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A palladium complex formed from a solution containing bromide ion and pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) (a good electron-pair donor), is found on elemental analysis to contain 37.6\(\%\) bromine, 28.3\(\%\) carbon, 6.60\(\%\) nitrogen, and 2.37\(\%\) hydrogen by mass. The compound is slightly soluble in several organic solvents; its solutions in water or alcohol do not conduct electricity. It is found experimentally to have a zero dipole moment. Write the chemical formula, and indicate its probable structure.

Short Answer

Expert verified
The empirical formula of the palladium complex is PdBrC₅H₅N. Its probable square-planar structure consists of a central palladium atom coordinated with two bromine atoms above and below and two pyridine ligands on adjacent sides, creating a symmetrical and non-polar structure with a zero dipole moment: Br | N - Pd - N | Br

Step by step solution

01

Calculate moles of each element

To find the moles of each element, we will use the given mass percentage and divide by the respective molar mass. Let's assume a 100 g sample, so the mass% will be equal to the mass of the element in the sample. Moles of Br = (37.6 g) / (79.90 g/mol) = 0.4706 mol Moles of C = (28.3 g) / (12.01 g/mol) = 2.36 mol Moles of N = (6.60 g) / (14.01 g/mol) = 0.471 mol Moles of H = (2.37 g) / (1.01 g/mol) = 2.35 mol
02

Calculate mole ratios of each element

Now, let's find the mole ratio for each element by dividing the moles of each element by the smallest number found in the previous step. Mole ratio of Br = 0.4706 mol / 0.4706 mol = 1 Mole ratio of C = 2.36 mol / 0.4706 mol = 5 Mole ratio of N = 0.471 mol / 0.4706 mol = 1 Mole ratio of H = 2.35 mol / 0.4706 mol = 5
03

Determine Empirical Formula

Using the mole ratios from Step 2, we can now write the empirical formula for the palladium complex as: BrC₅H₅N However, we also need to include Palladium in the formula. Since we know the whole complex is neutral and pyridine could coordinate with the metal as a ligand similar to ammonia, we can write the empirical formula as: PdBrC₅H₅N
04

Propose Probable Structure

The information that the compound has a zero dipole moment suggests that the complex has a symmetrical structure. Based on this, we can propose the following probable square-planar structure for the palladium complex: Br | N - Pd - N | Br In this structure, the palladium atom is in the center with bromine atoms above and below and pyridine ligands replacing two adjacent hydrogen atoms in the aromatic ring on both sides. This arrangement results in a symmetrical and non-polar structure with a zero dipole moment.

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Key Concepts

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

Chemical Formula
The chemical formula of a compound gives us a precise representation of the elements present and their respective quantities within a single molecule of that compound. When referring to the palladium complex in this exercise, the chemical formula is derived not just from its components, but from how they interact with each other.
In the case of this palladium complex, it was determined to have the following elements: bromine, carbon, nitrogen, and hydrogen, forming part of a larger structure with palladium at its core.
Using both the elemental percentages found in the compound and considering the neutral charge required of the complex, we establish the empirical presence of each element within a molecular unit. This gives us the formula: PdBrC₅H₅N. This notation is a concise way of summarizing our palladium complex, describing the balance and coordination of multiple ligands around the central metal atom.
Empirical Formula
An empirical formula is a simplified kind of chemical formula. It gives the simplest whole number ratio of the elements in a compound. For the palladium complex described, this kind of formula is extremely useful to determine how the compound is constructed on a basic level.
By calculating the mole ratios from the mass percentages of the elements, we can pin down the approach to find the simplest form. We examined the sample's composition, considering each element's molar mass to find the number of moles present.
We found the ratios of bromine, carbon, nitrogen, and hydrogen to be 1:5:1:5 respectively. With palladium added as a central element, not governed by a percentage in this formula, it transforms the empirical formula into PdBrC₅H₅N. This compact visualization represents the base ratio in the overall chemical composition.
Symmetrical Structure
The symmetry of a molecular structure significantly influences its physical properties, such as the dipole moment. In our context, the noted zero dipole moment suggests the palladium complex must be symmetrical.
A symmetrical structure implies that the molecule is evenly balanced, without regions of unequal charge distribution. Thus, when we propose the structure for the palladium complex, symmetry suggests a probable square-planar arrangement.
In this configuration, the central palladium atom is surrounded symmetrically by bromine and pyridine ligands in a square-planar geometry. This not only accounts for the observed zero dipole moment but also explains the distribution of elements in a stable, equally set formation.
Elemental Analysis
Elemental analysis is the process of determining the internal constitution of a chemical compound based on its elemental composition. For the palladium complex discussed, this method allowed for the calculation of empirical and subsequently chemical formulas.
By analyzing the mass percentages of bromine, carbon, nitrogen, and hydrogen, we could convert these percentages into moles, revealing the underlying ratios required for the complex's empirical formula.
This analysis is invaluable because it informs chemists about the amount and type of each element, paving the way for further analysis of the structural and chemical characteristics of the compound. It acts as a foundational step in drawing out the chemical and empirical formulas accurately.

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Most popular questions from this chapter

Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the \(+3\) rather than in the \(+2\) oxidation state (for metals that form stable \(+3\) ions in the first place). Suggest an explanation, keeping in mind the Lewis acid-base nature of the metal-ligand bond.

The square-planar complex \(\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\right]\) only forms in one of two possible geometric isomers. Which isomer is not observed: cis or trans?

For each of the following molecules or polyatomic ions, draw the Lewis structure and indicate if it can act as a monodentate ligand, a bidentate ligand, or is unlikely to act as a ligand at all: (a) ethylamine, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) , (b) trimethylphosphine, \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3},\) (c) carbonate, \(\mathrm{CO}_{3}^{2-},\) \((\mathbf{d})\) ethane \(, \mathrm{C}_{2} \mathrm{H}_{6}.\)

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(Il) complex; deoxyhemoglobin, without the 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) \(\mathrm{A} 15\) -minute exposure to air containing 400 \(\mathrm{ppm}\) 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) \(\mathrm{CO}\) is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

For a given metal ion and set of ligands, is the crystal-field splitting energy larger for a tetrahedral or an octahedral geometry?

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