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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 Fe鲁鈦� 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 脳 ChlorineOxidationState = TotalCharge x + 4 脳 (-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: Fe鲁鈦�: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}\)
03

Determine the unpaired electrons

The electron configuration of the Fe鲁鈦� 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: 鈫戔啈鈫戔啈鈫� (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

Use standard reduction potentials to calculate \(\mathscr{C}^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Fe}^{2+}\) (high and low spin) b. \(\mathrm{Fe}^{3+}(\text { high spin })\) c. \(\mathrm{Ni}^{2+}\)

Until the discoveries of Alfred Werner, it was thought that carbon had to be present in a compound for it to be optically active. Werner prepared the following compound containing \(\mathrm{OH}^{-}\) ions as bridging groups and separated the optical isomers. a. Draw structures of the two optically active isomers of this compound. b. What are the oxidation states of the cobalt ions? c. How many unpaired electrons are present if the complex is the low-spin case?

The complex ion \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\operatorname{Ru}(\text { phen })_{3}^{2+} ?\) b. \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+} \)). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

Qualitatively draw the crystal field splitting of the \(d\) orbitals in a trigonal planar complex ion. (Let the \(z\) axis be perpendicular to the plane of the complex.)
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