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Chapter 8: Problem 20

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{V}^{3+}\), (c) \(\mathrm{Ni}^{2+}\), (d) \(\mathrm{Pt}^{2+}\), (e) \(\mathrm{Ge}^{2-}\), (f) \(\mathrm{Ba}^{2+}\).

Short Answer

Expert verified
The electron configurations of the given ions are as follows: (a) \(\mathrm{Fe}^{2+} : [Ar] 3d^6\), (b) \(\mathrm{V}^{3+} : [Ar] 3d^0\), (c) \(\mathrm{Ni}^{2+} : [Ar] 3d^8\), (d) \(\mathrm{Pt}^{2+} : [Xe] 4f^{14} 5d^7\), (e) \(\mathrm{Ge}^{2-} : [Ar] 4s^2 3d^{10} 4p^4\), and (f) \(\mathrm{Ba}^{2+} : [Xe]\). Among these ions, only V鲁鈦� and Ba虏鈦� have noble-gas configurations.

Step by step solution

01

1. Write the electron configuration of the neutral atoms

We use periodic table to find the atomic number(Z) of each given element and then write their corresponding electron configurations.
02

2. Write the electron configuration for each ion

We remove/add the required number of electrons as per their given charges (loss of electrons for positive ions and gain of electrons for negative ions).
03

3. Compare with noble gas configurations

If the electron configuration of an ion matches with the electron configuration of noble gas element, then it has a noble gas configuration. Now let's use these steps for the given ions: (a) \(\mathrm{Fe}^{2+}\): Step 1: Electron configuration of neutral Fe (Z=26) is \([Ar] 4s^2 3d^6\). Step 2: Removing 2 electrons from Fe atom to form Fe虏鈦� ion. Its configuration is: \([Ar] 3d^6\). Step 3: The electron configuration does not match any noble gas, so Fe虏鈦� does not have a noble-gas configuration. (b) \(\mathrm{V}^{3+}\): Step 1: Electron configuration of neutral V (Z=23) is \([Ar] 4s^2 3d^3\). Step 2: Removing 3 electrons from V atom to form V鲁鈦� ion. Its configuration is: \([Ar] 3d^0\). Step 3: The electron configuration matches Argon, so V鲁鈦� has a noble-gas configuration. (c) \(\mathrm{Ni}^{2+}\): Step 1: Electron configuration of neutral Ni (Z=28) is \([Ar] 4s^2 3d^8\). Step 2: Removing 2 electrons from Ni atom to form Ni虏鈦� ion. Its configuration is: \([Ar] 3d^8\). Step 3: The electron configuration does not match any noble gas, so Ni虏鈦� does not have a noble-gas configuration. (d) \(\mathrm{Pt}^{2+}\): Step 1: Electron configuration of neutral Pt (Z=78) is \([Xe] 6s^2 4f^{14} 5d^9\). Step 2: Removing 2 electrons from Pt atom to form Pt虏鈦� ion. Its configuration is: \([Xe] 4f^{14} 5d^7\). Step 3: The electron configuration does not match any noble gas, so Pt虏鈦� does not have a noble-gas configuration. (e) \(\mathrm{Ge}^{2-}\): Step 1: Electron configuration of neutral Ge (Z=32) is \([Ar] 4s^2 3d^{10} 4p^2\). Step 2: Adding 2 electrons to Ge atom to form Ge虏鈦� ion. Its configuration is: \([Ar] 4s^2 3d^{10} 4p^4\). Step 3: The electron configuration does not match any noble gas, so Ge虏鈦� does not have a noble-gas configuration. (f) \(\mathrm{Ba}^{2+}\): Step 1: Electron configuration of neutral Ba (Z=56) is \([Xe] 6s^2\). Step 2: Removing 2 electrons from Ba atom to form Ba虏鈦� ion. Its configuration is: \([Xe]\). Step 3: The electron configuration matches Xe, so Ba虏鈦� has a noble-gas configuration. So, the ions with noble-gas configurations are V鲁鈦� and Ba虏鈦�.

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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 configurations are sought-after electronic arrangements for ions because they represent a stable state. Noble gases are at the far right of the periodic table and have filled electron shells, making them inherently stable. This stability is due to their full outer electron shells, which do not typically engage in chemical reactions.

