$$\eqalign{ & Gd\left( {Z = 64} \right) \Rightarrow 4{f^7}5{d^1}6{s^2} \cr & G{d^{3 + }} = 4{f^7} \cr} $$
$$G{d^{3 + }}$$  is stable due to the presence of exactly half-filled $$4f$$ - subshell.

$$NO_3^ - \,ion$$   are reduced by nascent hydrogen
$$\eqalign{ & {\text{Metal}} + HN{O_3} \to {\text{Metal nitrate}} + \left[ H \right] \cr & HN{O_3} + 8\left[ H \right] \to {N_2}O + 5{H_2}O \cr} $$

Electronic configuration of gadolinium is $$\left[ {Xe} \right]4{f^7}5{d^1}6{s^2}$$

Lanthanoid ion with no unpaired electron is diamagnetic in nature.
$$C{e_{58}} = \left[ {Xe} \right]4{f^2}5{d^0}6{s^2}$$
$$C{e^{2 + }} = \left[ {Xe} \right]4{f^2}$$     (two unpaired electrons)
$$S{m_{62}} = \left[ {Xe} \right]4{f^6}5{d^0}6{s^2}$$
$$S{m^{2 + }} = \left[ {Xe} \right]4{f^6}$$     (six unpaired electrons)
$$E{u_{63}} = \left[ {Xe} \right]4{f^7}5{d^0}6{s^2}$$
$$E{u^{2 + }} = \left[ {Xe} \right]4{f^7}$$     (seven unpaired electrons)
$$Y{b_{70}} = \left[ {Xe} \right]4{f^{14}}5{d^0}6{s^2}$$
$$Y{b^{2 + }} = \left[ {Xe} \right]4{f^{14}}$$     (no unpaired electrons)
Because of the absence of unpaired electrons, $$Y{b^{2 + }}$$  is diamagnetic.

$$M{n^{3 + }} = 3{d^4} = 4$$    unpaired electrons, $$C{r^{3 + }} = 3{d^3} = 3$$    unpaired electrons, $${V^{3 + }} = 3{d^2} = 2$$    unpaired electrons. $$C{r^{3 + }}$$  is most stable out of these in aqueous solution because it has half-filled $${t_{2g}}$$  level $$\left( {{\text{i}}{\text{.e}}{\text{.}},\,t_{2g}^3} \right).$$

No explanation is given for this question. Let's discuss the answer together.

$$C{u^ + }$$  ions undergo disproportionation, $$2C{u^ + } \to C{u^{2 + }} + Cu$$

$$C{u^{2 + }} + 4N{H_4}OH \to \mathop {{{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]}^{2 + }}}\limits_{{\text{deep blue comp}}} + 4{H_2}O$$

$$C{r^{2 + }}:{d^4}$$     ($$4$$  unpaired electrons)
$$M{n^{2 + }}:{d^5}$$     ($$5$$  unpaired electrons)
$$F{e^{2 + }}:{d^6}$$     ($$4$$  unpaired electrons)
$$C{o^{2 + }}:{d^7}$$     ($$3$$  unpaired electrons)
∴ $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$   has minimum number of unpaired electrons and thus, shows minimum paramagnetic behaviour.
Higher the unpaired $${e^ - }.$$
Higher the magnetic moment
$$\mu = \sqrt {n\left( {n + 2} \right)} $$
$$n=$$  Number of unpaired $${e^ - }$$

$$\eqalign{ & CuS{O_4} + 2KCN \to Cu{\left( {CN} \right)_2} + {K_2}S{O_4} \cr & 2Cu{\left( {CN} \right)_2} \to C{u_2}{\left( {CN} \right)_2} + {\left( {CN} \right)_2}\left( {{\text{Cyanogen}}} \right) \cr & C{u_2}{\left( {CN} \right)_2} + 6KCN \to 2{K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right] \cr} $$