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27 October, 01:06

In k4[fe (cn) 6], how many 3d electrons does the iron atom have?

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  1. 27 October, 01:26
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    First let's find out the oxidation number of Fe in K₄[Fe (CN) ₆] compound.

    The oxidation number of cation, K is + 1. Hence, the total charge of the anion, [Fe (CN) ₆] is - 4. CN has charge has - 1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.

    Sum of the oxidation numbers of each element = Charge of the compound

    a + 6 x (-1) = - 4

    a - 6 = - 4

    a = + 2

    Hence, oxidation number of Fe in [Fe (CN) ₆]⁴⁻ is + 2.

    Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26.

    Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²

    When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.

    Hence, the electronconfiguration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

    = [Ar] 3d⁶

    Hence, the number of 3d electrons of Fe in K₄[Fe (CN) ₆] compound is 6.
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