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Chapter 23: Problem 16

How many electrons are in the valence \(d\) orbitals in these transition-metal ions? $(\mathbf{a}) \mathrm{Ru}^{3+},(\mathbf{b}) \mathrm{Pd}^{2+},(\mathbf{c}) \mathrm{Ti}^{2+},\( (d) \)\mathrm{W}^{6+}$.

Short Answer

Expert verified
The number of electrons in the valence d orbitals for the given transition-metal ions are: \(Ru^{3+}\) has 5 electrons, \(Pd^{2+}\) has 9 electrons, \(Ti^{2+}\) has 1 electron, and \(W^{6+}\) has 1 electron.

Step by step solution

01

Determine the atomic numbers and electron configurations of the neutral atoms

The atomic numbers of the elements indicate the number of electrons in the neutral atom. Using a periodic table, we can determine the atomic numbers of the elements Ru (Ruthenium), Pd (Palladium), Ti (Titanium), and W (Tungsten): - Ru: atomic number 44 - Pd: atomic number 46 - Ti: atomic number 22 - W: atomic number 74 Now, let's find the electron configurations of the neutral atoms. We will be particularly interested in the d orbitals: - Ru: [Kr] 4d⁷ 5s¹ - Pd: [Kr] 4d¹⁰ - Ti: [Ar] 3d² 4s² - W: [Xe] 4f¹⁴ 5d⁴ 6s²
02

Determine the electron configurations of the ions

Now that we have the electron configurations of the neutral atoms, we can determine the electron configurations of the ions by removing the appropriate number of electrons. Remember that electrons are first removed from the s orbitals: - Ru³⁺: [Kr] 4d⁵ (3 electrons removed from 5s¹ and 4d⁷) - Pd²⁺: [Kr] 4d⁹ (2 electrons removed from 4d¹⁰) - Ti²⁺: [Ar] 3d¹ (2 electrons removed from 4s² and 3d²) - W⁶⁺: [Xe] 4f¹⁴ 5d¹ (6 electrons removed from 6s² and 5d⁴)
03

Count the electrons in the d orbitals of the ions

Finally, we count the number of electrons in the d orbitals of each ion to find the number of electrons in the valence d orbitals: - Ru³⁺: 5 electrons in the 4d orbital - Pd²⁺: 9 electrons in the 4d orbital - Ti²⁺: 1 electron in the 3d orbital - W⁶⁺: 1 electron in the 5d orbital Thus, the number of electrons in the valence d orbitals for each of the given transition-metal ions are as follows: - Ru³⁺: 5 electrons - Pd²⁺: 9 electrons - Ti²⁺: 1 electron - W⁶⁺: 1 electron

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Most popular questions from this chapter

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the molecule (a) Would you expect pyridine to act as a monodentate or bidentate ligand? (b) For the equilibrium reaction $$ \left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy} $$ would you predict the equilibrium constant to be larger or smaller than one?

Among the period 4 transition metals \((\mathrm{Sc}-\mathrm{Zn}),\) which elements do not form ions where there are partially filled \(3 d\) orbitals?

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

Four-coordinate metals can have either a tetrahedral or a square-planar geometry; both possibilities are shown here for $\left[\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right] .(\mathbf{a})$ What is the name of this molecule? (b) Would the tetrahedral molecule have a geometric isomer? (c) Would the tetrahedral molecule be diamagnetic or paramagnetic? (d) Would the square- planar molecule have a geometric isomer? (e) Would the square-planar molecule be diamagnetic or paramagnetic? (f) Would determining the number of geometric isomers help you distinguish between the tetrahedral and square-planar geometries? (g) Would measuring the molecule's response to a magnetic field help you distinguish between the two geometries? [Sections \(23.4-23.6]\)

A Zn electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and $1.00 \mathrm{M}\( in \)\mathrm{NH}_{3}$. When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+1.04 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)
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