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)

Question

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 BM. Choose the correct answer from the options given below.

Accepted Answer

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

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