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Write the electron configuration for each of the following ions. and determine which ones possess noble-gas configurations: (a) \(\mathrm{Be}^{2+}\) (b) \(\mathrm{Mn}^{2+},(\mathbf{c}) \mathrm{Cd}^{2+}\) (d) \(\mathrm{Fe}^{3+}\), (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{At}^{-}\).

Short Answer

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(a) Be虏鈦 and (f) At鈦 achieve noble-gas configurations.

Step by step solution

01

Write Neutral Atom Configurations

First, write the electron configurations for the neutral atoms corresponding to each ion from the periodic table. - Beryllium (Be): **Configuration**: \(\text{Be: } 1s^2 2s^2\) - Manganese (Mn): **Configuration**: \(\text{Mn: } [Ar] 3d^5 4s^2\) - Cadmium (Cd): **Configuration**: \(\text{Cd: } [Kr] 4d^{10} 5s^2\) - Iron (Fe): **Configuration**: \(\text{Fe: } [Ar] 3d^6 4s^2\) - Thallium (Tl): **Configuration**: \(\text{Tl: } [Xe] 4f^{14} 5d^{10} 6s^2 6p^1\) - Astatine (At): **Configuration**: \(\text{At: } [Xe] 4f^{14} 5d^{10} 6s^2 6p^5\)
02

Determine Ion Configurations

Next, remove (for cations) or add (for anions) the corresponding number of electrons to obtain the electron configurations of the ions. - \(\text{Be}^{2+}\): Remove two electrons **Configuration**: \(1s^2\) - \(\text{Mn}^{2+}\): Remove two electrons from 4s orbitals first **Configuration**: \[\text{Mn}^{2+}: [Ar] 3d^5\] - \(\text{Cd}^{2+}\): Remove two electrons from 5s orbitals **Configuration**: \[\text{Cd}^{2+}: [Kr] 4d^{10}\] - \(\text{Fe}^{3+}\): Remove three electrons; 2 from 4s orbital and 1 from 3d orbital **Configuration**: \[\text{Fe}^{3+}: [Ar] 3d^5\] - \(\text{Tl}^{+}\): Remove one electron from 6p orbital **Configuration**: \[\text{Tl}^{+}: [Xe] 4f^{14} 5d^{10} 6s^2\] - \(\text{At}^{-}\): Add one electron to 6p orbital **Configuration**: \[\text{At}^{-}: [Xe] 4f^{14} 5d^{10} 6s^2 6p^6\]
03

Identify Noble-Gas Configurations

Compare the ion configurations to the noble-gas configurations to identify any matches.- Noble gases: - Helium (He): \(1s^2\) - Neon (Ne): \[1s^2 2s^2 2p^6\] - Argon (Ar): \[1s^2 2s^2 2p^6 3s^2 3p^6\] - Krypton (Kr): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\] - Xenon (Xe): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\] - \(\text{Be}^{2+} ightarrow \ ext{He} (Noble gas)\)- \(\text{At}^{-} ightarrow \text{Rn} ([Xe] 4f^{14} 5d^{10} 6s^2 6p^6) \ ext{(Noble gas)}\) Only \(\text{Be}^{2+} \ ext{and } \text{At}^{-}\) ions have full noble-gas configurations.

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Key Concepts

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

Noble-Gas Configuration
Noble-gas configuration refers to the arrangement of electrons in an atom or ion that mirrors the electron arrangement of the inert or noble gases. Noble gases, like helium (He), neon (Ne), and argon (Ar), have full outer electron shells. The stability of these full electron shells contributes to the inert nature of noble gases, meaning they do not readily engage in chemical reactions.
Understanding noble-gas configuration is crucial for identifying whether a particular ion has attained a stable electron state akin to that of a noble gas. This is accomplished by either losing or gaining electrons to fill or empty the outermost electron shell.
  • For instance, a beryllium ion (B拾蛦虨虩拾虈虩拾虈虩丿賴B拾蛦虨虩), which loses 2 electrons (configuration: B拾蛦虨虩), reaches the configuration of helium (B拾蛦虨虩), a noble gas.
  • Astatine ion (B拾蛦虨虩拾虈虩丿賴B拾蛦虨虩), by gaining 1 electron, achieves the configuration of radon (B拾蛦虨虩B拾蛦虨虩), another noble gas.
Acquiring a noble-gas configuration usually increases an element's stability and explains why ion formation often results in noble-gas-like configurations.
Ion Electron Configuration
Ion electron configuration involves determining the correct arrangement of electrons when an atom loses or gains electrons to form an ion. When an atom becomes an ion, it loses its neutral charge due to the loss or gain of electrons, which are negatively charged. To find the electron configuration of an ion, you must adjust the electron count according to its charge.
  • For cations (positively charged ions like B拾蛦虨虩), you remove electrons. For example, B拾蛦虨虩 (formed from B拾蛦虨虩拾虈虩亘蕼炭虛) involves removing 2 electrons from B拾蛦虨虩 as shown: B拾蛦虨虩 B拾蛦虨虩
  • Conversely, anions (negatively charged ions, such as B拾蛦虨虩) gain electrons. For B拾蛦虨虩, an electron is added to B拾蛦虨虩, leading to the altered configuration B拾蛦虨虩.
By comparing the ion's electron configuration to neutral atoms, you can determine if it matches a noble-gas configuration, reflecting greater stability.
Periodic Table
The periodic table is an essential tool for understanding the properties of elements and their electron configurations. Each element is placed according to its atomic number, and its position gives insights into its chemical properties and the likely electron configuration. The arrangement of the periodic table into periods and groups corresponds to the filling of electron shells and subshells.
When dealing with electron configurations, the periodic table helps predict both the configuration of neutral atoms and the resulting configurations of ions:
  • As you move across a period, electrons are added one by one, filling the subshells in the order of increasing energy levels (like 1s, 2s, 2p, etc.).
  • The vertical groups, on the other hand, share similar chemical traits, resulting from having a similar valence shell electron configuration.
By knowing the position of an element on the periodic table, you can efficiently write its electron configuration and understand how it will form ions. This is pivotal for understanding whether an ion attains a noble-gas configuration upon losing or gaining electrons.

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Most popular questions from this chapter

By referring only to the periodic table, select \((\mathbf{a})\) the most electronegative element in group \(13 ;(\mathbf{b})\) the least electronegative element in the group As, Se and Br; (c) the most electronegative element in the group K, Mg, Al and In; (d) the element in the group \(\mathrm{Na}\), Be, \(\mathrm{Si}\), Ar, that is most likely to form an ionic compound with \(B \mathrm{r}\).

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, \(e\).

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround them: \((\mathbf{a}) \mathrm{HCl},(\mathbf{b}) \mathrm{ICl}_{5},(\mathbf{c}) \mathrm{NO},\) (d) \(\mathrm{CF}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{I}_{3}^{-}\).

The substance chlorine monoxide, \(\mathrm{ClO}(g)\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D},\) and the \(\mathrm{Cl}-\mathrm{O}\) bond length is \(160 \mathrm{pm} .(\mathbf{a})\) Determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in units of the electronic charge, \(e .(\mathbf{b})\) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion \(\mathrm{ClO}^{-}\) exists. What is the formal charge on the Cl for the best Lewis structure for \(\mathrm{ClO}^{-}\) ?

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and Cl, (d) Li and \(O\). (b) \(\mathrm{Mg}\) and \(\mathrm{O},(\mathbf{c}) \mathrm{Zn}\) and \(\mathrm{Cl},\)

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