NCERT INTEXT QUESTIONS: d &f -BLOCK ELEMENTS
Q 4.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?
Ans: Ag is considered as a transition element because it has 4d9(incompletely filled) configuration in Ag2+ oxidation state
Explanation:Transition element is defined as the element which has incompletely filled d – sub-shell in its ground state or any of its stable oxidation state. Configuration of Ag: [Kr]4d105s1 and Ag2+: [Kr]4d95s0.
Q 4.2 In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol–1. Why?
Ans: Enthalpy of atomization depends on the strength of metallic bond which further based on the number of unpaired electrons. Zinc has fully filled orbitals(3d10), which do not participate in metallic bonding, resulting in weak metallic bonds and the lowest enthalpy of atomization in the series.
Explanation: Enthalpy of atomization is the energy required to separate all the atoms in solid state to into gaseous state. If the metallic bond is weak then metal needs less energy to separate into gaseous atoms and the strength of metallic bond depends on the number of d – electrons available for bonding. If d – sub-shell is completely filled then electrons are not easily available bonding.
Q 4.3 Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Ans: In the 3d series of elements Manganese(Mn) exhibits maximum number of oxidation states from +2 to +7, due to the more number of unpaired electrons.
Explanation: Electronic configuration of Mn is [Ar]3d54s2. Two electrons from 4s and all five unpaired electrons are available for bonding to show oxidation states from +2 to +7 (Mn2+, Mn3+, Mn4+, Mn5+, Mn6+ and Mn7+)
Q 4.4 The Eo(M2+/M) value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high ∆aHo and low ∆hydHo)
Ans: In the first transition series only Cu has a positive value because high energy required for the transformation Cu → Cu2+ is not compensated by its hydration enthalpy.
Explanation: Eo(M2+/M) value depends on enthalpy of atomization, first and second ionization enthalpies of Cu and Enthalpy of hydration of Cu2+. Due to high second ionization enthalpy, hydration enthalpy(difficult remove electron from 3d10) not sufficient to compensate ionization enthalpy.
Q 4.5 How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?
Ans: Irregular ionization enthalpies is due the exceptionally stable half filled and fully filled configurations(example: Cr : [Ar]3d54s1; Cu : [Ar]3d104s1)
Explanation: Generally from Sc → Zn ionization enthalpies increases due to increase in nuclear charge. Due to stable half filled d5 configuration of Cr and fully filled d10 configuration of Cu caused high second ionization enthalpies because it requires more energy to remove electrons from stable configurations.
Q 4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Ans: Florine and Oxygen are first and second highest electronegative elements which form strong bonds with metal and stabilizes highest oxidation state of transition metal.
Explanation: Fluorine and oxygen are most electronegative and attract electrons from metal and enables metal to lose electrons and will adopt highest oxidation states. These elements form strong bonds to stabilize highest oxidation state.
Q 4.7 Which is a stronger reducing agent Cr2+ or Fe2+ and why ?
Ans: Cr2+ is stronger reducing agent than Fe2+ because Cr2+ is unstable and it oxidizes to Cr3+ which has a stable half-filled t2g configuration.
Explanation: Configuration of Cr2+ is [Ar]3d4, when it oxidizes it changes to Cr3+ with configuration [Ar]3d3. In aqueous solution (H2O is ligand), Cr3+ has stable half filled configuration:[1] t2g3. Therefore, Cr2+ oxidizes to Cr3+. In case of Fe2+(3d6), if it oxidizes it gets stable half filled configuration(3d5). But Fe2+ is more stable in aqueous solutions, due to its high Eo(Fe3+/Fe2+).
Q 4.8 Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).
Ans: Spin-only magnetic moment can be calculated by using the following formula in which n = number of unpaired electrons.
Atomic number with Z = 27 is Cobalt. Therefore, electronic configuration of Co2+ = [Ar]3d7, which contain 3 unpaired electrons. Therefore, n = 2. Substitute this n value in the above equation.
Q 4.9Explain why Cu+ ion is not stable in aqueous solutions?
Ans: In water, Cu+ is not stable because it disproportionates to Cu2+ and Cu because of high hydration enthalpy of Cu2+.
Explanation: In water, has a tendency to convert into more stable forms due to thermodynamic favorability and high hydration enthalpy. Reaction: Cu+ → Cu2+ + Cu.
Q 4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Ans: In Actinoids, 5f sub-shell is progressively filled which is poor shielding than 4f -sub-shell in lanthanoids.
Explanation: 5f sub-shell is more diffused and less shielding of the outer electrons which causes more contraction in size than the case of lanthanoids.