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Chapter 22: Problem 22

Which one of the following exhibits the large number of oxidation states? (a) Ti (22) (b) V (23) (c) \(\operatorname{Cr}(24)\) (d) \(\operatorname{Mn}(25)\)

Short Answer

Expert verified
Manganese (Mn) exhibits the largest number of oxidation states.

Step by step solution

01

Understand Oxidation States

The oxidation state of an element corresponds to the number of electrons an atom can gain, lose, or share when forming chemical bonds. Transition metals often exhibit a variety of oxidation states since they have multiple oxidation states due to the availability of d-orbitals.
02

Analyze Each Element

Examine each element's electron configuration and standard oxidation states. - Ti: [Ar] 3d虏 4s虏, has oxidation states of +2, +3, +4. - V: [Ar] 3d鲁 4s虏, has oxidation states of +2, +3, +4, +5. - Cr: [Ar] 3d鈦� 4s鹿, has oxidation states of +2, +3, +6. - Mn: [Ar] 3d鈦� 4s虏, has oxidation states of +2, +3, +4, +6, +7.
03

Identify the Element with the Most Oxidation States

Listing the number of oxidation states: - Ti: 3 states (+2, +3, +4) - V: 4 states (+2, +3, +4, +5) - Cr: 3 states (+2, +3, +6) - Mn: 5 states (+2, +3, +4, +6, +7) Mn has the most oxidation states among the given options.

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Key Concepts

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

Transition Metals
Transition metals are elements found in the center block of the periodic table. They are known for their unique properties, such as exhibiting a wide range of oxidation states, which are crucial for various chemical reactions. This trait arises because they can lose different numbers of d or s electrons. Transition metals are particularly distinct as they often form colorful compounds and can exhibit magnetism.
  • Found between groups 3 to 12 in the periodic table.
  • Include elements like Titanium (Ti), Vanadium (V), Chromium (Cr), and Manganese (Mn).
  • They play a significant role in catalysis and in biological systems, such as in hemoglobin.
Understanding transition metals and their oxidation states is essential for mastering inorganic chemistry and its applications.
Electron Configuration
Electron configuration is a method of describing the distribution of electrons in an atom's orbitals. For transition metals, it's vital for determining how an element behaves in a chemical reaction. Their electrons are arranged in shells and subshells, usually denoted by the levels s, p, d, and f.
  • Transition metals have electron configurations that involve d orbitals, such as Ti: \([\text{Ar}]\, 3d^2\, 4s^2\) and Mn: \([\text{Ar}]\, 3d^5\, 4s^2\).
  • This configuration allows several oxidation states, as electrons can be lost from both d and s orbitals.
  • The d orbitals can hold up to 10 electrons, which gives rise to multiple oxidation states.
Understanding electron configurations helps predict the properties of an element, including which oxidation states it might exhibit.
Chemical Bonds
Chemical bonds are forces that hold atoms together in a molecule or compound. They are formed by the attraction of atoms to each other through sharing, transferring, or pooling electrons. Transition metals form chemical bonds uniquely due to their d-electrons, leading to their versatile chemistry.
  • Types of chemical bonds include covalent, ionic, and metallic bonds.
  • Transition metals often form coordination bonds by accepting electron pairs from other atoms or molecules.
  • They can form complex compounds with unique shapes and properties, beneficial in industrial applications.
Recognizing how transition metals create chemical bonds highlights their role in catalysis and material science, where their varied oxidation states enhance reactivity and bond formation. Whether in creating alloys or acting as catalysts, the ability to form different chemical bonds is key to their widespread applications.

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Most popular questions from this chapter

Which of the following combination will produce \(\mathrm{H}_{2}\) gas? (a) Fe metal and conc. \(\mathrm{HNO}_{3}\) (b) Cu metal and conc. \(\mathrm{HNO}_{3}\) (c) \(\mathrm{Zn}\) metal and \(\mathrm{NaOH}(\mathrm{aq})\) (d) Au metal and \(\mathrm{NaCN}(\mathrm{aq})\) in the presence of air

Consider the following list of reagents: [Adv. 2014] Acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), alkaline \(\mathrm{KMnO}_{4}, \mathrm{CuSO}_{4}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cl}_{2}, \mathrm{O}_{3}, \mathrm{FeCl}_{3}, \mathrm{HNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) The total number of reagents that can oxidise aqueous iodide to iodine is

Match the following, choosing one item from column \(\mathbf{X}\) and one from column Y. (i) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (ii) \(\left(\mathrm{NaPO}_{3}\right)_{n}\) (iii) \(\mathrm{NO}_{3}^{-}\) (iv) \(\mathrm{SnO}_{2}\) (v) \(\quad \mathrm{KCl} . \mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (vi) \(\mathrm{AgNO}_{3}\) \(\mathbf{X} \quad \mathbf{Y}\) (a) cassiterite (b) lunar caustic (c) producer gas (d) water softener (e) brown ring test (f) carnallite (vii) \(\mathrm{CO}+\mathrm{N}_{2}\) (g) calomel

Fusion of \(\mathrm{MnO}_{2}\) with \(\mathrm{KOH}\) in presence of \(\mathrm{O}_{2}\) produces a salt \(W\). Alkaline solution of \(W\) upon electrolytic oxidation yields another salt\(X\). The manganese containing ions present in \(\mathrm{W}\) and \(X\), respectively are \(Y\) and \(Z\). Correct statement(s) is (are) (a) In both \(Y\) and \(Z\), p-bonding occurs between \(p\)-orbitals of oxygen and \(d\) orbitals of manganese (b) In aqueous acidic solution, \(Y\) undergoes dispro-portionation reaction to give \(Z\) and \(\mathrm{MnO}_{2}\) (c) Both \(Y\) and \(Z\) are coloured and have tetrahedral shape (d) \(Y\) is diamagnetic in nature while \(Z\) is paramagnetic

What happens when : (i) aqueous ammonia is added dropwise to a solution of copper sulphate till it is in excess. (ii) \(\mathrm{CrCl}_{3}\) solution is treated with sodium hydroxide and then with hydrogen peroxide.
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