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Among the following series of transition metal ions, the one where all metal ions have \(3 \mathrm{~d}^{2}\) electronic configuration is (a) \(\mathrm{Ti}^{+}, \mathrm{V}^{4+}, \mathrm{Cr}^{6+}, \mathrm{Mn}^{7+}\) (b) \(\mathrm{Ti}^{2+}, \mathrm{V}^{3+}, \mathrm{Cr}^{4+}, \mathrm{Mn}^{3+}\) (c) \(\mathrm{Ti}^{3+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{4+}\) (d) \(\mathrm{Ti}^{4+}, \mathrm{V}^{3+}, \mathrm{Cr}^{2+}, \mathrm{Mn}^{3+}\)

Short Answer

Expert verified
None of the options have all ions with a 3d虏 configuration.

Step by step solution

01

Understand the Electronic Configuration

For an element in the periodic table, the atomic number informs us about the distribution of electrons across orbitals. Transition metals primarily fill their 3d orbitals.
02

Determine the Neutron State

The series question specifies, for each ion, checking the electron configuration to ensure the 3d sublevel contains exactly 2 electrons.
03

Analyze Option (a)

For Ti鈦, the atomic number is 22, so the neutral atom is [Ar] 3d虏 4s虏, losing one becomes [Ar] 3d虏 4s鹿, but Ti鈦 simplifies to 3d虏. V鈦粹伜, atomic number 23, follows [Ar] 3d鲁 4s虏, losing four yields [Ar]. Cr鈦垛伜 follows atomic 24, from [Ar] 3d鈦 4s鹿 loses six to [Ar]. Mn鈦封伜 follows atomic 25 from [Ar] 3d鈦 4s虏 loses seven to [Ar]. All but Ti don't have 3d虏, eliminating this option.
04

Analyze Option (b)

Ti虏鈦 from [Ar] 3d虏 4s虏 loses two to [Ar] 3d虏. V鲁鈦 from [Ar] 3d鲁 4s虏 loses three to [Ar] 3d虏. Cr鈦粹伜 goes from [Ar] 3d鈦 4s鹿 losing four to [Ar] 3d虏. Mn鲁鈦 goes from [Ar] 3d鈦 4s虏 losing three to [Ar] 3d鈦. Not all have 3d虏.
05

Analyze Option (c)

For Ti鲁鈦, [Ar] 3d虏 4s虏 becomes [Ar] 3d鹿. V虏鈦 goes from [Ar] 3d鲁 4s虏 to [Ar] 3d鲁. Cr鲁鈦 [Ar] 3d鈦 4s鹿 becomes [Ar] 3d鲁. Mn鈦粹伜 [Ar] 3d鈦 4s虏 becomes [Ar] 3d鲁. Not all have 3d虏.
06

Analyze Option (d)

Ti鈦粹伜 with [Ar] 3d虏 4s虏 to [Ar], zero d. V鲁鈦 from [Ar] 3d鲁 4s虏 becomes [Ar] 3d虏 after three lost. Cr虏鈦 [Ar] 3d鈦 4s鹿 becomes [Ar] 3d鲁. Mn鲁鈦 [Ar] 3d鈦 4s虏 becomes [Ar] 3d鈦. Does not meet the condition.
07

Confirm the Correct Series

Review all ions' electron configurations: only option (b) gives all elements with electronic configurations adjacent to having d orbitals remaining but slightly higher. Option (b) doesn't fit 3d虏 for all either.

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Key Concepts

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

Electronic Configuration
Electronic configuration refers to the arrangement of electrons in an atom's orbitals. It is primarily governed by the principles of quantum mechanics. For transition metals, electrons fill the 3d orbitals after the 4s orbitals. This means, for example, that in a neutral titanium atom, the electronic configuration would be
  • 1s2 2s2 2p6 3s2 3p6 3d2 4s2 or,
  • [Ar] 3d2 4s2,
where [Ar] represents the electron configuration of argon. As electrons are removed (or added), this configuration changes, which affects the chemical properties of the element. For ions, electrons are preferentially removed from the outermost shell first, which often means the 4s electrons are lost before the 3d electrons.
3d Orbitals
The 3d orbitals are a set of five orbitals in the d subshell with a principal quantum number of 3. These orbitals are filled with electrons after the 4s orbital in the atoms of the transition metals.
Each d orbital can hold a maximum of two electrons, so a complete 3d subshell can hold up to ten electrons. The filling of these orbitals influences the geometry and magnetic properties of the atoms.
When transition metal ions are formed, the electrons in the 3d subshell play a significant role. For example, for the transition metal ions mentioned in the exercise:
  • Ti2+ has an electron configuration of [Ar] 3d2, indicating that the two 3d electrons remain after losing the 4s electrons.
  • The role of d orbitals becomes even more significant in the bonding and chemical behavior of the elements.
These metal ions exhibit various oxidation states due to the involvement of these d electrons.
Ionization States
Transition metal atoms can lose different numbers of electrons to form positive ions or cations, resulting in different ionization states. The oxidation state or ionization state is the total number of electrons removed.
These states are pivotal in determining the chemical reactivity and coordination properties of the ions. For example, in the transition metal series question, the ions Ti2+, V3+, Cr4+, and Mn3+ exhibit different positive ionization states depending on how many electrons have been removed.
  • The ionization states can complicate the electron configuration since electrons from the 3d orbitals and 4s orbitals can be involved.
  • Higher ionization states generally mean more electrons have been removed, typically from the 4s before the 3d orbitals.
By understanding ionization states, students can predict and explain the behavior of transition metals in chemical reactions and distinguish between different ions of the same element based on their electronic configurations.

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Most popular questions from this chapter

Which among the following paramagnetic pairs rare? (1) \(\left[\mathrm{BaO}_{2}, \mathrm{NO}_{2}\right]\) (2) \(\left[\mathrm{KO}_{2}, \mathrm{NO}\right]\) (3) \(\left[\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{NO}\right]\) (4) \(\left[\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}, \mathrm{CuCl}_{2}\right]\) (a) 3,4 only (b) 1,3 only (c) 2,4 only (d) \(1,2,4\)

Which of the following compound is both paramagnetic and coloured? (a) \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{TiCl}_{6}\right]\) (b) \(\mathrm{VOSO}_{4}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\)

Which of the following factors may be regarded as the main cause of lanthanide contraction? \(\quad\) (a) poor shielding of one of \(4 \mathrm{f}\) electrons by another in the subshell (b) effective shielding of one of \(4 \mathrm{f}\) electrons by another in the subshell (c) poor shielding of 5 d-electrons by 4 f electrons (d) greater shielding of 5 d-electrons by \(4 \mathrm{f}\) electrons

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What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) are formed (c) \(\mathrm{CrO}_{4}^{2-}\) is oxidized to \(+7\) state to \(\mathrm{Cr}\) (d) \(\mathrm{CrO}_{4}^{2-}\) is reduced to \(+3\) state of \(\mathrm{Cr}\)

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