Chemistry Class 12 NCERT Solutions: Chapter 8 The d and f Block Elements Part 10

Download PDF of This Page (Size: 171K)

Q: 1. Silver atom has completely filled in its ground state. How can you say that it is a transition element?


has a completely filled orbital in its ground state. Now, silver display two oxidation states . In the oxidation state, an electron is removed from the orboital. However, in the oxidation state, an electron is removed from the orbital. Thus, the orbital now becomes incomplete . Hence, it is a transition element.

Q: 2. In the series , the enthalpy of atomization of zinc is the lowest, i.e., . Why?


The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except , electronic configuration: ), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in and as a result, it has the least enthalpy of atomization.

Q: 3. Which of the series of the transition metals exhibits the largest number of oxidation states and why?


has the maximum number of unpaired electrons present in the (5 electrons). Hence, exhibits the largest number of oxidation states, ranging from to .

Q: 4. The value for copper is positive . What is possibly the reason for this? (Hint: consider it’s high and low )


value of a metal depends on the energy changes involved in the following:

1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

3. Hydration: The energy released when one mole of ions are hydrated.

Now, copper has a high energy of atomization and low hydration energy. Hence, the value for copper is positive.

Q: 5. How would your account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?


Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as . Since these states are exceptionally stable, their ionization enthalpies are very high.

In case of first ionization energy, has low ionization energy. This is because after losing one electron, it attains the stable configuration . On the other hand, has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals .

Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like and have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration . Hence, taking out one electron more from this stable configuration will require a lot of energy.

Q: 6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?


Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.

Q: 7. Which is a stronger reducing agent and why?


The following reactions are involved when act as reducing agents.

The value is is V. This means that can be easily oxidized to , but does not get oxidized to easily. Therefore, is a better reducing agent that .

Q: 8. Calculate the ‘spin only’ magnetic moment of .


i.e., unpaired electrons