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Chapter 23: Problem 61

For each of the following metals, write the electronic configuration of the atom and its \(2+\) ion: (a) \(\mathrm{Mn},(\mathbf{b}) \mathrm{Ru},(\mathbf{c}) \mathrm{Rh}\). Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(2+\) ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

Short Answer

Expert verified
The electronic configurations of Mn2+, Ru2+, and Rh2+ ions are: Mn2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\) Ru2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\) Rh2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^7\) In an octahedral strong-field complex, the d orbitals are split into two groups: \(t_{2g}\) and \(e_{g}\), as the energy levels increase. Placement of d electrons follows Hund's rule: Mn2+: one unpaired electron. Ru2+: two unpaired electrons. Rh2+: one unpaired electron.

Step by step solution

01

Find the electronic configuration of the atoms

Using the periodic table, find the atomic number of the given elements Mn, Ru, and Rh. Then, write their electronic configuration using the Aufbau principle: Mn has atomic number 25, so its electronic configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\] Ru has atomic number 44, so its electronic configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^6\] Rh has atomic number 45, so its electronic configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7\]
02

Find the electronic configuration of the 2+ ions

Now, we will remove two electrons to get the electronic configuration of the 2+ ions. We should remove electrons from the outermost energy levels: Mn2+ has electronic configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\] Ru2+ has electronic configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\] Rh2+ has electronic configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^7\]
03

Draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex

Since we assume a strong-field complex, the energy levels of the d orbitals will split into two groups: the lower-energy \(t_{2g}\) orbitals (dxy, dxz, dyz) and the higher-energy \(e_{g}\) orbitals (dz^2 and dx^2 - y^2).
04

Place the d electrons in the energy-level diagram for each 2+ ion

Using the electronic configuration of each 2+ ion, place the d electrons in the energy-level diagram following Hund's rule (maximize spin multiplicity): For Mn2+ (5 d electrons): four electrons in the \(t_{2g}\) orbitals and the last electron in the \(e_{g}\) set. For Ru2+ (6 d electrons): four electrons in the \(t_{2g}\) orbitals and two electrons in the \(e_{g}\) set. For Rh2+ (7 d electrons): six electrons in the \(t_{2g}\) and \(e_{g}\) orbitals, and the last electron in the \(t_{2g}\) set.
05

Find the number of unpaired electrons for each 2+ ion

Now, count the number of unpaired electrons in the energy-level diagram for each 2+ ion: Mn2+ has one unpaired electron. Ru2+ has two unpaired electrons. Rh2+ has one unpaired electron.

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Most popular questions from this chapter

A classmate says, "A weak-field ligand usually means the complex is high spin." Is your classmate correct? Explain.

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