When an ion achieves a noble-gas configuration, it adopts a similar stability as a noble gas. To check if an ion has achieved this, we compare the electron configuration of the ion with that of a noble gas.
  • If they match, the ion has a noble-gas configuration.
  • If they do not, the ion does not achieve this stable form.
In the given exercise, examples of ions that achieve noble-gas configurations are V鲁鈦�, which is isoelectronic with Argon, and Ba虏鈦�, which resembles Xenon.
Transition Metal Ions
Transition metals are elements found in the d-block of the periodic table. They often form positive ions, or cations, by losing electrons. A unique feature of these metals is their tendency to lose their outer s electrons before they lose their d electrons. This is contrary to the Aufbau principle typically followed by elements.

For transition metal ions like Fe虏鈦�, V鲁鈦�, Ni虏鈦�, and Pt虏鈦�, electron removal initially occurs from the 4s orbital before 3d, despite having been filled last.
  • Fe虏鈦� results from losing two electrons, leading to a configuration of [Ar] 3d鈦�.
  • V鲁鈦� loses three electrons, resulting in [Ar] 3d鈦�, which matches Argon.
  • Nickel's ion, Ni虏鈦�, similarly has two electrons removed, resulting in [Ar] 3d鈦�.
Thus, understanding electron removal in transition metals is vital for predicting their behavior and properties.
Atomic Number
The atomic number of an element is the number of protons in its nucleus, which also equals the number of electrons in a neutral atom. This number defines the identity of the element and its position on the periodic table. The atomic number is crucial for determining the electron configuration of an element.

For instance, Germanium has an atomic number of 32, meaning it has 32 electrons in its neutral state. This allows us to determine its electron configuration as [Ar] 4s虏 3d鹿鈦� 4p虏.
  • Transition metals like Vanadium (Z=23) possess atomic numbers which help in identifying their electron arrangement and subsequent ion configurations.
Reliable predictions about elemental reactivity and ion formation can be made based on this fundamental concept.
Removal of Electrons
The process of removing electrons to form ions is typical in transition metals, where positive ions or cations are created. In doing so, the outermost electrons, often from the s orbital, are removed first.

For the ions in the original exercise, this concept was applied to:
  • Fe虏鈦� is formed by removing two electrons from its 4s subshell.
  • Pt虏鈦� loses two electrons, also altering its electron configuration from [Xe] 6s虏 4f鹿鈦� 5d鈦� to [Xe] 4f鹿鈦� 5d鈦�.
The resulting electron configuration reflects the positive charge on the ion. As electrons are removed, the atom's overall charge increases, and its radius generally decreases.
Addition of Electrons
Adding electrons typically results in the formation of negatively charged ions, or anions. During this process, electrons are added to the outermost shell, following the Aufbau principle, where electrons fill the lowest available energy levels first.

For example, with Ge虏鈦�, two electrons are added to its neutral configuration [Ar] 4s虏 3d鹿鈦� 4p虏, resulting in [Ar] 4s虏 3d鹿鈦� 4p鈦�. This added electron density increases the ion's negative charge.
  • Electrons added to atomic orbitals must account for electron-electron repulsions, potentially altering the size of the ion and its reactivity.
Understanding how electrons are added allows us to predict the properties and interactions of anions in various chemical contexts.

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Most popular questions from this chapter

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}, 1.83 \% \mathrm{H}\), \(64.30 \% \mathrm{Cl}\), and \(13.35 \% \mathrm{O}\) by mass, and has a molar mass of \(165.4 \mathrm{~g} / \mathrm{mol} .(\mathbf{a})\) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single \(C\) atom and that there are a \(\mathrm{C}-\mathrm{C}\) bond and two \(\mathrm{C}-\mathrm{O}\) bonds in the compound.

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance structures are possible that satisfy the octet rule? (c) On the basis of the formal charges, which Lewis structure is expected to be dominant for \(\mathrm{BeCl}_{2}\) ?

Consider the formate ion, \(\mathrm{HCO}_{2}^{-}\), which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ion. The \(\mathrm{H}\) and the two O atoms are bonded to the central C atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2}\) ?

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: \((\mathbf{a}) \mathrm{OCS},(\mathbf{b}) \mathrm{SOCl}_{2}(\mathrm{~S}\) is the central atom), (c) \(\mathrm{BrO}_{3}^{-}\), (d) \(\mathrm{HClO}_{2}(\mathrm{H}\) is bonded to \(\mathrm{O})\).

Ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), is a very soluble salt in water. (a) Draw the Lewis structures of the ammonium and chloride ions. (b) Is there an \(\mathrm{N}-\mathrm{Cl}\) bond in solid ammonium chloride? (c) If you dissolve \(14 \mathrm{~g}\) of ammonium chloride in \(500.0 \mathrm{~mL}\) of water, what is the molar concentration of the solution? (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?
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