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Chapter 23: Problem 47

Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left[\mathrm{ZnCl}_{4}\right]^{2-}\) (b) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (c) \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\)

Short Answer

Expert verified
(a) \(\left[\mathrm{ZnCl}_{4}\right]^{2-}\) is diamagnetic. (b) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is diamagnetic. (c) \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is diamagnetic. (d) \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\) is paramagnetic.

Step by step solution

01

Identify the metal and its oxidation state

Zn is the central metal atom in this complex. As it is in a tetrahedral arrangement with 4 Cl atoms, and each Cl has a -1 charge, it balances the charge with a +2 oxidation state for a total charge of -2 on the complex.
02

Determine the electronic configuration

Zinc has an atomic number of 30, so its general electronic configuration is [Ar]4s^2 3d^10. When it loses 2 electrons to form the Zn^2+ ion, the electronic configuration becomes [Ar]3d^10, which means all electrons are paired.
03

Identify as diamagnetic or paramagnetic

As all electrons are paired in the Zn^2+ ion, the complex is diamagnetic. For (b) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\):
04

Identify the metal and its oxidation state

Pd is the central metal atom in this complex, and it is surrounded by 2 NH3 and 2 Cl. Both NH3 and Cl are neutral, so the oxidation state of Pd is 0.
05

Determine the electronic configuration

Palladium has atomic number 46, so it has an electronic configuration of [Kr]5s^0 4d^10. As it has no charge, this configuration remains the same in the complex.
06

Identify as diamagnetic or paramagnetic

Since all electrons in Pd are paired, the complex is also diamagnetic. For (c) \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\):
07

Identify the metal and its oxidation state

V is the central metal atom in this complex. Its oxidation state is +3, as indicated by the 3+ charge on the complex and the neutral charges of the six H2O ligands.
08

Determine the electronic configuration

Vanadium has an atomic number of 23, and its electronic configuration is [Ar]4s^2 3d^3. When it loses 3 electrons to become V^3+, the electronic configuration is [Ar]3d^0, which means all electrons are paired.
09

Identify as diamagnetic or paramagnetic

As there are no unpaired electrons in the V^3+ ion, the complex is diamagnetic. For (d) \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\):
10

Identify the metal and its oxidation state

Ni is the central metal atom in this complex. As it is surrounded by three neutral en ligands (ethylenediamine), its oxidation state is +2, as indicated by the 2+ charge on the complex.
11

Determine the electronic configuration

Nickel has atomic number 28, with an electronic configuration of [Ar]4s^2 3d^8. When it loses 2 electrons to form Ni^2+, the electronic configuration is [Ar]3d^8, which has 2 unpaired electrons.
12

Identify as diamagnetic or paramagnetic

As there are 2 unpaired electrons in the Ni^2+ ion, the complex is paramagnetic. In summary: (a) \(\left[\mathrm{ZnCl}_{4}\right]^{2-}\) is diamagnetic. (b) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) is diamagnetic. (c) \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is diamagnetic. (d) \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\) is paramagnetic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coordination Complexes
Coordination complexes form when a central metal atom or ion bonds to surrounding molecules or ions, known as ligands. These ligands are typically Lewis bases because they donate electron pairs to the metal, which is the Lewis acid. Coordination complexes often have remarkable colors, variable reactivity, and can be either diamagnetic (all electrons are paired) or paramagnetic (contains unpaired electrons).

For instance, in a coordination complex like \[\mathrm{ZnCl}_{4}\]^{2-}, zinc (Zn) acts as the central metal ion, and four chloride ions (Cl-) serve as the ligands, creating a tetrahedral geometry. Understanding the structure and composition of such complexes is key to predicting properties like magnetic behavior, which in this case, is diamagnetic due to all electrons being paired.
Electronic Configuration
Electronic configuration refers to the distribution of electrons in atomic or molecular orbitals. The arrangement of electrons within these orbitals determines the magnetic properties of atoms and complexes. For example, the zinc ion in \[\mathrm{ZnCl}_{4}\]^{2-} loses two electrons to become \[\mathrm{Zn}^{2+}\], resulting in a fully filled d-orbital (3d^10). This configuration lacks unpaired electrons, thus the complex exhibits diamagnetism. In contrast, \[\mathrm{Ni}(\mathrm{en})_{3}\]^{2+} has two unpaired electrons in its 3d orbitals after losing two electrons, showing paramagnetism as a result.
Oxidation States
The oxidation state of a central metal in a coordination complex indicates the degree of electron loss or gain relative to the atom's neutral state. It is essential for determining the complex's electronic configuration and overall charge. For example, in the complex \[\mathrm{V}(\mathrm{H}_{2} \mathrm{O})_{6}\]^{3+}, vanadium (V) is in a +3 oxidation state, which influences its electronic configuration and consequently, its magnetic properties. The accurate determination of oxidation states is a crucial step in the examination of coordination complexes.
Ligand Field Theory
Ligand Field Theory (LFT) explains the electronic structure and properties of coordination complexes. It describes how the ligands' electric fields affect the d-orbitals of the central metal ion, leading to energy splitting of these orbitals. In cases like \[\mathrm{Ni}(\mathrm{en})_{3}\]^{2+}, the field created by the ethylenediamine (en) ligands causes the d-orbitals to split into different energy levels. As electrons fill these orbitals, the existence of unpaired electrons can result in paramagnetic behavior. LFT provides a framework for understanding the color, magnetism, and reactivity of coordination compounds.

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Most popular questions from this chapter

The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a \(0.100\) - L sample of the water with a solution of EDTA \({ }^{4-}\). The EDTA \({ }^{4-}\) chelates the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+\left[\mathrm{EDTA}^{4-}\right.& \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second \(0.100-L\) sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) [EDTA] ]- Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in mg/I.

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\mathrm{e}}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\mathrm{a}}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (both octahedral), and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in ectahedral ones, assign a color to each complex.

Which of the complexes shown here are chiral? [Section 23.4] $$ \mathrm{Cr} \quad \mathrm{Cr}=\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} \quad \Theta=\mathrm{Cl} \quad \theta=\mathrm{NH}_{3} $$ (1) ( (3) (4)

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) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}(o-\mathrm{phen})\right] \mathrm{Cl}_{3}\) (b) $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \
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