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Write the Lewis dot symbols of the reactants and products in the following reactions. (First balance the equations.) (a) \(\mathrm{Sr}+\mathrm{Se} \longrightarrow \mathrm{SrSe}\) (b) \(\mathrm{Ca}+\mathrm{H}_{2} \longrightarrow \mathrm{CaH}_{2}\) (c) \(\mathrm{Li}+\mathrm{N}_{2} \longrightarrow \mathrm{Li}_{3} \mathrm{~N}\) (d) \(\mathrm{Al}+\mathrm{S} \longrightarrow \mathrm{Al}_{2} \mathrm{~S}_{3}\)

Short Answer

Expert verified
Each reaction is already balanced or requires balancing, with electrons transferred as shown in the Lewis structures.

Step by step solution

01

Balance the Chemical Equation - Reaction (a)

The reaction is \( \mathrm{Sr} + \mathrm{Se} \rightarrow \mathrm{SrSe} \). Both reactants and products are balanced, as 1 strontium (Sr) and 1 selenium (Se) are present on both sides. Thus, the balanced equation is \( \mathrm{Sr} + \mathrm{Se} \rightarrow \mathrm{SrSe} \).
02

Write Lewis Dot Symbols - Reaction (a)

Strontium (Sr) has 2 valence electrons, and selenium (Se) has 6 valence electrons. Therefore, the Lewis dot symbols are \[ \mathrm{Sr} : \cdot \cdot \quad \mathrm{Se} : \cdot \cdot \cdot \cdot \cdot \cdot \]. In \( \mathrm{SrSe} \), Sr loses 2 electrons to form \( \mathrm{Sr}^{2+} \) and Se gains 2 electrons to form \( \mathrm{Se}^{2-} \), so the Lewis structure becomes complete.
03

Balance the Chemical Equation - Reaction (b)

The initial reaction is \( \mathrm{Ca} + \mathrm{H}_{2} \rightarrow \mathrm{CaH}_{2} \). Calcium provides 2 electrons to form calcium hydride (CaH鈧), which matches the number of hydrogen atoms, so the balanced equation remains the same: \( \mathrm{Ca} + \mathrm{H}_{2} \rightarrow \mathrm{CaH}_{2} \).
04

Write Lewis Dot Symbols - Reaction (b)

Calcium (Ca) has 2 valence electrons and hydrogen (H鈧) consists of a bond with an electron each. So the Lewis dot symbols are \( \mathrm{Ca} : \cdot \cdot \quad \mathrm{H} :\cdot \cdot \). In \( \mathrm{CaH}_{2} \), calcium gives up its 2 electrons to two hydrogen atoms, forming \( \mathrm{Ca}^{2+} \) and two \( \mathrm{H}^{-} \) ions.
05

Balance the Chemical Equation - Reaction (c)

The starting equation \( \mathrm{Li} + \mathrm{N}_{2} \rightarrow \mathrm{Li}_{3} \mathrm{N} \) is not balanced. We need 3 lithium atoms to bond with one nitrogen atom in \( \mathrm{Li}_{3} \mathrm{N} \), and 2 nitrogen atoms from \( \mathrm{N}_{2} \), so we balance it as follows: \( 6\mathrm{Li} + \mathrm{N}_{2} \rightarrow 2 \mathrm{Li}_{3} \mathrm{N} \).
06

Write Lewis Dot Symbols - Reaction (c)

Lithium (Li) has 1 valence electron and Nitrogen (N) has 5 valence electrons. The Lewis symbols are \( \mathrm{Li} :\cdot \quad \mathrm{N} : \cdot \cdot \cdot \cdot \cdot \). In \( \mathrm{Li}_{3} \mathrm{N} \), three Li atoms each give up one electron to form \( \mathrm{Li}^+ \), and nitrogen captures these electrons forming \( \mathrm{N}^{3-} \).
07

Balance the Chemical Equation - Reaction (d)

The unbalanced reaction is \( \mathrm{Al} + \mathrm{S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} \). Aluminum and sulfur have a 3:2 ratio in the product, requiring 2 aluminum atoms to bond with 3 sulfur atoms. Thus, we balance the equation as: \( 2\mathrm{Al} + 3\mathrm{S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} \).
08

Write Lewis Dot Symbols - Reaction (d)

Aluminum (Al) has 3 valence electrons, and sulfur (S) has 6. The Lewis representations are \( \mathrm{Al} : \cdot \cdot \cdot \quad \mathrm{S} : \cdot \cdot \cdot \cdot \cdot \cdot \). In \( \mathrm{Al}_{2} \mathrm{S}_{3} \), each Al loses 3 electrons to become \( \mathrm{Al}^{3+} \), and sulfur captures 2 electrons each forming \( \mathrm{S}^{2-} \).

