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Chapter 21: Problem 14

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

Short Answer

Expert verified
The stability of the tetrahedral Co^2+ complex can be rationalized using a d-orbital splitting diagram. The Co^2+ ion has 7 d-electrons, filling 3 electrons in the lower-energy e orbitals, and 4 electrons in the higher-energy t2 orbitals. The Crystal Field Stabilization Energy (CFSE) for the Co^2+ tetrahedral complex is calculated to be \(0.6\Delta_{tet}\), which is positive, indicating that the complex is energetically favored and stable.

Step by step solution

01

Obtain the electron configuration of Co^2+

Cobalt, Co, has an atomic number of 27, and its electron configuration is [Ar] 3d^7 4s^2. When Co loses two electrons to form Co^2+, its electron configuration becomes [Ar] 3d^7.
02

Understand d-orbital splitting in a tetrahedral complex

In a tetrahedral complex, the d-orbitals split into two groups due to the ligands' effect. The d_xy, d_xz, and d_yz orbitals form a lower-energy group (called e set), while the d_z^2 and d_x^2-y^2 orbitals form a higher-energy group (called t2 set). The energy difference between these two groups is denoted as Δ_tet.
03

Calculate the Crystal Field Stabilization Energy (CFSE)

Crystal Field Stabilization Energy (CFSE) is the energy difference between the electron configuration situation in a free metal ion and the respective complex ion. For the tetrahedral complex, the stabilization energy per electron in the e orbitals is \(-\frac{2}{5}\Delta_{tet}\), and in the t2 orbitals, it is \(+\frac{3}{5}\Delta_{tet}\). To compute the CFSE, we will multiply the stabilization energies with the number of electrons in each orbital group and sum them up.
04

Create the d-orbital splitting diagram for Co^2+ tetrahedral complex

Since the Co^2+ ion has 7 d-electrons, we will fill in the available d-orbitals following Hund's rule and the Aufbau principle: - 3 electrons in the e orbitals (lower-energy group) - 4 electrons in the t2 orbitals (higher-energy group)
05

Calculate CFSE for the Co^2+ tetrahedral complex

We can now compute the CFSE for the Co^2+ tetrahedral complex using the stabilization energies and the number of electrons in each group: CFSE=\((-3\cdot \frac{2}{5} + 4\cdot \frac{3}{5})\Delta_{tet}\) = \(0.6\Delta_{tet}\) Since the CFSE value is positive, it indicates that the tetrahedral Co^2+ complex is stabilized relative to the free Co^2+ ion.
06

Conclusion

The stability of the tetrahedral Co^2+ complex can be rationalized using a d-orbital splitting diagram. The calculated CFSE for the complex is positive, which suggests that it is energetically favored and leads to the complex's stability.

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Most popular questions from this chapter

Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution: $$\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \stackrel{\mathrm{Basic}}{\longrightarrow} \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)$$ Balance this equation by using the half-reaction method.

When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. $\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\( Would you expect \)\mathrm{Hg} \mathrm{I}_{4}^{2-}$ to form colored solutions? Explain.

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) b. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}$ c. \(\mathrm{K}_{2}\left[\mathrm{PtC}_{4}\right]\) d. \(\mathrm{K}_{4}\left[\mathrm{Pt} \mathrm{C}_{6}\right]\) e. $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ f. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

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.)

Name the following complex ions. a. \(\operatorname{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. $\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}$ d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{55} \mathrm{NO}_{2}^{2+}\) e. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) f. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}\) g. \(\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}\) h. $\mathrm{Co}(\mathrm{SCN})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{+}$
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