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What is the Lewis symbol for each of the following atoms or ions? (a) \(\mathrm{K},\) (b) As, (c) \(\mathrm{Sn}^{2+},(\mathbf{d}) \mathrm{N}^{3-}\) .

Short Answer

Expert verified
The Lewis symbols for the given atoms and ions are: (a) \(\text{K}\bullet\) , (b) \[\begin{array}{c}\cdot \text{As} \cdot \\ | \end{array}\], (c) \(\bullet \text{Sn}^{2+} \bullet\), and (d) \[\begin{array}{c}\cdot \text{N}^{3-} \cdot \\ | \end{array}\].

Step by step solution

01

Determine valence electrons for each atom/ion

(a) Potassium (K) - As it belongs to the first group in the periodic table, it has only 1 valence electron. (b) Arsenic (As) - Being a member of Group 15 in the periodic table, it has 5 valence electrons. (c) Tin(II) ion (Sn虏鈦) - Tin, as a Group 14 element, has 4 valence electrons. However, due to the 2+ charge, which means it loses 2 electrons, that leaves it with 2 valence electrons. (d) Nitride ion (N鲁鈦) - Nitrogen, as a Group 15 element, has 5 valence electrons. Since the ion has a 3- charge, it gains 3 electrons, thus giving it a total of 8 valence electrons.
02

Represent elements and ions with their symbols and valence electrons

(a) Potassium (K) - \[\text{K}\bullet\] (b) Arsenic (As) - \[\begin{array}{c}\cdot \text{As} \cdot \\ | \end{array}\] (c) Tin(II) ion (Sn虏鈦) - \[\bullet \text{Sn}^{2+} \bullet\] (d) Nitride ion (N鲁鈦) - \[\begin{array}{c}\cdot \text{N}^{3-} \cdot \\ | \end{array}\] The Lewis symbols for the given atoms and ions are: (a) K鈥 , (b) 鈭橝s鈭檤 , (c) 鈥n虏鈦衡 , and (d) 鈭橬鲁鈦烩垯| .

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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 in an atom, which are crucial in determining how atoms interact with one another. These electrons play a significant role in chemical bonds, as they are the ones involved in gaining, losing, or sharing to form compounds.
  • Valence electrons dictate how an element behaves chemically.
  • They are located in the atom's highest energy level or shell.
  • The number of valence electrons in an atom is often represented in Lewis symbols.
For example, potassium (K) has only one valence electron because it belongs to the first group of the periodic table. Arsenic (As), on the other hand, has five valence electrons since it is part of group 15.
The tin(II) ion (Sn虏鈦) originally has four valence electrons, but due to its 2+ charge, it loses two. Thus, it has two valence electrons left. Nitride ion (N鲁鈦) gains three electrons, bringing its valence count to eight.
Understanding valence electrons enables us to visualize the interaction possibilities and predict chemical reactions.
Electron Configuration
The electron configuration of an atom describes how its electrons are distributed in different atomic orbitals. This configuration helps us understand the atom's characteristics and chemical behavior.
  • Electrons are arranged into shells and subshells (s, p, d, f) based on their energy levels.
  • Each shell follows a specific order based on increasing energy levels, usually noted as 1s, 2s, 2p, 3s, and so on.
  • The number of valence electrons stems from the electron configuration, impacting how Lewis symbols are drawn.
For example, the electron configuration for potassium is \\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \), showing that the single valence electron is in the 4s subshell. For arsenic, the configuration \\( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^3 \) places the five valence electrons in the 4s and 4p subshells.
This electron arrangement explains why some atoms are more likely to form certain types of bonds.
Periodic Table Groups
The periodic table is organized into vertical columns known as groups or families, each possessing specific chemical properties determined by their electron configuration. These groups are essential for predicting chemical behavior and properties, including the number of valence electrons.
  • Elements in the same group have the same number of valence electrons.
  • This similarity leads to comparable chemical properties among group elements.
  • Groups are numbered from 1 to 18, sometimes using old notations like IA, IIA, etc.
For instance, potassium (K) belongs to group 1 (alkali metals), sharing a single valence electron trended feature with its group members. Arsenic (As) is part of group 15, meaning it carries five valence electrons, akin to all pnictogens. Tin (Sn), in group 14, has four valence electrons initially, following the carbon group pattern, although its ion form adjusts due to electron loss. Analyzing groups helps in swiftly referencing the basic behavior and potential interactions of elements, bringing a better context to their Lewis symbols representation.

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Most popular questions from this chapter

Which of these elements are unlikely to form covalent bonds? S, H, K, Ar, Si.

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) , in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. (a) What is the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What is the predicted dominant Lewis structure if achieving the best formal charges is the top criterion?

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

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 CaO is large enough to overcome these processes. (b) 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 CaO is small. (e) CaO is stable to atmospheric conditions.

Consider the lattice energies of the following Group 2 \(\mathrm{A}\) compounds: \(\mathrm{Be} \mathrm{H}_{2}, 3205 \mathrm{kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{CaH}_{2}, 2410 \mathrm{kJ} / \mathrm{mol} ; \mathrm{SrH}_{2}, 2250 \mathrm{kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{kJ} / \mathrm{mol}\) (a) What is the oxidation number of \(\mathrm{H}\) in these compounds? (b) Assuming that all of these compounds have the same three-dimensional arrangement of ions in the solid, which of these compounds has the shortest cation-anion distance? (c) Consider BeH \(_{2} .\) Does it require 3205 kJ of energy to break one mole of the solid into its ions, or does breaking up one mole of solid into its ions release 3205 \(\mathrm{kJ}\) of energy? (d) The lattice energy of \(\mathrm{ZnH}_{2}\) is 2870 \(\mathrm{kJ} / \mathrm{mol}\) . Considering the trend in lattice enthalpies in the Group 2 \(\mathrm{A}\) compounds, predict which Group 2 \(\mathrm{A}\) element is most similar in ionic radius to the \(\mathrm{Zn}^{2+}\) ion.

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