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Write electron configurations for the most stable ion formed by each of the elements Te, Cl, Sr, and Li (when in stable ionic compounds).

Short Answer

Expert verified
The electron configurations for the most stable ions of the given elements when in stable ionic compounds are: - Te虏鈦 (Tellurium): \(\mathrm{[Kr] \: 4d^{10} \: 5s^2 \: 5p^6}\) - Cl鈦 (Chlorine): \(\mathrm{[Ne] \: 3s^2 \: 3p^6}\) - Sr虏鈦 (Strontium): \(\mathrm{[Kr]}\) - Li鈦 (Lithium): \(\mathrm{1s^2}\)

Step by step solution

01

Te (Tellurium)

The atomic number of Te is 52. Its electron configuration for the neutral atom is: \[\mathrm{[Kr] \: 4d^{10} \: 5s^2 \: 5p^4}\]
02

Cl (Chlorine)

The atomic number of Cl is 17. Its electron configuration for the neutral atom is: \[\mathrm{[Ne] \: 3s^2 \: 3p^5}\]
03

Sr (Strontium)

The atomic number of Sr is 38. Its electron configuration for the neutral atom is: \[\mathrm{[Kr] \: 5s^2}\]
04

Li (Lithium)

The atomic number of Li is 3. Its electron configuration for the neutral atom is: \[\mathrm{1s^2 \: 2s^1}\] Step 2: Determine the charges of the most stable ions
05

Te (Tellurium)

To achieve a stable electron configuration, Te needs to gain 2 electrons to fill its 5p subshell. Therefore, the charge of the most stable Te ion is -2.
06

Cl (Chlorine)

To achieve a stable electron configuration, Cl needs to gain 1 electron to fill its 3p subshell. Therefore, the charge of the most stable Cl ion is -1.
07

Sr (Strontium)

To achieve a stable electron configuration, Sr needs to lose its 2 electrons in the 5s subshell to expose the filled 4d subshell. Therefore, the charge of the most stable Sr ion is +2.
08

Li (Lithium)

To achieve a stable electron configuration, Li needs to lose its 1 electron in the 2s subshell to expose the filled 1s subshell. Therefore, the charge of the most stable Li ion is +1. Step 3: Write the electron configurations for the stable ions
09

Te虏鈦 (Tellurium)

The electron configuration for the Te虏鈦 ion is: \[\mathrm{[Kr] \: 4d^{10} \: 5s^2 \: 5p^6}\]
10

Cl鈦 (Chlorine)

The electron configuration for the Cl鈦 ion is: \[\mathrm{[Ne] \: 3s^2 \: 3p^6}\]
11

Sr虏鈦 (Strontium)

The electron configuration for the Sr虏鈦 ion is: \[\mathrm{[Kr]}\]
12

Li鈦 (Lithium)

The electron configuration for the Li鈦 ion is: \[\mathrm{1s^2}\]

