Chapter 8 The d and f Block Elements Solutions
Question - 1 : - Silver atom has completely filled d orbitals (4d10) in its ground state. How can you saythat it is a transition element?
Answer - 1 : -
Ag has a completely filled 4d orbital (4d10 5s1) in its ground state. Now, silverdisplays two oxidation states (+1 and +2). In the +1 oxidation state, anelectron is removed from the s-orboital.However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9).Hence, it is a transition element.
Question - 2 : - Silver atom has completely filled d-orbitals (4d10) in itsground state. How can you say that it is a transition element?
Answer - 2 : -
Silver (Z = 47) can exhibit +2 oxidation statewherein it will have incompletely filled d-orbitals (4d), hence it is atransition element.
Question - 3 : - In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation ofzinc is the lowest, i.e., 126 kJ mol-1. Why?
Answer - 3 : -
In the formation of metallic bonds, noelectrons from 3d-orbitals are involved in case of zinc, while in all othermetals of the 3d-series, electrons from the d-orbitals are always involved inthe formation of metallic bonds.
Question - 4 : - Which of the 3d-series of the transition metals exhibits the largestnumber of oxidation states and why?
Answer - 4 : -
Manganese (Z = 25), as its atom has themaximum number of unpaired electrons.
Question - 5 : - The E°(M2+/M) value for copper is positive (+0.34 V). What ispossibly the reason for this? (Hint: consider its high ∆aH° and low∆hydH°)
Answer - 5 : -
The high energy to transform Cu(s) toCu2+(aq) is not found balanced by its hydrationenergy. Hydration energy and lattice energy of Cu2+ is morethan Cu.
Question - 6 : - How would you account for the irregular variation of ionisationenthalpies (first and second) in the first series of the transition elements?
Answer - 6 : -
Irregular variation of ionisation enthalpiesis mainly attributed to varying degree of stability of different 3d –configurations (e.g., d°, d5, d10 are exceptionallystable).
Question - 7 : - Why is the highest oxidation state of a metal exhibited in its oxide orfluoride only?
Answer - 7 : -
Because of small size and highelectronegativity oxygen or fluorine can oxidise the metal to its highestoxidation state.
Question - 8 : - Which is a stronger reducing agent Cr2+ or Fe2+ andwhy?
Answer - 8 : -
Cr2+ is stronger reducingagent than Fe2+ Reason: d4 → d3 occursin case of Cr2+ to Cr2+. But d6 → d5 occursin case of Fe2+ to Fe2+.
In a medium (like water) d’1 is more stable as compared to d5.
Question - 9 : - Calculate the ‘spin only’ magnetic moment of M2+(aq) ion(Z = 27).
Answer - 9 : - 
Question - 10 : - Explain why Cu+ ion is not stable in aqueous solutions?
Answer - 10 : -
Cu+ in aqueous solutionundergoes disproportionation, i.e.,
2Cu+(aq) → Cu2+(aq) + Cu(s)
The stability of Cu2+(aq) rather than Cu+(aq) isdue to the much more negative ∆hyd H°of Cu2+(aq) thanCu+, which more than compensates for the second ionisation enthalpyof Cu.