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Chapter 23: Problem 33

The \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) ion, which has a square- planar geometry, is diamagnetic, whereas the \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) ion, which has a tetrahedral geometry, is paramagnetic. Show the crystal field splitting diagrams for those two complexes.

Short Answer

Expert verified
The reason for \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) being diamagnetic (not attracted to magnetic fields) and \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) being paramagnetic (attracted to magnetic fields) comes from Crystal Field Theory (CFT). The difference in electromagnetic fields from their respective ligands causes the d orbital electrons to split into different energy levels. The exact pattern of splitting and filling up of these levels results in the \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) complex having all paired electrons (and thus being diamagnetic), while the \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) complex has one unpaired electron (and thus being paramagnetic).

Step by step solution

01

Knowing the Crystal Field Theory

Crystal Field Theory (CFT) is a model that describes the bonding, levels of energy, and the properties of transition metal complexes. It's a tool to understand the magnetic properties, colors, and structures of these complexes. According to the CFT, when ligands approach the central metal ion, degenerate (same energy) orbitals of the metal ion split into different energy levels.
02

Configuring the electron for Nickel ion (Ni^{2+})

Nickel in its neutral state has the electron configuration of [Ar] 3d8 4s2. In this exercise, we are dealing with the Ni^{2+} ion, which means two electrons are removed. That results in the electron configuration [Ar] 3d8. So, Ni^{2+} has eight d electrons.
03

Crystal Field Splitting in Square Planar Complex

In a square planar complex such as \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\), the d orbitals split into two levels due to the crystal field of the ligands. The more directly facing dxy, dxz, and dyz orbitals have higher energy, and dz^2 and dx^2-y^2 have lower energy. For the Ni^{2+} ion with eight d electrons, these electrons will fill the lower energy orbitals first (due to the Aufbau principle), resulting in a full lower energy level and a half-filled higher energy level. Consequently, this renders the complex diamagnetic - it has all paired electrons.
04

Crystal Field Splitting in Tetrahedral Complex

For a tetrahedral complex such as \(\left[\mathrm{NiCl}_{4}\right]^{2-}\), the d orbitals split into two energy levels. But this time, the dx^2-y^2 and dz^2 have higher energy, while dxy, dxz, and dyz have lower energy. Filling these with the 8 d-electrons of the Ni^{2+} ion, the result still is a fully filled lower energy level, but now an unpaired electron is in the higher energy level. This unpaired electron makes the complex paramagnetic - it's attracted to magnetic fields.

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Most popular questions from this chapter

A student in 1895 prepared three coordination compounds containing chromium, with the following properties: $$ \begin{array}{llc} \hline & & \begin{array}{l} \mathrm{Cl}^{-} \text {Ions in } \\ \text { Solution per } \\ \text { Formula Unit } \end{array} \\ \text { Formula } & \text { Color } & 3 \\ \hline \text { (a) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Violet } & 3 \\ \text { (b) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Light green } & 2 \\ \text { (c) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Dark green } & 1 \\ \hline \end{array} $$ Write modern formulas for these compounds and suggest a method for confirming the number of \(\mathrm{Cl}^{-}\) ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates and \(\mathrm{Cr}\) has a coordination number of 6 in all the compounds.

Consider the following two ligand exchange reactions: \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+6 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O}\) $$ \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+3 \mathrm{en} \rightleftharpoons\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} $$ (a) Which of the reactions should have a larger \(\Delta S^{\circ} ?\) (b) Given that the Co-N bond strength is approximately the same in both complexes, which reaction will have a larger equilibrium constant? Explain your choices.

Write the electron configurations of the following ions: \(\mathrm{V}^{5+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Cu}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}\)

In a dilute nitric acid solution, \(\mathrm{Fe}^{3+}\) reacts with thiocyanate ion (SCN \(^{-}\) ) to form a dark-red complex: $$ \mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+} $$ The equilibrium concentration of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, \(1.0 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) was mixed with \(1.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{KSCN}\) and \(8.0 \mathrm{~mL}\) of dilute \(\mathrm{HNO}_{3}\). The color of the solution quantitatively indicated that the \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) concentration was \(7.3 \times 10^{-5} M .\) Calculate the formation constant for \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)

Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammin eaquachlorocobalt(III) chloride.
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