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

Short Answer

Expert verified
Silicon (Si) is unlikely to form ionic bonds.

Step by step solution

01

Understand Ionic Bond Formation

Ionic bonds are formed between a metal and a non-metal. Metals lose electrons to become positively charged cations, while non-metals gain electrons to become negatively charged anions. Examine the given elements to categorize them as metals or non-metals.
02

Classify Each Element

Identify the nature of each element: - Mg is a metal. - Al is a metal. - Si is a metalloid (neither a true metal nor non-metal). - Br is a non-metal. - I is a non-metal.
03

Determine Likelihood of Ionic Bond Formation

- Mg (metal) can form ionic bonds with non-metals such as Br and I. - Al (metal) can also form ionic bonds with non-metals such as Br and I. - Si, being a metalloid, typically does not form ionic bonds but rather covalent bonds due to its shared characteristics. - Br and I are both non-metals more likely to form ionic bonds with metals than with each other or with Si.
04

Conclusion on Ionic Bond Likelihood

Identify elements unlikely to form ionic bonds based on the analysis. Si, being a metalloid with a preference for covalent bonding, is the least likely to form an ionic bond, particularly with the non-metal elements given.

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Key Concepts

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

Metals and Non-Metals
Metals and non-metals play a crucial role in the formation of ionic bonds. Ionic bonds occur between metals and non-metals because of their complementary electrical charges when they form ions. Here's how they work:

Metals, such as magnesium (Mg) and aluminum (Al), readily lose electrons during chemical reactions. This loss of electrons results in the formation of positively charged ions, or cations. On the other hand, non-metals, including bromine (Br) and iodine (I), have a tendency to gain electrons, forming negatively charged ions known as anions.

This exchange of electrons from metals to non-metals allows the creation of strong ionic bonds. The opposite charges (positive and negative) attract each other, leading to the formation of a stable ionic compound.

  • Metals: lose electrons, form cations
  • Non-metals: gain electrons, form anions
  • Ionic bonds result from the attraction between cations and anions
Understanding the nature of the elements and their ability to exchange electrons is key to identifying likely ionic bond formations.
Metalloids
Metalloids like silicon (Si) possess characteristics of both metals and non-metals, positioning them uniquely on the periodic table. These elements are not completely inclined towards ion formation like true metals or non-metals. Instead, metalloids usually prefer to share electrons and form covalent bonds rather than losing or gaining electrons to form ionic bonds.

Silicon, for example, typically forms covalent bonds by sharing its electrons with other elements. This shared electron approach helps them achieve a stable electronic configuration similar to that of noble gases, an essential aspect of bond formation.

  • Properties of both metals and non-metals
  • Prefer covalent bonding by sharing electrons
  • Less likely to engage in ionic bonding
Recognizing the nature of metalloids helps understand why they seldom take part in ionic bond formation, unless under unique circumstances.
Covalent Bonds
Covalent bonds are fundamentally different from ionic bonds. They emerge from the sharing of electrons between atoms, typically between non-metals or between a non-metal and a metalloid.

This type of bonding occurs because neither atom is sufficiently electronegative to completely remove an electron from the other. Rather, they achieve stability by pooling their electrons in a kind of mutual exchange. Silicon (Si), being a metalloid, often engages in covalent bonding.

Elements such as Br and I, though non-metals, prefer ionic bonds with metals. However, they may also form covalent bonds with metalloids like Si or with each other when no metals are available to provide the required electrons for ionic bonding.

  • Involves the sharing of electron pairs
  • Occurs mainly among non-metals and some metalloids
  • Essential for achieving stable electronic configurations
Understanding covalent bonds helps explain the likelihood of Si forming covalent rather than ionic bonds in certain circumstances.

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Most popular questions from this chapter

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance structures are possible that satisfy the octet rule? (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for \(\mathrm{BeCl}_{2} ?\)

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 \(2: 69.59 \%\) 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 M? (b) What is the empirical formula for Compound 2 and the identity of 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. \((\mathbf{e})\) Are compounds 1 and 2 ionic or molecular?

The iodine monobromide molecule, IBr, has a bond length of \(249 \mathrm{pm}\) and a dipole moment of \(1.21 \mathrm{D}\). (a) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, \(e\).

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{BF}_{3}\) from \(\mathrm{B}\) and \(\mathrm{F}\) atoms, showing valence- shell electrons. (a) How many valence electrons does B have initially? (b) How many bonds F has to make in order to achieve an octet? (c) How many valence electrons surround the B in the \(\mathrm{BF}_{3}\) molecule? (d) How many valence electrons surround each F in the \(\mathrm{BF}_{3}\) molecule? (e) Does \(\mathrm{BF}_{3}\) obey the octet rule?

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \% \mathrm{~N}\). Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} / \mathrm{mol}\). The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(S-S\) distance in the \(S_{8}\) ring is 205 pm.) \((\mathbf{d})\) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} / \mathrm{mol}^{-1} . \Delta H_{f}^{\circ}\) of \(\mathrm{S}(g)\) is \(222.8 \mathrm{~kJ} / \mathrm{mol}\). Estimate the average bond enthalpy in the compound.

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