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Chapter 23: Problem 23

A complex is written as \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{2}\), (a) What is the oxidation state of the Ni atom in this complex? (b) What is the likely coordination number for the complex? (c) If the complex is treated with excess \(\mathrm{AgNO}_{3}\left(\mathrm{aq}_{2}\right)\), how many moles of AgBr will precipitate per mole of complex?

Short Answer

Expert verified
The oxidation state of Ni in the complex is +2. The likely coordination number for the complex is 8. Per one mole of the complex, there will be 2 moles of AgBr that will precipitate.

Step by step solution

01

(a) Determine the oxidation state of the Ni atom in the complex.

Since \(\mathrm{NiBr}_2\) is a neutral molecule, the oxidation state of Br is known to be -1. Therefore, let's write the oxidation states: Ni = x, Br = -1, NH3 = 0 Ni + 2 × (-1) = 0 Ni - 2 = 0 Solve for x: x = +2 So, the oxidation state of Ni in the complex is +2.
02

(b) Determine the likely coordination number for the complex.

The coordination number is the number of ligand (Lewis base) donor atoms bonded to the central metal ion. In \(\mathrm{NiBr}_{2}\cdot 6\mathrm{NH}_{3}\), Ni is the central metal ion and the ligands are Br and \(\mathrm{NH}_{3}\). Each Br ion is capable of donating one pair of electrons while each \(\mathrm{NH}_{3}\) molecule can donate one pair of electrons through its nitrogen atom. In total, there are 2 Br ions and 6 \(\mathrm{NH}_{3}\) molecules in the complex, therefore: Coordination number = 2 (Br) + 6 (NH3) = 8 The likely coordination number for this complex is 8.
03

(c) Calculate the moles of AgBr that will precipitate per mole of complex.

To find how many moles of AgBr will precipitate per mole of complex when treated with excess \(\mathrm{AgNO}_{3}(a q)\), we need to use stoichiometry. The balanced equation for the reaction between the complex and \(\mathrm{AgNO}_{3}\) is: \(\mathrm{NiBr}_2\cdot 6\mathrm{NH}_{3} + 2\mathrm{AgNO}_{3} \rightarrow \mathrm{Ni(NO}_{3})_{2} \cdot 6\mathrm{NH}_{3} + 2\mathrm{AgBr (s)}\) From this balanced equation, we can see that one mole of complex \(\mathrm{NiBr}_2\cdot 6\mathrm{NH}_{3}\) reacts with 2 moles of \(\mathrm{AgNO}_{3}\) to produce 2 moles of AgBr precipitate. Thus, the ratio between the moles of complex and moles of precipitate is 1:2. So, per one mole of the complex, there will be 2 moles of AgBr that will precipitate.

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Key Concepts

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

Oxidation State
Understanding the oxidation state of elements within a compound is fundamental in coordination chemistry. It represents the number of electrons an atom either gains, loses, or appears to use when joining with other atoms in compounds. In the exercise, we determined the oxidation state of Nickel (Ni) in NiBr2 · 6NH3 by considering the overall charge as well as the known oxidation state of the Bromine (Br), which is -1. As Ammonia (NH3) is a neutral ligand, it does not affect the oxidation state of Ni.

The calculation went as follows: Ni, with an unknown oxidation state represented by x, is balanced by two Br ions: Ni + 2×(-1) = 0, which led to the oxidation of Ni being +2. This principle can be applied generally to deduce oxidation states in coordination complexes, ensuring that the total charge is balanced by the sum of the oxidation states.
Coordination Number
The coordination number is a count of how many atoms of ligands (Lewis bases) are directly bonded to the central metal ion. In our example, the complex NiBr2·6NH3 has Nickel (Ni) as the central metal with Bromine (Br) and Ammonia (NH3) as ligands. Each Br ion contributes one coordination site and each NH3 contributes another. With 2 Br ions and 6 NH3 molecules, the coordination number totals 8.

This calculation is crucial in predicting the shape and properties of the complex. Generally, coordination numbers range from 2 to 12, with 4 and 6 being the most common in stable complexes. The coordination number helps predict the possible geometries of the complex, crucial in understanding its reactivity and interaction with other substances.
Ligand
Ligands are ions or molecules that can donate a pair of electrons to the central metal atom or ion in a coordination complex. They are Lewis bases by definition. In the given complex, Br− and NH3 are acting as ligands. Ammonia acts as a monodentate ligand, meaning it bonds through a single pair of electrons donated from its nitrogen atom.

