Match List - I with List - II.List - I : (a) [Fe(CN)_6]^3- (b) [Fe(H_2O)_6]^3+ (c) [Fe(CN)_6]^4- (d) [Fe(H_2O)_6]^2+List - II : (i) 5.92 BM (ii) 0 BM (iii) 4.90 BM (iv) 1.73 BMChoose the correct answer from the options given below.

1. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

2. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

3. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

4. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

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Best answer

Answer: 4. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Key idea: find unpaired electrons \(n\) and use \(\mu=\sqrt{n(n+2)}\) BM.

(a) \(\mathrm{[Fe(CN)_6]^{3-}}\): \(\mathrm{Fe^{3+}}\) \(d^5\), strong field, low spin, n = 1: 1.73 BM (iv)
(b) \(\mathrm{[Fe(H_2O)_6]^{3+}}\): \(d^5\), weak field, n = 5: 5.92 BM (i)
(c) \(\mathrm{[Fe(CN)_6]^{4-}}\): \(\mathrm{Fe^{2+}}\) \(d^6\), low spin, n = 0: 0 BM (ii)
(d) \(\mathrm{[Fe(H_2O)_6]^{2+}}\): \(d^6\), high spin, n = 4: 4.90 BM (iii)

So (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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