$$C{u^{2 + }}\left[ {Ar} \right]3{d^9},T{i^{4 + }}\left[ {Ar} \right]3{d^0},$$       $$C{o^{2 + }}\left[ {Ar} \right]3{d^7},F{e^{2 + }}\left[ {Ar} \right]3{d^6}$$       (i), (iii), (iv) are coloured ions hence the answer is B.

$$2AgN{O_3} + {K_2}Cr{O_4} \to \mathop {A{g_2}Cr{O_4}}\limits_{{\text{Red precpitate}}} + 2KN{O_3}$$

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

$$\mu = \sqrt {n\left( {n + 2} \right)} \,\,B.M.$$
Electronic configuration of ion $$\left( {25} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^2}$$
No. of unpaired electrons $$\left( n \right) = 5$$
$$\eqalign{ & \mu = \sqrt {5\left( {5 + 2} \right)} \cr & \,\,\,\,\, = \sqrt {35} \cr & \,\,\,\,\, = 5.9\,B.M. \cr} $$

Hypo solution $$\left( {N{a_2}{S_2}{O_3}} \right) $$   is used in photography to remove the unaffected $$AgBr$$  in the form of soluble complex.
$$AgBr + 2N{a_2}{S_2}{O_3} \to \mathop {N{a_3}\left[ {Ag{{\left( {{S_2}{O_3}} \right)}_2}} \right]}\limits_{{\text{Sod}}{\text{.}}\,{\text{argentothiosulphate}}} + NaBr$$

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

This can be understood on the basis of $${E^ \circ }$$  values for $${M^{2 + }}/M.$$
\[\underset{{{M}^{2+}}/M}{\mathop{{{E}^{\circ }}/V}}\,\,\,\,\underset{\begin{smallmatrix} \\ \\ -0.90 \end{smallmatrix}}{\mathop{Cr}}\,\,\,\,\,\underset{\begin{smallmatrix} \\ \\ -1.18 \end{smallmatrix}}{\mathop{Mn}}\,\,\,\,\,\underset{\begin{smallmatrix} \\ \\ -0.44 \end{smallmatrix}}{\mathop{Fe}}\,\,\,\,\,\underset{\begin{smallmatrix} \\ \\ -0.28 \end{smallmatrix}}{\mathop{Co}}\,\]
$${E^ \circ }$$ value for $$Mn$$  is more negative than expected from general trend due to extra stability of half-filled electronic configuration of $$M{n^{2 + }}\,ion.$$
Thus, the correct order should be $$Mn > Cr > Fe > Co$$
An examination of $${{E^ \circ }}$$ values for redox couple $${M^{3 + }}/{M^{2 + }}$$  shows that $$C{r^{2 + }}$$  is strong reducing agent $$\left( {E_{{M^{3 + }}/{M^{2 + }}}^ \circ = 0.41V} \right)$$     and liberates $${H_2}$$ from dilute acids.
$$2C{r^{2 + }}\left( {aq} \right) + 2{H^ + }\left( {aq} \right) \to $$      $$2C{r^{3 + }}\left( {aq} \right) + {H_2} \uparrow \left( g \right)$$
∴ The correct order is $$Mn > Fe > Cr > Co.$$

Actinoid series has elements from atomic no. 90 to 103. Thulium $$(Tm)$$  has atomic no. 69.

Stainless steel is resistant to rusting. It contains 73% iron, 18% chromium and 8% nickel. Stainless steel is used in utensils, cycle, cutlery and automobile parts.

$$C{r_2}O_7^{2 - } + 14{H^ + } + 3S{n^{2 + }} \to $$      $$2C{r^{3 + }} + 3S{n^{4 + }} + 7{H_2}O$$