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Chapter 21: Problem 65

How many unpaired electrons are present in the tetrahedral ion \(\mathrm{FeCl}_{4}^{-} ?\)

Short Answer

Expert verified
In the tetrahedral ion \(\mathrm{FeCl}_{4}^{-}\), the central iron (Fe) atom has an oxidation state of +3. The electron configuration of the Fe³⁺ ion is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}\), and in the 3d subshell, there are 5 unpaired electrons. Therefore, there are 5 unpaired electrons present in the tetrahedral ion \(\mathrm{FeCl}_{4}^{-}\).

Step by step solution

01

Identify the central atom and oxidation state

The central atom in the complex ion \(\mathrm{FeCl}_{4}^{-}\) is the iron (Fe) atom. To determine the oxidation state, we will use the fact that the total charge of the ion must equal the sum of the oxidation states of each element in it. Considering that a chlorine (Cl) atom has a charge of -1, we can set up the equation: IronOxidationState + 4 × ChlorineOxidationState = TotalCharge x + 4 × (-1) = -1 Solve for x: x = -1 + 4 = 3 Thus, the oxidation state of the Fe atom in the complex ion is +3.
02

Write the electron configuration

Now, we need to write down the electron configuration for Fe. In its neutral state, Fe has an atomic number of 26, which means it has 26 electrons. The electronic configuration of a neutral Fe atom is: Fe: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\) However, in the complex ion \(\mathrm{FeCl}_{4}^{-}\), the Fe atom has an oxidation state of +3 (it is losing 3 electrons). Therefore, we need to modify the electron configuration for the Fe ion: Fe³⁺: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}\)
03

Determine the unpaired electrons

The electron configuration of the Fe³⁺ ion is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}\). The number of unpaired electrons can be found by checking the occupancy of the subshells in the configuration that contain unpaired electrons. In this case, the 3d subshell is the not fully occupied one: 3d: ↑↑↑↑↑ (5 unpaired electrons) Therefore, there are 5 unpaired electrons in the tetrahedral ion \(\mathrm{FeCl}_{4}^{-}\).

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Most popular questions from this chapter

Draw all geometrical and linkage isomers of square planar \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\) .

Draw all the geometrical isomers of \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+1}\) . Which of these isomers also have an optical isomer? Draw the various isomers.

Would it be better to use octahedral \(\mathrm{Ni}^{2+}\) complexes or octahe- dral \(\mathrm{Cr}^{2+}\) complexes to determine whether a given ligand is a strong-field or weak-field ligand by measuring the number of unpaired electrons? How else could the relative ligand field strengths be determined?

Give formulas for the following. a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide

Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.
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