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Key Concepts

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

Stable Ions
Stable ions are forms of elements that have gained or lost electrons to achieve a full outer electron shell. Elements tend to form ions by either shedding or acquiring electrons to reach a more stable, lower-energy state known as the "octet rule." This rule suggests that atoms are most stable when they have eight electrons in their valence shell鈥攕imilar to the electron configuration of noble gases.
- **Te (Tellurium):** Gains 2 electrons to form Te虏鈦, achieving a full 5p subshell. - **Cl (Chlorine):** Gains 1 electron to form Cl鈦, completing its 3p subshell. - **Sr (Strontium):** Loses 2 electrons to form Sr虏鈦, revealing a stable 4d subshell. - **Li (Lithium):** Loses 1 electron to form Li鈦, resulting in a full 1s subshell.
By rearranging their electron configurations in this way, ions attain stability, mimicking the electron configurations of noble gases, which are inherently stable.
Atomic Number
The atomic number of an element defines the number of protons in an atom's nucleus and is crucial for determining its place on the periodic table. It also equals the number of electrons in a neutral atom, shaping the atom's basic electron configuration. Understanding an element's atomic number helps identify how it might form stable ions.
- **Te (Tellurium):** Atomic number 52, with its electron configuration adjusted when forming ions. - **Cl (Chlorine):** Atomic number 17, requiring just one more electron to fill its outer shell. - **Sr (Strontium):** Atomic number 38, losing two electrons to match the stable configuration. - **Li (Lithium):** Atomic number 3, shedding an electron to simplify its structure.
These atomic numbers guide us in predicting the movement of electrons, whether they will be gained or lost, and thus what stable ions will form.
Electron Configuration Notation
Electron configuration notation is a concise way of depicting the distribution of electrons among various atomic orbitals. It can be broken into subshells designated by letters (s, p, d, f) and numbers representing the quantity of electrons in each.
- **Te (Tellurium):** Originally \([Kr] \: 4d^{10} \: 5s^2 \: 5p^4\), gains electrons to become \([Kr] \: 4d^{10} \: 5s^2 \: 5p^6\).- **Cl (Chlorine):** Initially \([Ne] \: 3s^2 \: 3p^5\), gains an electron for \([Ne] \: 3s^2 \: 3p^6\).- **Sr (Strontium):** Originally \([Kr] \: 5s^2\), loses electrons to show \([Kr]\).- **Li (Lithium):** From \([1s^2 \: 2s^1\) to \([1s^2\) by losing one electron.
These configurations demonstrate how atoms shift from their neutral ground states to stable ionic states.
Periodic Table
The periodic table is an organized chart of elements arranged by increasing atomic number, showcasing recurring chemical properties. It serves as a useful reference for predicting how elements will react and the types of ions they will form.
- **Position:** Elements in the same column usually share similar valence electron configurations, leading to comparable chemical behavior. - **Groups:** Elements like halogens (Cl) tend to gain electrons, while alkali metals (Li) often lose them.
By referring to the periodic table, we can quickly determine the likely electron arrangements and the resulting stable ions like Cl鈦 and Li鈦. It is an invaluable tool in both understanding and predicting chemical interactions.
Studying the table helps visualize why elements gain or lose electrons as they strive for the electron clouds of their more stable, noble gas neighbors.

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Most popular questions from this chapter

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)

Consider the following energy changes: $$\begin{array}{ll} \text {} & \quad { \Delta H} \\ \text {} & {(k J / m o l)} \\ \hline \\ {\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-}} & {735} \\ {\mathrm{Mg}^{+}(g) \rightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-}} & {1445} \\ {\mathrm{O}(g)+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(g)} & {-141} \\ {\mathrm{O}^{-}(g)+\mathrm{e}^{-} \rightarrow 0^{2-}(g)} & {878}\end{array}$$ Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) Explain.

An ionic compound made from the metal \(M\) and the diatomic gas \(X_{2}\) has the formula \(M_{a} X_{b},\) in which \(a=1\) or 2 and \(b=1\) or \(2 .\) Use the data provided to determine the most likely values for \(a\) and \(b,\) along with the most likely charges for each of the ions in the ionic compound. Data (in units of \(\mathrm{kJ} / \mathrm{mol} )\) Successive ionization energies of \(\mathrm{M} : 480 ., 4750\) . Successive electron affinity values for \(\mathrm{X} :-175,920\) . Enthalpy of sublimation for \(\mathrm{M}(s) \rightarrow \mathrm{M}(g) : 110\) . Lattice energy for MX \(\left(\mathrm{M}^{+} \text { and } \mathrm{X}^{-}\right) :-1200\) . Lattice energy for \(\mathrm{MX}_{2}\left(\mathrm{M}^{2+} \text { and } \mathrm{X}^{-}\right) :-3500\) . Lattice energy for \(\mathrm{M}_{2} \mathrm{X}\left(\mathrm{M}^{+} \text { and } \mathrm{X}^{2-}\right) :-3600\) . Lattice energy for \(\mathrm{MX}\left(\mathrm{M}^{2+} \text { and } \mathrm{X}^{2-}\right) :-4800\) .

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}, \mathrm{V}^{5+}\) b. \(\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Cs}^{+}\) c. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}\) d. \(\mathrm{P}, \mathrm{P}^{-}, \mathrm{P}^{2-}, \mathrm{P}^{3-}\) e. \(\mathrm{O}^{2-}, \mathrm{S}^{2-}, \mathrm{Se}^{2-}, \mathrm{Te}^{2-}\)

Given the following information: Heat of sublimation of \(\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}\) Ionization energy of \(\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}\) Lattice energy of LiCl(s) \(=-829 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g)$$

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