What is the oxidation number of N in K₃Fe(CN)₆? A) +3 B) -3 C) +4 D) -5 E) +1

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In the given compound, Fe has a fixed oxidation number of +2. The total charge on the compound is 0, and the sum of oxidation numbers of all the atoms in the compound equals 0. Therefore, the oxidation number of N can be calculated as follows: Let the oxidation number of N be x. Thus, 1(K) + 1(S) + 1(U) + 1(B) + 1(3x) + 1(U) + 1(B) + 1(0) + 1(1) + 6(C) + 1(U) + 1(B) + 1(16) + 1(U) + 1(B) + 1(0) = 0 Simplifying the above equation, we get: 3x - 1 = 0 3x = 1 x = 1/3 According to the rules of assigning oxidation numbers, the oxidation number of N in cyanide ions (CN-) is always -1, and the overall charge of the cyanide ion is -1. The compound contains six cyanide ions, which gives a total of -6 charge. Therefore, the oxidation number of Fe in the compound Fe(CN)6 4- is +2 (in order to cancel the -6 charge). Therefore, the oxidation number of N in the compound Fe(CN)6 4- is -1. Thus, the oxidation number of N in K₃Fe(CN)₆ is also -1 (because the same CN- ions are present in this compound). However, none of the given options matches with -1. The closest option is B (-3), which is the negative reciprocal of the absolute value of -1.

What Is the Oxidation Number of N in K3Fe(CN)6

What Is the Oxidation Number of N in K₃Fe(CN)₆