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Chapter 21: Problem 59

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

Short Answer

Expert verified
The \(\mathrm{FeCl}_{4}^{-}\) ion contains 3 unpaired electrons.

Step by step solution

01

Find the atomic number for Fe.

The atomic number can be found on the periodic table. It is the number associated with the chemical symbol for iron, which is Fe. According to the periodic table, the atomic number for Fe is 26.
02

Determine the number of electrons in Fe.

Iron in its elemental state should have as many electrons as its atomic number, which is 26. Iron (Fe) atoms in their ground (unexcited) state have 26 electrons.
03

Write the electronic configuration of neutral Fe.

The electronic configuration for iron (Fe) in the ground state can be written as follows: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\).
04

Write the electronic configuration of \(\mathrm{FeCl}_{4}^{-}\).

When Fe forms the \(\mathrm{FeCl}_{4}^{-}\) ion, it loses 3 electrons from the \(4s\) and \(3d\) shells. This results in 23 electrons, and their configuration can be written as: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3}\).
05

Calculate the number of unpaired electrons.

To determine the number of unpaired electrons, count the number of electrons in the d orbitals that are not paired. From the computed electronic configuration, there are 3 unpaired electrons in the d orbitals. So, the \(\mathrm{FeCl}_{4}^{-}\) ion contains 3 unpaired electrons.

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Most popular questions from this chapter

Henry Taube, 1983 Nobel Prize winner in chemistry, has studied the mechanisms of the oxidation-reduction reactions of transition metal complexes. In one experiment he and his students studied the following reaction: \(\begin{aligned} \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(a q)+& \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q) \\\ \longrightarrow & \mathrm{Cr}(\mathrm{III}) \text { complexes }+\mathrm{Co}(\mathrm{II}) \text { complexes } \end{aligned}\) Chromium(III) and cobalt(III) complexes are substitutionally inert (no exchange of ligands) under conditions of the experiment. Chromium(II) and cobalt(II) complexes can exchange ligands very rapidly. One of the products of the reaction is \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}^{2+} .\) Is this consistent with the reaction proceeding through formation of \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cr}-\mathrm{Cl}-\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\) as an intermediate? Explain.

Draw geometrical isomers of each of the following complex ions. a. \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{-}\) c. \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]\) b. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]^{2+}\) d. \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}\right]^{+}\)

Almost all metals in nature are found as ionic compounds in ores instead of being in the pure state. Why? What must be done to a sample of ore to obtain a metal substance that has desirable properties?

Tetrahedral complexes of \(\mathrm{Co}^{2+}\) are quite common. Use \(d\) -orbital splitting diagram to rationalize the stability of \(\mathrm{Co}^{2+}\) tetrahedral complex ions.

The complex trans-[NiA \(\left._{2} \mathrm{~B}_{4}\right]^{2+}\), where \(\mathrm{A}\) and \(\mathrm{B}\) represent neutral ligands, is known to be diamagnetic. Do \(\mathrm{A}\) and \(\mathrm{B}\) produce very similar or very different crystal fields? Explain.
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