Comprehension

Transition metals have incomplete d-subshells either in the neutral atom or in their ions. The presence of partly filled d-orbitals in their atoms makes transition elements different from non-transition elements. With partly filled d-orbitals, these elements exhibit certain characteristic properties such as:
  • Display of a variety of oxidation states
  • Formation of coloured ions
  • Entering into complex formation with a variety of ligands

The transition metals and their compounds also exhibit catalytic properties and paramagnetic behaviour. The transition metals are very hard and have low volatility. An examination of the E0 M2+/M values shows the varying trends:
 Transition metals have incomplete d-subshell

Question: 1

On what basis can we say that Cu is a transition element but Zn is not? (Atomic number: Cu = 29, Zn = 30)

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- Transition elements typically have an incomplete d-orbital in one or more oxidation states. - Non-transition elements like Zn have a full d-orbital in all of their oxidation states.
Updated On: Feb 25, 2025
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Solution and Explanation

Cu is a transition element because it has an incomplete d-orbital in its +2 oxidation state, whereas Zn has a fully filled d-orbital in its ground state as well as in +2 oxidation state, making it a non-transition element. Transition elements must have at least one oxidation state with an incomplete d-orbital.
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Question: 2

Why do transition elements show a variety of oxidation states?

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- Transition metals can exhibit multiple oxidation states due to the participation of both d and s electrons in bonding. - The presence of unpaired electrons in the d-orbitals allows for flexibility in oxidation states.
Updated On: Feb 25, 2025
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Solution and Explanation

Transition elements show a variety of oxidation states because both (n-1)d and ns subshell electrons take part in bond formation due to their comparable energies. This allows for the formation of multiple oxidation states. The presence of unpaired electrons in d-orbitals also contributes to this variability.
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Question: 3

Why do $\text{E\(^0\)}_{M^{2+}/M }$ values show irregular trends from Vanadium to Zinc?

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- Irregular trends in E\(^0\) values are often due to ionization energies, sublimation enthalpies, and other thermodynamic factors. - The transition elements show varying trends due to these complex thermodynamic effects.
Updated On: Feb 25, 2025
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Solution and Explanation

The irregular trends in $\text{E\(^0\)}_{M^{2+}/M }$ values from Vanadium to Zinc are due to irregular values of \((\Delta H(_1) + \Delta H(_2))\) and sublimation enthalpies. These values are influenced by factors like the ionization energies and the differences in the stability of the ions formed.

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Question: 4

How is the variability in oxidation states of transition metals different from that of non-transition elements?

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- Transition metals have more flexibility in oxidation states because of the availability of d-orbitals. - Non-transition elements tend to show fewer oxidation states, typically differing by 2.
Updated On: Feb 25, 2025
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Solution and Explanation

In transition metals, oxidation states vary by +1, whereas in non-transition metals, oxidation states differ by +2. This is due to the involvement of d-orbitals in transition metals, which allows for a greater variety of oxidation states.
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Question: 5

Of the d\(^4\) species, Cr\(^{2+}\) is strongly reducing while Mn\(^{3+}\) is strongly oxidizing. Why? (Atomic number: Cr = 24, Mn = 25)

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- Chromium in the +2 oxidation state is reducing, while Mn in the +3 oxidation state is oxidizing due to their electronic configurations. - The stability of half-filled d-orbitals drives these oxidation and reduction tendencies.
Updated On: Feb 25, 2025
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Solution and Explanation

Cr\(^{2+}\) is strongly reducing because it will be converted to Cr\(^{3+}\), which has a more stable half-filled t\(_2\)g configuration. On the other hand, Mn\(^{3+}\) is strongly oxidizing because it changes to Mn\(^{2+}\), which has a more stable half-filled d\(^5\) configuration. The stability of these configurations makes Cr\(^{2+}\) more likely to donate electrons, and Mn\(^{3+}\) more likely to accept electrons.
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Question: 6

Complete the following ionic equation:
2MnO4 + H2O + I

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- The reaction shows the reduction of MnO\(_4^-\) and the oxidation of I\(^-\). - The balance of charges and atoms is important when completing ionic equations.
Updated On: Feb 25, 2025
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Solution and Explanation

The ionic equation is: \[ 2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^- \]
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