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Chapter 2: Problem 113

For each pair of elements, (i) through (vii), (a) Determine whether an ionic compound, a molecular compound, or no compound would form. (b) Write an appropriate formula for each compound you expect to form and name the compound. (i) \(\quad\) Chlorine and bromine (ii) \(\quad\) Lithium and tellurium (iii) Sodium and argon (iv) \(\quad\) Magnesium and fluorine (v) Nitrogen and bromine (vi) \(\quad\) Indium and sulfur (vii) Selenium and bromine

Short Answer

Expert verified
(i) No compound; (ii) Li鈧俆e, lithium telluride; (iii) No compound; (iv) MgF鈧�, magnesium fluoride; (v) NBr鈧�, nitrogen tribromide; (vi) In鈧係鈧�, indium sulfide; (vii) SeBr鈧�, selenium dibromide.

Step by step solution

01

Determine Types of Elements

Identify whether each element in the pair is a metal, non-metal, or metalloid. Metals tend to form ionic compounds with non-metals, while pairs of non-metals can form molecular compounds.
02

Analyzing Pair (i): Chlorine and Bromine

Both chlorine and bromine are non-metals from the halogen group. They will form a molecular (covalent) compound due to their similar electronegativity. No stable compound is typically formed as they are both halogens. So, the answer is no compound would form.
03

Analyzing Pair (ii): Lithium and Tellurium

Lithium is a metal, and tellurium is a metalloid with non-metallic characteristics and can accept electrons. Lithium and tellurium will form an ionic compound. The lithium ion is Li鈦�, and tellurium typically forms the Te虏鈦� ion, so the formula is Li鈧俆e, named lithium telluride.
04

Analyzing Pair (iii): Sodium and Argon

Sodium is a metal, but argon is a noble gas. Noble gases do not usually form compounds under normal conditions. Therefore, no compound is formed between sodium and argon.
05

Analyzing Pair (iv): Magnesium and Fluorine

Magnesium is a metal, and fluorine is a non-metal. They form an ionic compound because magnesium loses electrons and fluorine gains electrons. Magnesium ion is Mg虏鈦� and fluorine ion is F鈦�, leading to the formula MgF鈧�, named magnesium fluoride.
06

Analyzing Pair (v): Nitrogen and Bromine

Both nitrogen and bromine are non-metals. They can form molecular compounds. One common compound is nitrogen tribromide, where nitrogen shares electrons with bromine, with the formula NBr鈧�.
07

Analyzing Pair (vi): Indium and Sulfur

Indium is a metal, and sulfur is a non-metal. They form an ionic compound. Indium forms a 3+ ion as In鲁鈦�, and sulfur forms a 2- ion as S虏鈦�. The formula is In鈧係鈧�, named indium sulfide.
08

Analyzing Pair (vii): Selenium and Bromine

Both selenium and bromine are non-metals. They can form a molecular compound. The typical compound is selenium dibromide with the formula SeBr鈧�.

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Key Concepts

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

Ionic Compounds
Ionic compounds are formed between metals and non-metals. They are created when metals lose electrons to become positively charged ions, while non-metals gain those electrons to become negatively charged ions. These oppositely charged ions attract each other, resulting in a stable ionic bond.
For example:
  • Magnesium fluoride, which forms from magnesium, a metal losing electrons, and fluorine, a non-metal gaining electrons, results in the ionic compound MgF鈧�.
  • Lithium telluride is another ionic compound forming from the metal lithium (Li鈦�) and metalloid tellurium (Te虏鈦�). The compound formula is Li鈧俆e.
In ionic compounds, the total positive charge must balance the total negative charge. This balance is what determines the subscripts in the chemical formula.
Molecular Compounds
Molecular compounds are also known as covalent compounds. They usually form between non-metal elements, which share electrons rather than transferring them. This sharing creates a stable covalent bond. Unlike ionic compounds, molecular compounds do not consist of ions but neutral molecules held by these shared electrons.
For instance:
  • Nitrogen tribromide (NBr鈧�) is a molecular compound where nitrogen and bromine, both non-metals, share electrons.
  • An example seen in selenium dibromide (SeBr鈧�) where selenium and bromine form covalent bonds by sharing electrons.
Molecular compounds often have lower melting and boiling points compared to ionic compounds because the intermolecular forces holding them together are relatively weaker.
Chemical Formulas
Chemical formulas represent the types and numbers of atoms in a compound. They are vital in understanding a compound鈥檚 composition and properties. For ionic compounds, chemical formulas are determined by the need to balance positive and negative charges. For example, in MgF鈧�, each magnesium ion (Mg虏鈦�) is balanced by two fluoride ions (F鈦�).
For molecular compounds, chemical formulas depict the number of each type of atom present in a molecule. A good case is NBr鈧�, where three bromine atoms are covalently bonded to one nitrogen atom. The subscript numbers in formulas indicate these ratios.
Element Pairing
Understanding which elements can pair to form compounds is crucial in chemistry.
Some key principles include:
  • Metals and non-metals typically form ionic compounds due to the transfer of electrons.
  • Pairs of non-metals form molecular compounds by sharing electrons.
  • Elements like noble gases rarely form compounds because of their complete outer electron shells.
In exercises analyzing element pairs, observe whether the elements involved can achieve a stable electron arrangement, which drives compound formation.
Chemistry Education
Chemistry education helps students understand the world at a molecular level. Core concepts like ionic and molecular bonding, interpreting chemical formulas, and recognizing element pairing create foundational knowledge.
To enhance learning, students should:
  • Engage in hands-on experiments to see theoretical concepts in action.
  • Utilize visual aids and models to grasp complex topics more easily.
  • Practice by solving diverse exercises to reinforce theoretical understanding.
A strong grasp of these concepts equips students with problem-solving skills that are applicable across various scientific fields.

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Most popular questions from this chapter

Although not a transition element, lead can form two cations: \(\mathrm{Pb}^{2+}\) and \(\mathrm{Pb}^{4+}\). Write the formulas for the compounds of these ions with the chloride ion.

Predict which compounds are ionic. Explain your answers. (a) \(\mathrm{CF}_{4}\) (b) \(\mathrm{SrBr}_{2}\) (c) \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{SiO}_{2}\) (e) \(\mathrm{KCN}\) (f) \(\mathrm{SCl}_{2}\)

A mixture contains only \(\mathrm{MgSO}_{4}\) and \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} .\) If the mass percent of \(\mathrm{MgSO}_{4}\) in the mixture is \(32.0 \%,\) what is the mass percent of sulfate in the mixture?

Chalky, white crystals in mineral collections are often labeled borax, which has the molecular formula \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O},\) when actually they are partially dehydrated samples with the molecular formula \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) which is more stable under the storage conditions. Real crystals of borax are colorless and transparent. (a) Calculate the percent mass that the mineral has lost when it partially dehydrates. (b) Is the percent boron by mass the same in both compounds?

Which sets of values are possible? Why are the others not possible? Explain your reasoning. $$ \begin{array}{lrrrr} \hline & \begin{array}{l} \text { Mass } \\ \text { Number } \end{array} & \begin{array}{l} \text { Atomic } \\ \text { Number } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Protons } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Neutrons } \end{array} \\ \hline \text { (a) } & 19 & 42 & 19 & 23 \\ \text { (b) } & 235 & 92 & 92 & 143 \\ \text { (c) } & 53 & 131 & 131 & 79 \\ \text { (d) } & 32 & 15 & 15 & 15 \\ \text { (e) } & 14 & 7 & 7 & 7 \\ \text { (f) } & 40 & 18 & 18 & 40 \\ \hline \end{array} $$
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