In $${}_{23}V = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^3}$$  subshell.
$${}_{24}Cr = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^1}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^5}$$  subshell.
$${}_{26}Fe = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^6}$$  subshell.
$${}_{25}Mn = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^2}$$
Third electron which is removed to give third ionisation potential, belongs to $$3{d^5}$$  subshell.
In all elements shell and subshells are same. Required amount of energy (enthalpy) is based upon the stability of $$d$$ - subshell.
The $$3{d^5}$$ -subshell has highest stability in all because it is half-filled subshell. So, $$Mn$$  shows highest third ionisation potential.

$$Np$$  and $$Pu$$  in $${\left[ {Np{O_6}} \right]^{5 - }}$$  and $${\left[ {Pu{O_6}} \right]^{5 - }}$$  oxoanions show +7 oxidation states which are not so stable.

$$L \to M$$  charge transfer spectra. $$KMn{O_4}$$  is colored because it absorbs light in the visible range of electromagnetic radiation. The permanganate ion is the source of color, as a ligand to metal, $$\left( {L \to M} \right)$$   charge transfer takes place between oxygen's $$p$$  orbitals and the empty $$d - $$ orbitals on the metal. This charge transfer takes place when a photon of light is absorbed, which leads to the purple color of the compound.

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$$\eqalign{ & S{m^{2 + }}\left( {Z = 62} \right) \cr & \left[ {Xe} \right]4{f^6}6{s^2} - 6\,{\text{unpaired}}\,{e^ - }\,E{u^{2 + }}\left( {Z = 63} \right) \cr & \left[ {Xe} \right]4{f^7}6{s^2} - 7\,{\text{unpaired}}\,{e^ - }\,Y{b^{2 + }}\left( {Z = 70} \right) \cr & \left[ {Xe} \right]4{f^{14}}6{s^2} - 0\,{\text{unpaired}}\,{e^ - }\,C{e^{2 + }}\left( {Z = 58} \right) \cr & \left[ {Xe} \right]4{f^1}5{d^1}6{s^2} - 2\,{\text{unpaired}}\,{e^ - } \cr & {\text{Only }}Y{b^{2 + }}\,{\text{is diamagnetic}}{\text{.}} \cr} $$

Copper sulphate when react with $$KCN$$  first give precipitate of cupric cyanide which reduce into $$C{u_2}C{N_2}$$   and dissolve in excess of $$KCN$$  to give soluble $${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$   complex salt

\[\begin{align} & \,\,\,\,\,\,\,[CuS{{O}_{4}}+2KCN\to \underset{\begin{smallmatrix} \text{Cupric} \\ \text{cyanide} \end{smallmatrix}}{\mathop{Cu{{\left( CN \right)}_{2}}}}\,+{{K}_{2}}S{{O}_{4}}]\times 2 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2Cu{{\left( CN \right)}_{2}}\to C{{u}_{2}}{{\left( CN \right)}_{2}}+\underset{\text{Cyanogen}}{\mathop{NC-CN}}\, \\ & C{{u}_{2}}{{\left( CN \right)}_{2}}+6KCN\to \underset{\text{Soluble}\,\,\text{complex}\,\,\text{salt}}{\mathop{2{{K}_{3}}\left[ Cu{{\left( CN \right)}_{4}} \right]}}\, \\ & \overline{2CuS{{O}_{4}}+10KCN\to 2{{K}_{3}}\left[ Cu{{\left( CN \right)}_{4}} \right]+2{{K}_{2}}S{{O}_{4}}+{{\left( CN \right)}_{2}}} \\ \end{align}\]

NOTE: The main reason for exhibiting larger number of oxidation states by actinoids as compared to lanthanoids is lesser energy difference between $$5f$$  and $$6d$$  orbitals as compared to that between $$4f$$  and $$5d$$  orbitals.
In case of actinoids we can remove electrons from $$5f$$  as also from $$6d$$  and due to this actinoids exhibit larger number of oxidation state than lanthanoids. Thus the correct answer is option (B).

Ionic radii decrease across lanthanide series due to lanthanide contraction. As all ions are in $$ + 3\,O.S.,$$  ionic radii will follow the trend of atomic radii.
$$\therefore \mathop {L{a^{3 + }}}\limits_{\left( {Z = 57} \right)} > \mathop {C{e^{3 + }}}\limits_{\left( {Z = 58} \right)} > \mathop {P{m^{3 + }}}\limits_{\left( {Z = 61} \right)} > \mathop {Y{b^{3 + }}}\limits_{\left( {Z = 70} \right)} $$

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