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

Short Answer

Expert verified
The d-orbital splitting diagrams for the octahedral complex ions are as follows: a. Zn虏鈦: No splitting occurs as all d-orbitals are completely filled (d鹿鈦 configuration). b. Co虏鈦: - High spin: 4 orbitals with one electron and 1 orbital with two electrons. - Low spin: 3 orbitals with one electron and 1 orbital with three electrons. c. Ti鲁鈦: One electron in the lowest energy d-orbital and four empty d-orbitals.

Step by step solution

01

a. Zn虏鈦 d-orbital Splitting Diagram

Step 1: Determine the electron configuration of Zn虏鈦. Zinc has an atomic number of 30, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2\] In the Zn虏鈦 ion, it loses two electrons from the 4s orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\] Step 2: Draw the d-orbital splitting diagram. Zn虏鈦 ion is a d鹿鈦 metal, meaning that all five d-orbitals are completely filled. Since there is no interaction between the d-electrons and ligands, there is no splitting in the Zn虏鈦 d-orbitals.
02

b. Co虏鈦 (High and Low Spin) d-Orbital Splitting Diagram

Step 1: Determine the electron configuration of Co虏鈦. Cobalt has an atomic number of 27, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\] In the Co虏鈦 ion, it loses two electrons, one from the 4s and one from the 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\] Step 2: Draw the d-orbital splitting diagram for high spin. In a high-spin complex, the crystal field splitting energy (螖) is small, and the electrons will fill all five d-orbitals before any of the orbitals are doubly occupied. The diagram will show 4 orbitals with one electron and 1 orbital with two electrons. Step 3: Draw the d-orbital splitting diagram for low spin. In a low-spin complex, the crystal field splitting energy (螖) is large, and the electrons will first completely fill the lower energy d-orbitals before moving to the higher energy orbitals. The diagram will show 3 orbitals with one electron and 1 orbital with three electrons.
03

c. Ti鲁鈦 d-Orbital Splitting Diagram

Step 1: Determine the electron configuration of Ti鲁鈦. Titanium has an atomic number of 22, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2\] In the Ti鲁鈦 ion, it loses two electrons from the 4s and one from the 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\] Step 2: Draw the d-orbital splitting diagram for Ti鲁鈦. The Ti鲁鈦 ion is a d鹿 metal, meaning it has only one unpaired electron in its d-orbitals. The octahedral splitting diagram will show one electron in the lowest energy d-orbital, and the other four orbitals will be empty.

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Most popular questions from this chapter

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. \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}\right]^{+}\)

a. Calculate the molar solubility of AgBr in pure water. \(K_{\text {sp }}\) for AgBr is \(5.0 \times 10^{-13}\) b. Calculate the molar solubility of AgBr in \(3.0 M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\), that is, \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}\) c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{~mL}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3}\) ? e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and \(\mathrm{b}\) ?

There are three salts that contain complex ions of chromium and have the molecular formula \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\). Treating \(0.27 \mathrm{~g}\) of the first salt with a strong dehydrating agent resulted in a mass loss of \(0.036 \mathrm{~g}\). Treating \(270 \mathrm{mg}\) of the second salt with the same dehydrating agent resulted in a mass loss of \(18 \mathrm{mg}\). The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to \(100.0-\mathrm{mL}\) portions of \(0.100 M\) solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 \(\mathrm{mg} \mathrm{AgCl} ;\) another, \(2870 \mathrm{mg} \mathrm{AgCl}\); the third, \(4300 \mathrm{mg} \mathrm{AgCl}\). Two of the salts are green and one is violet. Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

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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