CBSE Class 12 Chemistry Chapter 15: The d and f Block Elements NCERT Solutions

Solution:

An element is classified as a transition element if it possesses incompletely filled d orbitals either in its ground state or in any of its stable oxidation states. While a silver atom in its ground state has a completely filled 4d orbital (\(4d^{10} 5s^1\)), it exhibits a +2 oxidation state. In the \(Ag^{2+}\) ion, one electron is removed from the 5s orbital and another from the 4d orbital, resulting in an incomplete 4d orbital configuration (\(4d^9\)). Therefore, due to the presence of incomplete d orbitals in the \(+2\) oxidation state, silver is considered a transition element.

Solution:

The enthalpy of atomization is largely determined by the strength of metallic bonding between atoms in the solid state. Metallic bonding is stronger when there are more unpaired electrons available for delocalization. In the first transition series (Sc to Zn), most elements have unpaired electrons in their d orbitals, which contribute significantly to strong metallic bonding. Zinc, however, has the electronic configuration \(3d^{10} 4s^2\). Since all the d orbitals are completely filled and there are no unpaired electrons in the d subshell, the inter-atomic metallic bonding in zinc is the weakest among these elements. Consequently, less energy is required to break these bonds and convert the element into individual atoms, resulting in the lowest enthalpy of atomization (126 kJ mol^-1).

Solution:

Manganese (Mn), with atomic number Z = 25, exhibits the largest number of oxidation states among the 3d transition metals. Its electronic configuration is \(3d^5 4s^2\). Manganese has five unpaired electrons in the 3d subshell and two electrons in the 4s subshell. The ability to lose electrons from both the 4s and 3d orbitals allows manganese to exhibit a wide range of oxidation states, from +2 (by losing the two 4s electrons) up to +7 (by losing all seven valence electrons). This extensive variability in oxidation states is due to the presence of a large number of unpaired electrons.

Solution:

The standard electrode potential (\(E^{\theta}(M^{2+}/M)\)) for a metal reflects the overall energy change involved in the process of converting the solid metal to its aqueous ions (\(M_{(s)} \rightarrow M^{2+}_{(aq)}\)). This process can be broken down into three main energy steps:

1. Sublimation: The energy required to convert the solid metal into gaseous atoms (\(M_{(s)} \rightarrow M_{(g)}\)), represented by \(\Delta_s H^{\theta}\) (enthalpy of atomization).

   $M\_\{(s)\}\; \backslash longrightarrow\; M\_\{(g)\}\; \backslash qquad\; \backslash Delta\_s\; H^\{\backslash theta\}$

2. Ionization: The energy required to remove electrons from gaseous atoms to form gaseous ions (\(M_{(g)} \rightarrow M^{2+}_{(g)}\)), represented by \(\Delta_i H^{\theta}\) (sum of first and second ionization enthalpies).

   $M\_\{(g)\}\; \backslash longrightarrow\; M^\{2+\}\_\{(g)\}\; \backslash qquad\; \backslash Delta\_i\; H^\{\backslash theta\}$

3. Hydration: The energy released when gaseous ions dissolve in water to form aqueous ions (\(M^{2+}_{(g)} \rightarrow M^{2+}_{(aq)}\)), represented by \(\Delta_{hyd} H^{\theta}\) (enthalpy of hydration).

   $M^\{2+\}\_\{(g)\}\; \backslash longrightarrow\; M^\{2+\}\_\{(aq)\}\; \backslash qquad\; \backslash Delta\_\{hyd\}\; H^\{\backslash theta\}$

The overall enthalpy change is \(\Delta_{total} H^{\theta} = \Delta_s H^{\theta} + \Delta_i H^{\theta} + \Delta_{hyd} H^{\theta}\). A positive \(E^{\theta}(M^{2+}/M)\) value indicates that the overall process is endothermic or requires significant energy input. For copper, the enthalpy of atomization (\(\Delta_s H^{\theta}\)) is notably high, and the hydration energy (\(\Delta_{hyd} H^{\theta}\)) is relatively low compared to the energy required for sublimation and ionization. This unfavorable combination of high energy input (atomization and ionization) and less energy release (hydration) results in a positive \(E^{\theta}(Cu^{2+}/Cu)\) value (+0.34 V).

Solution:

The ionization enthalpies of the first transition series (Sc to Zn) generally increase from left to right due to the gradual increase in nuclear charge and the poor shielding effect of the 3d electrons. However, the variation is not perfectly regular and shows some irregularities. These irregularities can be explained by considering the extra stability associated with certain electronic configurations:

• Half-filled d-orbitals (\(d^5\)): Configurations like \(3d^5\) are particularly stable. For example, Chromium (Cr) has the configuration \(3d^5 4s^1\). It has a lower first ionization enthalpy than expected because losing one electron from the 4s orbital results in a stable \(3d^5\) configuration.
• Completely filled d-orbitals (\(d^{10}\)): Configurations like \(3d^{10}\) are also exceptionally stable. For instance, Zinc (Zn) has the configuration \(3d^{10} 4s^2\). It exhibits an exceptionally high first ionization enthalpy because the electron must be removed from the 4s orbital, leaving behind the very stable, completely filled \(3d^{10}\) configuration.

Similarly, elements with \(d^0\) configurations (like Sc) also have specific ionization energy characteristics. The presence of these stable configurations disrupts the smooth trend of increasing ionization enthalpies across the series, leading to the observed irregularities.