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Question 106 of 106

Q.(i) Write a reaction for preparation of Nylon-6.

(ii) The salt of Sc3+^{3+} ion is colourless and the salt of Mn3+^{3+} ion is coloured. Explain. [Z of Sc = 21 and Z of Mn = 25]
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★
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Nylon-6 forms by ring-opening polymerisation of caprolactam; colour in transition-metal ions requires partially-filled d orbitals for d–d transitions, which Sc3+^{3+} lacks but Mn3+^{3+} has.

(i) Preparation of Nylon-6: Caprolactam (a cyclic amide) undergoes ring-opening (hydrolytic) polymerisation on heating in the presence of a small amount of water, at high temperature, to form Nylon-6.

n (Caprolactam)→ΔH2O−[NH−(CH2)5−CO]n− (Nylon-6)n\,(Caprolactam) \xrightarrow[\Delta]{H_2O} -[NH-(CH_2)_5-CO]_n-\ (Nylon\text{-}6)

(ii) Colour of Sc3+^{3+} vs Mn3+^{3+}:

Sc (Z = 21): [Ar]3d14s2[Ar]3d^14s^2. Sc3+Sc^{3+} loses all 3 valence electrons: [Ar]3d0[Ar]3d^0 — no d electrons at all, so no d–d electronic transition is possible; hence Sc3+Sc^{3+} salts are colourless.

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