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Chapter 8: Problem 32

Which of these elements are unlikely to form ionic bonds? \(\mathrm{Mg}, \mathrm{Al}, \mathrm{Si}, \mathrm{Br}, \mathrm{I}\).

Short Answer

Expert verified
Al, Si, and I are unlikely to form ionic bonds with the other given elements, as their electronegativity differences with the other elements are lower than the threshold typically required for ionic bond formation.

Step by step solution

01

Identify metals and non-metals

In the given elements, Mg and Al are metals, while Si, Br, and I are non-metals.
02

Calculate electronegativity differences

For the formation of ionic bonds, a high electronegativity difference between elements is expected. Ionic bonding occurs with an electronegativity difference of around 1.7 or more. The electronegativity values for the given elements are: - Mg: 1.31 - Al: 1.61 - Si: 1.90 - Br: 2.96 - I: 2.66
03

Assess electrostatic attractions

We can use electronegativity differences to estimate the likelihood of electrostatic attraction between elements: 1. Mg & Al: \(|1.31 - 1.61| = 0.3\) 2. Mg & Si: \(|1.31 - 1.90| = 0.59\) 3. Mg & Br: \(|1.31 - 2.96| = 1.65\) 4. Mg & I: \(|1.31 - 2.66| = 1.35\) 5. Al & Si: \(|1.61 - 1.90| = 0.29\) 6. Al & Br: \(|1.61 - 2.96| = 1.35\) 7. Al & I: \(|1.61 - 2.66| = 1.05\) 8. Si & Br: \(|1.90 - 2.96| = 1.06\) 9. Si & I: \(|1.90 - 2.66| = 0.76\) 10. Br & I: \(|2.66 - 2.96| = 0.3\) Note that the highest electronegativity differences occur when Mg and Br (\(1.65\)), Mg and I (\(1.35\)), Al and Br (\(1.35\)), and Si and Br (\(1.06\)) are paired. However, only the Mg-Br pairing approaches the threshold (\(1.7\)) typically required for ionic bond formation.
04

Identify unlikely candidates for ionic bonding

From the differences in electronegativity, Mg with Br is closest to forming an ionic bond, though it doesn't reach the typical threshold. The other pairings have some similarities (Al-Br and Si-Br), but the electronegativity differences are lower, making them less likely to form ionic bonds. Thus, Al, Si, and I are unlikely to form ionic bonds with the other given elements.

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Key Concepts

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

Electronegativity
Electronegativity is a measure of how strongly atoms attract bonding electrons to themselves. It is an important factor when determining the type of bond formed between two atoms. In general, the greater the difference in electronegativity between two atoms, the more likely they are to form an ionic bond.
Ionic bonds typically form when the difference in electronegativity is around 1.7 or more. For the given elements:
  • Magnesium (Mg) has an electronegativity of 1.31.
  • Aluminum (Al) has an electronegativity of 1.61.
  • Silicon (Si) has an electronegativity of 1.90.
  • Bromine (Br) has an electronegativity of 2.96.
  • Iodine (I) has an electronegativity of 2.66.
These values allow us to calculate differences in electronegativity and assess the potential for ionic bonding.
Metals and Non-metals
Metals and non-metals have distinct characteristics that influence their ability to form ionic bonds. Metals, such as magnesium and aluminum, tend to lose electrons easily due to their low electronegativity. This makes them good candidates to form cations.
Non-metals like silicon, bromine, and iodine, on the other hand, have higher electronegativities and tend to gain electrons, forming anions. The transfer of electrons from a metal to a non-metal typically results in the formation of ionic bonds.
In the original exercise, Mg and Al are identified as metals, whereas Si, Br, and I are non-metals. This categorization helps to determine which elements are more likely to engage in ionic bonding.
Electrostatic Attraction
Electrostatic attraction is the force that holds ions together in an ionic bond. It occurs between oppositely charged ions, such as cations and anions, resulting from the transfer of electrons. The strength of this attraction depends on the magnitude of the charge and the distance between ions.
When evaluating the potential for ionic bonding, it's useful to examine electronegativity differences because they provide an indirect measure of electrostatic attraction potential. For example, when considering the pairs:
  • Magnesium and bromine: electronegativity difference of 1.65
  • Aluminum and bromine: electronegativity difference of 1.35
These differences suggest a stronger likelihood of forming ionic bonds with Br, but still remain below the common threshold of 1.7. Thus, while electrostatic attraction is essential for ionic bonds, the identified pairs do not exhibit sufficient potential based on the given data.

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Most popular questions from this chapter

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3}\), (b) \(\mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6}\), (c) \(\mathrm{PbCl}_{4}\) and RbCl.

A classmate of yours is convinced that he knows everything about electronegativity. (a) In the case of atoms X and Y having different electronegativities, he says, the diatomic molecule \(X-Y\) must be polar. Is your classmate correct? (b) Your classmate says that the farther the two atoms are apart in a bond, the larger the dipole moment will be. Is your classmate correct?

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound 1: \(40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \% \mathrm{M}\), \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of \(\mathrm{M} ?(\mathbf{b})\) What is the empirical formula for Compound 2 and the identity of \(\mathrm{M}\) ? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. (e) Are compounds 1 and 2 ionic or molecular?

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and Cl, (b) \(\mathrm{Mg}\) and \(\mathrm{O},(\mathbf{c}) \mathrm{Zn}\) and \(\mathrm{Cl}\), (d) \(\mathrm{Li}\) and \(\mathrm{O}\).

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround them: \((\mathbf{a}) \mathrm{HCl},(\mathbf{b}) \mathrm{ICl}_{5},\) (c) \(\mathrm{NO}\) (d) \(\mathrm{CF}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{I}_{3}^{-}\)
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