Problem 113
Question
The transition elements with some exceptions can show a large number of oxidation states. The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of \((\mathrm{n}-1) \mathrm{d}\) and outer \(\mathrm{ns}\)-electrons. For the first five elements of 3 d-transition series. The minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of \(4 \mathrm{~s}\) and \(3 \mathrm{~d}\)-electrons. The relative stability of various oxidation state of a given element can be explained on the basis of stability of \(\mathrm{d}^{0}, \mathrm{~d}^{5}\) and \(\mathrm{d}^{10}\) configurations. Identify the correct statement (a) \(\mathrm{Ti}^{4+}, \mathrm{Mn}^{2+}\) are stable oxidation states (b) The most common oxidation state of 3 d-series is \(+2\) (c) The lowest oxidation state of \(\mathrm{Cr}\) and \(\mathrm{Cu}\) is \(+1\) while for others it is \(+2\). (d) All of these
Step-by-Step Solution
VerifiedKey Concepts
Electronic Configuration
3d Transition Series
- These metals exhibit various oxidation states, which can be explained by the number of electrons available in the 3d and 4s orbitals.
- Through this series, metals can exhibit oxidation states ranging from +1 to +7.
Variable Oxidation States
Stability of Oxidation States
| For instance, configurations with half-filled ( d5) or fully filled (d10) d orbitals exhibit notable stability due to symmetrical electron distribution and exchange energy. |
| As a case, Ti4+ is stable because it achieves a d0 state, losing all 3d and 4s electrons. Similarly, Mn2+ is stable with a d5 configuration. |