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Chapter 4: Problem 61

Among the following series of transition metal ions, the one where all metal ion have \(3 \mathrm{~d}^{2}\) electronic configuration is (a) \(\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}\) (b) \(\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}\) (c) \(\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\) (d) \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{5+}\) (At. wt \(\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25)\)

Short Answer

Expert verified
Option (d) is correct; all ions have \(3d^2\) electron configuration.

Step by step solution

01

Understand the Problem

We need to find the series of transition metal ions where each metal ion has a \( 3d^2 \) electronic configuration. This requires knowing the initial electronic configuration of each metal and then determining the configuration after ionization.
02

Analyze Ti

Titanium (Ti) has an atomic number of 22, so its ground state electronic configuration is \([Ar]4s^23d^2\). - For \(Ti^{3+}\), it loses 3 electrons, resulting in \([Ar]3d^1\). - For \(Ti^{+}\), it loses 1 electron, resulting in \([Ar]4s^13d^2\). - For \(Ti^{4+}\), it loses 4 electrons, resulting in \([Ar]\). - For \(Ti^{2+}\), it loses 2 electrons, resulting in \([Ar]3d^2\). Ti \(2^+\) has \(3d^2\) configuration.
03

Analyze V

Vanadium (V) has an atomic number of 23, with a ground state electronic configuration of \([Ar]4s^23d^3\). - For \(V^{2+}\), it loses 2 electrons, resulting in \([Ar]3d^3\). - For \(V^{4+}\), it loses 4 electrons, resulting in \([Ar]3d^1\). - For \(V^{3+}\), it loses 3 electrons, resulting in \([Ar]3d^2\). V \(3^+\) has \(3d^2\) configuration.
04

Analyze Cr

Chromium (Cr) has an atomic number of 24, with a ground state electronic configuration of \([Ar]4s^13d^5\).- For \(Cr^{3+}\), it loses 3 electrons, resulting in \([Ar]3d^3\). - For \(Cr^{6+}\), it loses 6 electrons, resulting in \([Ar]\). - For \(Cr^{2+}\), it loses 2 electrons, resulting in \([Ar]3d^4\). - For \(Cr^{4+}\), it loses 4 electrons, resulting in \([Ar]3d^2\). Cr \(4^+\) has \(3d^2\) configuration.
05

Analyze Mn

Manganese (Mn) has an atomic number of 25, with a ground state electronic configuration of \([Ar]4s^23d^5\).- For \(Mn^{4+}\), it loses 4 electrons, resulting in \([Ar]3d^3\). - For \(Mn^{7+}\), it loses 7 electrons, resulting in \([Ar]\). - For \(Mn^{3+}\), it loses 3 electrons, resulting in \([Ar]3d^4\). - For \(Mn^{5+}\), it loses 5 electrons, resulting in \([Ar]3d^2\). Mn \(5^+\) has \(3d^2\) configuration.
06

Identify the Correct Series

Review the calculations: - Series (a): \(Ti^{3+} = 3d^1\), \(V^{2+} = 3d^3\), \(Cr^{3+} = 3d^3\), \(Mn^{4+} = 3d^3\). - Series (b): Incorrect as all are not \(3d^2\).- Series (c): \(Ti^{4+} = 3d^0\), \(V^{3+} = 3d^2\), \(Cr^{2+} = 3d^4\), \(Mn^{3+} = 3d^4\).- **Series (d)**: \(Ti^{2+} = 3d^2\), \(V^{3+} = 3d^2\), \(Cr^{4+} = 3d^2\), \(Mn^{5+} = 3d^2\). Series (d) is correct because all ions have \(3d^2\) configuration.

