In the Victor-Meyer's test, the colour given by 1°, 2° and 3° alcohols are respectively:

In the Victor-Meyer's test, alcohols are classified as primary (1°), secondary (2°), or tertiary (3°) based on the colour they produce. The test involves several steps:

First, the alcohol is converted to an alkyl iodide using iodine and red phosphorus. Then, the alkyl iodide reacts with silver nitrite to form a nitroalkane. The nitroalkane is treated with nitrous acid (generated from sodium nitrite and hydrochloric acid), and finally, the mixture is made alkaline with sodium hydroxide.

The colour observed depends on the type of alcohol:

For a primary alcohol (1°), the nitroalkane formed is primary. This primary nitroalkane reacts with nitrous acid to form a nitrolic acid. When treated with sodium hydroxide, the nitrolic acid gives a red colour.

For a secondary alcohol (2°), the nitroalkane formed is secondary. This secondary nitroalkane reacts with nitrous acid to form a pseudonitrol (a nitroso compound), which appears blue.

For a tertiary alcohol (3°), no reaction occurs in the initial steps because tertiary alcohols undergo dehydration to form alkenes instead of alkyl iodides. Therefore, no nitro compound is formed, and the mixture remains colourless.

So, the colours given by 1°, 2°, and 3° alcohols are red, blue, and colourless, respectively.

Now, comparing with the options:

Option A: Red, colourless, blue → Incorrect, as it assigns colourless to 2° and blue to 3°.

Option B: Red, blue, colourless → Correct, as it matches red for 1°, blue for 2°, and colourless for 3°.

Option C: Colourless, red, blue → Incorrect, as it assigns colourless to 1° and blue to 3°.

Option D: Red, blue, violet → Incorrect, as it assigns violet to 3° instead of colourless.

Hence, the correct answer is Option B.