In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a) \( Na_4[Fe(CN)_6]\)

(b) \( Fe_4[Fe(CN)_6]_3\)

(c) \( Fe_2[Fe(CN)_6]\)

(d) \( Fe_3[Fe(CN)_6]_4\)

Asked by Abhisek | 11 months ago |  147

1 Answer

Solution :-

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

Hence, the Prussian blue colour is due to the formation of Fe4[Fe(CN)6]3.

Answered by Pragya Singh | 11 months ago

Related Questions

Which of the following carbocation is most stable?

(a) \((CH_3)_3 C.C^{+}H_2\)

(b) \( (CH_3)_3\; C^{+}\)

(c) \( CH_3 CH_2C^{+}H_2\)

(d) \( CH_3 C^{+}HCH_2CH_3\)

Class 11 Chemistry Organic Chemistry Some Basic Principles and Techniques View Answer

The reaction:

\(CH_3 CH_2 I+KOH_{(aq)} \rightarrow CH_3 CH_2 OH +KI\) is classified as :

(a) electrophilic substitution

(b) nucleophilic substitution

(c) elimination

(d) addition

Class 11 Chemistry Organic Chemistry Some Basic Principles and Techniques View Answer

The best and latest technique for isolation, purification and separation of organic compounds is:

(a) Crystallisation

(b) Distillation

(c) Sublimation

(d) Chromatography

Class 11 Chemistry Organic Chemistry Some Basic Principles and Techniques View Answer

In the organic compound \( CH_2=CH–CH_2–CH_2–C≡CH\), the pair of hydridised orbitals involved in the formation of: \( C_2 – C_3\) bond is:

(a) sp – sp2

(b) sp – sp3

(c) sp2– sp3

(d) sp3– sp3

Class 11 Chemistry Organic Chemistry Some Basic Principles and Techniques View Answer