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Key Concepts

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

Electronic Configuration
To understand the behavior of transition metal ions, it's essential to start with their electronic configuration. The electronic configuration describes how electrons are distributed among the various atomic orbitals of an atom. For transition metals, these configurations involve the 3d orbitals, which gain electrons after the 4s orbitals are filled.
  • Typically, transition metals have their outer electron configurations in the form of \([Ar]4s^23d^n\), where n varies depending on the metal.
  • When these metals ionize, they lose electrons to form cations. Usually, electrons from the 4s orbital are lost first, followed by the 3d electrons in a way that leaves the ion with a stable configuration.
  • This rearranging of electrons through ionization alters the electronic configuration of the ion, and contributes to the chemical properties of the element.
Thus, by analyzing the electronic configuration, we can predict which ions form and determine their chemical behavior.
3d Orbitals
The term '3d orbitals' refers to a set of five orbitals within the d-block of the periodic table where electrons reside. These orbitals are crucial in understanding the properties of transition metals.
  • The 3d orbitals follow after the 4s in terms of energy, but during ionization, electrons are typically removed from the 4s orbital first.
  • The five d orbitals, named \(d_{xy}, d_{xz}, d_{yz}, d_{x^2-y^2}, \,\&\,, d_{z^2}\), hold a total of 10 electrons.
  • In transition metals, the 3d orbitals give rise to various electronic configurations resulting in multiple oxidation states, influencing their ability to form different ions.
This flexibility and variation in 3d orbital occupation help explain the diverse chemistry of transition elements, including their magnetic properties and how they participate in bond formation.
Ionization
Ionization refers to the process by which an atom or a molecule acquires a positive or negative charge by losing or gaining electrons. In the context of transition metals, it typically involves the metal losing electrons to form positive ions, known as cations.
  • Transition metals often lose their outermost \(4s\) electrons first during ionization, followed by the \(3d\) electrons, leading to various oxidation states.
  • The specific electrons that are removed will depend on the metal and its environment, affecting the resulting ion's electronic configuration.
  • The possibility of multiple ionization states among transition elements gives them characteristic properties like high melting points, complex formation, and colorful compounds.
The ability to form different cations means that transition metals can participate in a wide range of chemical reactions, making them extremely valuable in chemistry.
Transition Elements
Transition elements are found in the d-block of the periodic table and are characterized by the partially filled d orbitals in their atoms or ions. These elements are fundamental to the chemistry of the d-block.
  • These elements include metals like titanium (Ti), vanadium (V), chromium (Cr), and manganese (Mn) known for their high density and heat and electrical conductivity.
  • Transition metals often exhibit variable oxidation states, which arise from their ability to lose different numbers of 3d and 4s electrons, contributing to their versatility in forming compounds.
  • The unique properties of transition elements, such as forming colored compounds and acting as catalysts, stem from the electronic transitions between their d orbitals.
By understanding transition elements, we can better grasp how and why they play critical roles in many industrial and biological processes.

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Most popular questions from this chapter

The ratio of the difference between 2 nd and 3 rd Bohr's orbit energy to that between 3 rd and 4 th orbit energy is (a) \(7 / 20\) (b) \(20 / 7\) (c) \(27 / 9\) (d) \(9 / 27\)

The maximum number of 3 d electrons having \(\mathrm{s}=+1 / 2\) are (a) 10 (b) 5 (c) 14 (d) 7

Which one of the following forms a colourless solution in aqueous medium? (a) \(\mathrm{V}^{3+}\) (b) \(\mathrm{Cr}^{3+}\) (c) \(\mathrm{Ti}^{3+}\) (d) \(\mathrm{Sc}^{3+}\) Atomic number \(\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24\)

The correct order of number of unpaired electrons in the ion \(\mathrm{Cu}^{2+} \mathrm{Ni}^{2+}, \mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) is (a) \(\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Fe}^{3+}\) (b) \(\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}>\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}\) (c) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Ni}^{2+}>\mathrm{Cu}^{2+}\) (d) \(\mathrm{Fe}^{3+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Ni}^{2+}\)

Which of the following statement(s) are correct? (1) the electronic configuration of \(\mathrm{Cr}\) is \([\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{1}\) (atomic number of \(\mathrm{Cr}=24\) ) (2) the magnetic quantum number may have a negative value (3) in silver atom, 23 electrons have a spin of one type and 24 of the opposite type (atomic number of \(\mathrm{Ag}=47\) )
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