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A solution containing \(2.675 \mathrm{~g}\) of \(\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}\) (molar mass \(=267.5 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of \(\mathrm{AgNO}_{3}\) to give \(4.78 \mathrm{~g}\) of \(\mathrm{AgCl}\) (molar mass \(\left.=143.5 \mathrm{~g} \mathrm{~mol}^{-1}\right) .\) The formula of the complex is (Atomic mass of \(\mathrm{Ag}=108 \mathrm{u}\) ) [2010] (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{CoCl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}\) (c) \(\left[\mathrm{CoCl}_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]\) (d) \(\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right] \mathrm{Cl}_{2}\)

Short Answer

Expert verified
The formula of the complex is (a) \([\text{Co}(\text{NH}_3)_6]\text{Cl}_3\).

Step by step solution

01

Calculate Moles of AgCl

First, determine the moles of \(\text{AgCl}\) formed. Use the mass given in the problem and the molar mass of \(\text{AgCl}\). The formula to find moles is \( moles = \frac{mass}{molar\ mass} \). So, the moles of \(\text{AgCl}\) is \( \frac{4.78 \text{ g}}{143.5 \text{ g/mol}} \approx 0.0333 \text{ mol} \).
02

Find Moles of Cl⁻ Ion

Assuming all chlorine ends up as \(\text{AgCl}\), the moles of \(\text{Cl}^-\) ion would be the same as the moles of \(\text{AgCl}\), which is \(0.0333 \text{ mol}\).
03

Determine Moles of CoCl₃·6NH₃

Calculate the moles of \( \text{CoCl}_{3} \cdot 6 \text{NH}_{3} \) using its given mass and molar mass: \( moles = \frac{2.675 \text{ g}}{267.5 \text{ g/mol}} \approx 0.0100 \text{ mol} \).
04

Calculate Cl⁻ Ion per Mole of Compound

Each molecule of the complex releases chloride ions. Since \(0.0100 \text{ mol} \) of the compound gives \(0.0333 \text{ mol} \) of \(\text{Cl}^-\), the ratio is \( \frac{0.0333}{0.0100} = 3.33 \), implying that each molecule of the complex releases \(3\) chloride ions.
05

Determine the Formula of the Complex

Check which of the given options releases 3 chloride ions: (a) \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \) releases 3 \(\text{Cl}^-\) ions from its formula, matching our requirement of 3 \(\text{Cl}\) ions. Confirm that it fits.

