hydroxylamine does not contain C, so with `Na` metal. `CN^(-)` ion is not formed.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both the assertion (A) and the reason (R) provided. ### Step-by-Step Solution: 1. **Understanding Assertion (A)**: - Assertion (A) states that "Hydroxylamine (NH₂OH) contains N, and hence gives Prussian blue colour in Lassaigne's test." - In Lassaigne's test, the presence of nitrogen in an organic compound is tested by fusing it with sodium metal. If cyanide ions (CN⁻) are formed, they can react to give a Prussian blue color with ferric ions (Fe³⁺). - Hydroxylamine does contain nitrogen, but we need to check if it can give the Prussian blue color. 2. **Understanding Reason (R)**: - Reason (R) states that "Hydroxylamine does not contain C, so with Na metal, CN⁻ ion is not formed." - Hydroxylamine (NH₂OH) indeed does not contain carbon. Therefore, when it is fused with sodium, it cannot form sodium cyanide (NaCN), which is necessary for the Prussian blue test. 3. **Evaluating Assertion (A)**: - Since hydroxylamine does not contain carbon, it cannot form cyanide ions when fused with sodium. Therefore, it will not give the Prussian blue color in Lassaigne's test. - Thus, Assertion (A) is incorrect. 4. **Evaluating Reason (R)**: - Reason (R) correctly states that hydroxylamine does not contain carbon, which is why CN⁻ cannot be formed. - Therefore, Reason (R) is correct. 5. **Conclusion**: - Assertion (A) is incorrect, while Reason (R) is correct. - According to the options provided, this corresponds to option (d): "If (A) is incorrect and (R) is correct." ### Final Answer: The correct answer is (d): If (A) is incorrect and (R) is correct.
To solve the question, we need to analyze both the assertion (A) and the reason (R) provided. ### Step-by-Step Solution: 1. **Understanding Assertion (A)**: - Assertion (A) states that "Hydroxylamine (NH₂OH) contains N, and hence gives Prussian blue colour in Lassaigne's test." - In Lassaigne's test, the presence of nitrogen in an organic compound is tested by fusing it with sodium metal. If cyanide ions (CN⁻) are formed, they can react to give a Prussian blue color with ferric ions (Fe³⁺). - Hydroxylamine does contain nitrogen, but we need to check if it can give the Prussian blue color. ...
Promotional Banner

Topper's Solved these Questions

  • PURIFICATION OF ORGANIC COMPOUNDS AND QUALITATIVE AND QUANTITATIVE ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Single Correct Answer Type|40 Videos

  • PERIODIC CLASSIFICATION OF ELEMENTS AND GENERAL INORGANIC CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Archives )Subjective|4 Videos

  • REDOX REACTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives (Integers)|1 Videos

Similar Questions

Explore conceptually related problems

(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. ltbr. (e). If both (A) and (R) are incorrect. Q. Assertion (A): Essential oils are volatile and are isoluble in H_2O Reason (R): Essential oils are purified by steam ditillation.

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): Concentration of H_(2)O_(2) is expressed in volume Reason (R): Volume strength = normality xx5.6

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): H_(3)PO_(2) is mono basic acid. Reason (R): Two H-atoms are attached to pghosphorous (P).

(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is not the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. (e). If both (A) and (R) are incorrect. Q. Assertion (A): Dumas method is more applicable to nitrogen containing organic compounds than Kjeldahl's method. Reason (R): Kjeldahl's method does not give stisfactory results for compounds in which N is linked to O atom.

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): IN the reaction i_(2) is a oxidant. 2S_(2)O_(3)^(2-)+I_(2)toS_(4)O_(6)^(2-)+2I^(ɵ) Reason (R): During oxidation loss of electron takes place.

(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. (e). If both (A) and (R) are incorrect. Q. Assertion (A): In organic layer test, Cl_2 water is added to the sodium extract, which oxidises Br^(-) and I^(-) ions to Br_2 and I_2 , respectively. Reason (R): Reduction potential of Cl_2 is greater than that of Br_2 and I_2 .

(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. ltbr. (e). If both (A) and (R) are incorrect. Q. Assertion (A): Benzene (boiling point 353K) and methanol (boiling point 338K) are separated by simple distillation. Reason (R): Fractional distillation is used to separate two liquids from their mixture when their boiling points differ by 20^@ or so.

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): In the titration of strong acid and strong base, phenolphthalein is used as suitable indocator. Reason (R): IN the titration of strong acid and strong base, the equivalence points lies is the pH range of (3.0-10.5) and phenolphthalein have pH range of (8.0-9.8) .

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion (A): Equivalent mass of KMnO_(4) is equal to one-fifth of its molecular mass when it acts as oxidising agent in mild basic medium. Reason (R): Oxidation number of Mn in KMnO_(4) is +7 .

(a). IF both (A) and (R) are correct and (R) is the correct explanation of (A). (b). If both (A) and (R) are correct but (R) is not the correct explanation of (A). (c). If (A) is correct, but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. ltbr. (e) if both (A) and (R) are incorrect. Q. Assertion: 5.0 " mol of "ferrous oxalate are completely oxidised by 2.5 moles of acidic solution of K_(2)Cr_(2)O_(7) . Reason(R): n-factor of ferrous oxalate against dichromate is 3.