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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

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{V}^{3+}\), (c) \(\mathrm{Ni}^{2+}\), (d) \(\mathrm{Pt}^{2+}\), (e) \(\mathrm{Ge}^{2-}\), (f) \(\mathrm{Ba}^{2+}\).

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group 4. Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of Hf behave as valence orbitals? Which behave as core orbitals?

A common form of elemental phosphorus is the white phosphorus, where four \(\mathrm{P}\) atoms are arranged in a tetrahedron. All four phosphorus atoms are equivalent. White phosphorus reacts spontaneously with the oxygen in air to form \(\mathrm{P}_{4} \mathrm{O}_{6} .\) (a) How many valance electron pairs are in the \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule? (b) When \(\mathrm{P}_{4} \mathrm{O}_{6}\) is dissolved in water, it produces a \(\mathrm{H}_{3} \mathrm{PO}_{3}\), molecule. \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has two forms, \(\mathrm{P}\) forms 3 covalent bonds in the first form and \(\mathrm{P}\) forms 5 covalent bonds in the second form. Draw two possible Lewis structures of \(\mathrm{H}_{3} \mathrm{PO}_{3}\). (c) Which structure obeys the octet rule?

(a) True or false: The hydrogen atom is most stable when it has a full octet of electrons. (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3},\) how many electrons must it gain to achieve an octet?

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+},\) and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-}\). Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of \(\mathrm{CaO}\) is large enough to overcome these processes. (b) \(\mathrm{CaO}\) is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of \(\mathrm{CaO}\) is small. (e) \(\mathrm{CaO}\) is stable to atmospheric conditions.
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