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Key Concepts

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

Coordination Compounds
Coordination compounds are fascinating entities wherein a central metal atom or ion binds to surrounding molecules or ions, known as ligands. To fully understand coordination chemistry, we need to dive into several key concepts that define these unique compounds.
- **Central Metal Ion or Atom**: This refers to the core element around which the coordination occurs, often a transition metal like cobalt, iron, or nickel. These metals have empty orbitals that allow the attraction and bonding with ligands.
- **Ligands**: Ligands are the ions or molecules that donate a pair of electrons to the central atom, forming what are called "coordination bonds." Common ligands include water, ammonia, and chloride ions. In our original exercise problem, \(\mathrm{NH}_3\) acts as a ligand coordinating with cobalt.
- **Coordination Sphere**: The coordination sphere refers to the central metal ion and its attached ligands, typically enclosed in square brackets in chemical formulas, like \[\mathrm{Co}(\mathrm{NH}_3)_6\]. Outside the brackets, you have ions like \(\mathrm{Cl}^-\) that are not part of the coordination sphere.
- **Applications**: These compounds have immense applications ranging from catalysis in industrial processes to their role in biological systems like hemoglobin, which is responsible for oxygen transport in our bodies.
Molar Mass Calculations
Molar mass, a crucial term in chemistry, represents the mass of one mole of a chemical compound and is measured in grams per mole (g/mol). Understanding how to calculate molar mass is fundamental in solving problems related to coordination compounds.
- **Calculating Molar Mass**: To find the molar mass of a compound, simply add up the molar masses of all the atoms present in the formula. In our exercise, the molar mass of \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\) is given as \(267.5 \mathrm{~g} \mathrm{~mol}^{-1}\). This takes into account one cobalt atom, three chlorine atoms, and six ammonia molecules.
- **Importance of Precision**: It’s vital to use accurate atomic masses, especially when solving problems where precise measurement can impact results greatly, like determining the yield of product in a reaction.
- **Application in Coordination Complexes**: By knowing the molar mass, one can calculate quantities like the number of moles of a substance, which is essential when dealing with reactions involving these compounds. For example, converting grams to moles enables us to understand how much of a reactant is used or what quantity of a product is formed. This is pivotal when dealing with exercises similar to the one we've seen.
Stoichiometry
Stoichiometry plays a central role in chemistry by enabling the quantitative analysis of reactants and products in a chemical reaction. This concept is as applicable to coordination compounds as it is to other chemical problems.
- **Understanding Stoichiometry**: At its core, stoichiometry involves calculations that relate the quantities of reactants and products in chemical reactions. It's all about the mole-to-mole relationship dictated by balanced chemical equations.
- **Application to the Exercise**: In the context of coordination chemistry and our exercise, we used stoichiometry to compare the moles of different substances. Calculating the moles of \(\mathrm{AgCl}\), as done in step 1 of the solution, gives us essential information about the moles of \(\mathrm{Cl}^-\) ions released by the complex.
- **Relevance in Complexes**: Understanding stoichiometry is crucial for predicting the outcomes of reactions involving coordination compounds, such as precipitation reactions with \(\mathrm{AgNO}_3\), as seen in our exercise. This foundational concept allows chemists to design reactions and processes with precision.
Chemical Formulas in Coordination Chemistry
In coordination chemistry, chemical formulas provide a concise way to represent the complex interactions between metal ions and ligands. Understanding these formulas is imperative for solving problems related to coordination compounds.
- **Structure of Formulas**: The chemical formula of a coordination compound includes the central metal atom, ligands, and sometimes counter ions. These formulas, like \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\), need careful interpretation to decipher the coordination environment and stoichiometric ratios.
- **Interpreting Formulas**: Knowing how to read these formulas helps us understand the number of ligands attached to the metal, which influences the geometry and reactivity of the compound. In our practice problem, analyzing formulas helped us determine the number of chloride ions, which was crucial to find the correct answer from the given options.
- **Why It Matters**: Recognizing the intricacies of coordination formulas aids in predicting chemical behavior, understanding electronic properties, and designing compounds with desired functionalities. This aspect is vital for students who wish to master chemistry since it directly relates to how substances interact in real-world applications.

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Most popular questions from this chapter

In \(\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\), the isomerism shown is [2002] (a) optical (b) ionization (c) geometrical (d) ligand

For a complex \(\mathrm{MA}_{3} \mathrm{~B}_{3}\) possessing a trigonal prismatic geometry, the number of possible isomers are (a) 3 (b) 4 (c) 5 (d) 6

How many EDTA (ethylenediamine-tetraacetic acid) molecules are required to make an octahedral complex with a \(\mathrm{Ca}^{2+}\) ion? [2006] (a) \(\operatorname{six}\) (b) three (c) one (d) two

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) }\left[\mathrm{MnCl}_{6}\right]^{3-} & \text { (p) One unpaired electron } \\ \text { (b) }\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} & \text { (q) } \mathrm{d}^{2} \mathrm{sp}^{3} \\ \text { (c) }\left[\mathrm{CoF}_{6}\right]^{3-} & \text { (r) } \mathrm{sp}^{3} \mathrm{~d}^{2} \\ \text { (d) }\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2-} & \text { (s) Four unpaired electrons } \\ & \text { (t) Three unpaired electrons } \\ \hline \end{array} \end{aligned} $$

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) }\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} & \text { (p) Octahedral } \\ \text { (b) }\left[\mathrm{MnF}_{6}\right]^{4-} & \text { (q) Paramagnetic } \\\ \text { (c) }\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} & \text { (r) Square planar } \\ \text { (d) }\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} & \text { (s) } \mathrm{d}^{2} \mathrm{sp}^{3} \text { hybridization } \\ & \text { (t) Weak field ligand. } \\ \hline \end{array} \end{aligned} $$

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