Due to poor shielding of $$4f$$  electrons, bond strength is maximum for $$H{g^ + } - H{g^ + }.$$

$$\eqalign{ & T{i^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^0} \cr & {V^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},3{d^2},4{s^0} \cr & C{r^{4 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^0} \cr & M{n^{5 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^0} \cr} $$

Magnetic moment $$\mu = \sqrt {n\left( {n + 2} \right)} $$    where $$n = $$  number of unpaired electrons
$$\eqalign{ & \sqrt {15} = \sqrt {n\left( {n + 2} \right)} \cr & \therefore \,\,n = 3 \cr} $$

The oxidation state in both ( lanthanide and actinide ) is $$+3.$$  The property of actinide are very similar to those of lanthanide when both are in $$+3$$  state.

Metallic bonding in mercury is weak.

The overall decrease in atomic and ionic radii from $$L{a^{3 + }}$$  to $$L{u^{3 + }}$$  is called lanthanoid contraction. Hence, the correct order is $$Y{b^{3 + }} < P{m^{3 + }} < C{e^{3 + }} < L{a^{3 + }}$$

Lanthanum is a $$d$$ - block element which resembles lanthanides.

In case of transition metal oxides, the oxides with metals in lower oxidation states are basic in nature.
$$O.S.$$  of $$Mn$$  in $$M{n_2}{O_7} = + 7;V$$    in $${V_2}{O_3} = + 3;V$$    in $${V_2}{O_5} = + 5;Cr$$    in $$CrO = + 2;Cr$$    in $$C{r_2}{O_3} = + 3$$
Thus in $${V_2}{O_3},CrO$$   and $$C{r_2}{O_3}$$  transition metal ion is in lower oxidation state but $$C{r_2}{O_3}$$  is amphoteric in nature. Hence $${V_2}{O_3}$$  and $$CrO$$  are basic in nature.

We know that lanthanides $$La,Gd$$  shows $$+3,$$  oxidation state, while $$Eu$$  shows oxidation state of $$+2$$  and $$+ 3.$$  Am shows $$+3,$$  $$+4, +5$$   and $$+6$$  oxidation states. Therefore Americium $$(Am)$$  has maximum number of oxidation states.

$$2CuS{O_4} + {K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to \mathop {C{u_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + 2{K_2}S{O_4}}\limits_{{\text{Chocolate ppt}}{\text{.}}} $$