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding. They are the electrons involved in forming bonds between atoms. For example, strontium (Sr) possesses 2 valence electrons, whereas selenium (Se) has 6. In chemical reactions, atoms tend to gain, lose, or share valence electrons to achieve a stable electron configuration, often resembling the nearest noble gas. This is known as the octet rule, where atoms aim to have eight electrons in their valence shell. For ionic bonds, one atom will typically lose electrons to become a cation, while another atom gains electrons to become an anion, as seen in the formation of SrSe.
Ionic Compounds
Ionic compounds arise from the electrostatic attraction between oppositely charged ions. These ions form when atoms transfer valence electrons between each other, leading to charged forms. For example, in calcium hydride (CaH鈧), calcium (Ca) loses 2 electrons to form Ca虏鈦, while hydrogen atoms each gain one electron, forming two H鈦 ions. This transfer of electrons from calcium to hydrogen atoms creates ionic bonds, holding the ions in a lattice structure. Ionic compounds often display high melting and boiling points due to the strong attractions between charged ions. They typically dissolve in water, conducting electricity when in solution due to the mobility of their ions.
Chemical Equations
Chemical equations represent chemical reactions, showing the reactants and products with their respective quantities. Balancing these equations ensures that the same number of atoms for each element exists on both sides of the reaction. For example, in the formation of lithium nitride (Li鈧僋), balancing requires 6Li and N鈧 to ensure both sides have equal elements, leading to the equation: 6Li + N鈧 鈫 2Li鈧僋. This process respects the Law of Conservation of Mass, which states that mass in a closed system must remain constant. When writing chemical equations, it's important to also consider the stoichiometry, or the relative quantities of each substance, which reflects the actual ratios necessary for the reaction to proceed as indicated.
Bonding and Structure
In chemistry, understanding how atoms bond together and form structures is essential. Atoms can bond through ionic, covalent, or metallic bonds. Ionic bonds, for example, involve the transfer of electrons, resulting in oppositely charged ions that attract each other. This is common in reactions where metals like aluminum (Al) react with nonmetals like sulfur (S) to form compounds like Al鈧係鈧.
  • Covalent bonds, on the other hand, involve the sharing of electrons between non-metal atoms.
  • Metallic bonds involve a "sea of electrons" shared among positive metal ions, allowing for conductivity and malleability of metals.
The structure of a compound impacts its properties, such as melting point, conductivity, and hardness. For instance, ionic compounds typically have a crystalline lattice structure, which accounts for their high melting points and electrical conductivity when dissolved, due to the movement of ions.

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Most popular questions from this chapter

Draw Lewis structures for the following organic molecules: (a) tetrafluoroethylene \(\left(\mathrm{C}_{2} \mathrm{~F}_{4}\right),(\mathrm{b})\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right),(\mathrm{c})\) butadiene \(\left(\mathrm{CH}_{2} \mathrm{CHCHCH}_{2}\right),(\mathrm{d})\) propyne \(\left(\mathrm{CH}_{3} \mathrm{CCH}\right),\) (e) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\). (To draw \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH},\) replace an \(\mathrm{H}\) atom in benzene with a COOH group.)

The amide ion \(\left(\mathrm{NH}_{2}^{-}\right)\) is a Br酶nsted base. Use Lewis structures to represent the reaction between the amide ion and water.

The \(\mathrm{N}-\mathrm{O}\) bond distance in nitric oxide is \(115 \mathrm{pm}\), which is intermediate between a triple bond ( \(106 \mathrm{pm}\) ) and a double bond (120 pm). (a) Draw two resonance structures for \(\mathrm{NO},\) and comment on their relative importance. (b) Is it possible to draw a resonance structure having a triple bond between the atoms?

Write a Lewis structure for \(\mathrm{SbCl}_{5}\). Does this molecule obey the octet rule?

An ionic bond is formed between a cation \(\mathrm{A}^{+}\) and an anion \(\mathrm{B}^{-}\). Based on Coulomb's law $$ E \propto \frac{Q_{1} \times Q_{2}}{d} $$ how would the energy of the ionic bond be affected by the following changes: (a) doubling the radius of \(\mathrm{A}^{+}\) (b) tripling the charge on \(\mathrm{A}^{+},(\mathrm{c})\) doubling the charges on \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-},(\mathrm{d})\) decreasing the radii of \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-}\) to half their original values?

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