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NEET CHEMISTRYMedium

Match List-I with List-II. List-I: (a) [Fe(CN)6]3−[Fe(CN)_6]^{3-}, (b) [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+}, (c) [Fe(CN)6]4−[Fe(CN)_6]^{4-}, (d) [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}. List-II: (i) 5.92 BM, (ii) 0 BM, (iii) 4.90 BM, (iv) 1.73 BM. Choose 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)

Step-by-Step Solution

Magnetic moment, μ=n(n+2)\mu = \sqrt{n(n+2)} BM (where nn = number of unpaired electrons). For [Fe(CN)6]3−[Fe(CN)_6]^{3-}, n=1n=1, μ=1.73\mu=1.73 BM. For [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+}, n=5n=5, μ=5.92\mu=5.92 BM. For [Fe(CN)6]4−[Fe(CN)_6]^{4-}, n=0n=0, μ=0\mu=0 BM. For [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}, n=4n=4, μ=4.90\mu=4.90 BM.

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