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Chapter 23: Problem 58

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}$, (b) $\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]$, (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\), (d) \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4}\), , (e) $\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$

Short Answer

Expert verified
The number of valence d electrons associated with each central metal ion are: (a) Pt(II): 8 d electrons (b) Cu(II): 9 d electrons (c) Os(III): 5 d electrons (d) Cr(III): 3 d electrons (e) Cd(II): 10 d electrons

Step by step solution

01

Determine the oxidation state of the central metal ion (Pt)

In this case, the oxidation state of Pt can be found using the total charge of the complex, which is zero. We have: Pt + 2 (NH₃) + 2 (Cl) = 0 Since ammonia (NH₃) and chloride (Cl-) are neutral ligands, the oxidation state of the central metal ion Pt is: Pt(II)
02

Determine the number of valence d electrons associated with Pt(II)

The ground-state electron configuration of Pt is \([Xe] 4f^{14} 5d^9 6s^1\). When Pt loses two electrons to become Pt(II), its electron configuration will be \([Xe] 4f^{14} 5d^8\). The number of valence d electrons is thus 8. (b) \(\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]\)
03

Determine the oxidation state of the central metal ion (Cu)

The total charge of the complex is 2-, because each potassium ion has a charge of +1. We have: Cu + 2 (C₂O₄²⁻) = -2 Thus, the oxidation state of Cu is: Cu(II)
04

Determine the number of valence d electrons associated with Cu(II)

The ground-state electron configuration of Cu is \([Ar] 3d^{10} 4s^1\). When Cu loses two electrons to become Cu(II), its electron configuration will be \([Ar] 3d^9\). The number of valence d electrons is thus 9. (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\)
05

Determine the oxidation state of the central metal ion (Os)

The total charge of the complex is 3+, because each chloride ion has a charge of -1. We have: Os + 3 (en) = +3 Since ethylenediamine (en) is a neutral ligand, the oxidation state of the central metal ion Os is: Os(III)
06

Determine the number of valence d electrons associated with Os(III)

The ground-state electron configuration of Os is \([Xe] 4f^{14} 5d^6 6s^2\). When Os loses three electrons to become Os(III), its electron configuration will be \([Xe] 4f^{14} 5d^5\). The number of valence d electrons is thus 5. (d) \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4}\)
07

Determine the oxidation state of the central metal ion (Cr)

The total charge of the complex is 1-, as the sulfate ion has a charge of 2-. We have: Cr + (EDTA⁴⁻) = -1 Thus, the oxidation state of Cr is: Cr(III)
08

Determine the number of valence d electrons associated with Cr(III)

The ground-state electron configuration of Cr is \([Ar] 3d^5 4s^1\). When Cr loses three electrons to become Cr(III), its electron configuration will be \([Ar] 3d^3\). The number of valence d electrons is thus 3. (e) \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\)
09

Determine the oxidation state of the central metal ion (Cd)

The total charge of the complex is 2+, because each chloride ion has a charge of -1. We have: Cd + 6 (H₂O) = +2 Since water (H₂O) is a neutral ligand, the oxidation state of the central metal ion Cd is: Cd(II)
10

Determine the number of valence d electrons associated with Cd(II)

The ground-state electron configuration of Cd is \([Kr] 4d^{10} 5s^2\). When Cd loses two electrons to become Cd(II), its electron configuration will be \([Kr] 4d^{10}\). The number of valence d electrons is thus 10.

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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.

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?

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) molecule, is a high- spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about \(10 \%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin? (e) CO is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) hexaammineiron(II) nitrate (b) tetraaquadibromochromium(III) perchlorate (c) ammonium hexachloropalladate(IV) (d) diammineoxolatonickel(II) (e) Hexaamminemolybdenum(III) tetrachlorocuprate(II)

Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) \(\operatorname{Cr}(\mathrm{EDTA})^{-}\) (b) $\left[\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right] \mathrm{Br}_{2}$ (c) $\left[\mathrm{Ru}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{3}$ (d) \(\mathrm{K}_{2}\left[\mathrm{Fe}(\mathrm{O}\) -phen \()(\mathrm{CN})_{4}\right]\)
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