Ligands can be monodentate, like NH3, or polydentate, meaning they can bond through multiple sites—also known as chelating ligands. The nature and strength of the ligands affect the color, solubility, and magnetic properties of the final coordination compound. An understanding of ligands is imperative for chemists to manipulate and use these complexes in various applications such as catalysis, material science, and medicine.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. In coordination chemistry, it allows us to predict the outcomes of reactions involving complex compounds. The step-by-step solution applies stoichiometry to ascertain how many moles of AgBr precipitate for each mole of the NiBr2·6NH3 complex.

The balanced chemical equation showcases a 1:2 ratio of complex to AgBr precipitate, emphasizing the importance of stoichiometric coefficients. These coefficients are the numbers that appear before formulas in a balanced equation and are fundamental in stoichiometry, as they specify the proportional amounts of each compound involved in the reaction. Grasping stoichiometry is necessary for not only academic exercises but also practical lab work and industrial chemical processes.
Precipitation Reactions
Precipitation reactions occur when substances in solution combine to form an insoluble solid, known as the precipitate. In the exercise, we predict the formation of AgBr, a precipitate, upon the reaction of the NiBr2·6NH3 complex with excess AgNO3. The formation of a precipitate is a key sign of a chemical change during a reaction.

Recognizing precipitation reactions allows chemists to isolate substances from a mixture, purify compounds, or simply indicate the presence of various ions in a solution. Precipitation reactions play a significant role in water treatment, analytical chemistry, and the separation of substances. Understanding these reactions involves recognizing solubility rules, which provide insight into whether a substance will dissolve in a particular solvent or form a precipitate.

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Most popular questions from this chapter

The complex \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\)undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. In all cases, the ruthenium stays in the \(+3\) oxidation state and the ligands use a nitrogen donor atom to bind to the metal. $$ \left[\operatorname{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}+\mathrm{L} \longrightarrow[\operatorname{Ru}(\mathrm{EDTA}) \mathrm{L}]^{-}+\mathrm{H}_{2} \mathrm{O} $$ The rate constants for several ligands are as follows: (a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the Ru(III) in the rate-determining step, and then the ligand L binds to Ru(III) in a rapid second step. A second possible mechanism is that L approaches the complex, begins to form a new bond to the Ru(III), and displaces the water molecule, all in a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the results suggest about the relative donor ability of the nitrogens of the three ligands toward Ru(TII))? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?

Many trace metal ions exist in the blood complexed with amino acids or small peptides. The anion of the amine acid glycine (gly). NCC(=O)[O-] can act as a bidentate ligand, coordinating to the metal through nitrogen and oxygen atoms. How many isomers are possible for (a) \(\left[\mathrm{Zn}(\mathrm{gly})_{2}\right]\) (tetrahedral), (b) [ \(\left.\mathrm{Pt}(\mathrm{gly})_{2}\right]\) (square planar), (c) [Co(gly) 3\(]\) (octahedral)? Sketch all possible isomers. Use the symbol to represent the ligand.

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(Z \mathrm{n}^{2+}\) in its active site, is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogen- containing groups and a water molecule. The coordinated water molecule has a pK of 7.5, which is crucial for the enxyme's activity. (a) Draw the active site geometry for the Zn(II) center in carbonic anhydrasc, just writing " \(\mathrm{N}^{\text {" }}\) for the three neutral nitrogen ligands from the protein. (b) Compare the \(p K_{a}\) of carbonic anhydrase's active site with that of pure water, which species is more acidic?

The molecule methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at \(298 \mathrm{~K}\) for reactions of methylamine and en with \(\mathrm{Cd}^{2+}(a q)\) : $$ \begin{aligned} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons & {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) } \\ \Delta H^{\circ}=&\left.-57.3 \mathrm{~kJ} ; \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \Delta G^{\circ}=-37.2 \mathrm{k}\right] \\ & \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=&\left.-56.5 \mathrm{k} ; ; \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \Delta G^{\circ}=-60.7 \mathrm{k}\right] \end{aligned} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: $$ \begin{aligned} {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+} & 2 \mathrm{en}(a q) \rightleftharpoons \\ & {\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+(a q)}+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) } \end{aligned} $$ Based on the value of \(K\) in part (a). what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic \(\left(\Delta H^{\circ}\right)\) and the entropic \(\left(-T \Delta S^{\circ}\right)\) contributions to \(\Delta G^{\circ}\) for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the "A Closer Look" box on the chelate effect, predict the sign of \(\Delta H^{2}\) for the following hypothetical reaction: $$ \begin{aligned} {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+} & 4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ & {\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) } \end{aligned} $$

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?
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