Dilute nitric acid converts chromate into dichromate and $${H_2}O.$$
$$2{K_2}Cr{O_4} + 2HN{O_3} \to $$     $${K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O$$
$$\text{or,}\underset{\text{yellow}}{2Cr{O}_4^{2-}}\stackrel{H^{+}}{\rightleftharpoons }\underset{\text{orange}}{C{r}_2{O}_7^{2-}}+H_{2}O$$

\[{{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }\]     \[\underset{\left( \text{green} \right)}{\mathop{C{{r}_{2}}{{O}_{3}}}}\,+{{N}_{2}}+4{{H}_{2}}O\]
Hence green coloured powder blown in the air is $$C{r_2}{O_3}.$$

Oxidation state of $$Fe$$  in $$Fe{\left( {CO} \right)_5}$$  is zero because $$CO$$  is a neutral ligand and it shows zero oxidation state.
NOTE
In metal carbonyls, oxidation state of metal is always zero.

In $$CuS{O_4}\,.5{H_2}O,$$   four $${H_2}O$$  molecules are directly coordinated to the central metal ion while one $${H_2}O$$  molecule is hydrogen bonded.

$$F{e^{3 + }}$$  is easily hydrolysed than $$F{e^{2 + }}$$  due to more positive charge.

NOTE: More the distance between nucleus and outer orbitals, lesser will be force of attraction on them. Distance between nucleus and $$5f$$  orbitals is more as compared to distance between $$4f$$  orbital and nucleus. So actinoids exhibit more number of oxidation states in general than the lanthanoids.

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

$$\eqalign{ & {\text{Ionic}}\,{\text{radii}}\, \propto \frac{1}{z} \cr & {\text{Thus}},\,\frac{{{z_2}}}{{{z_1}}}\, \Rightarrow \frac{{1.06}}{{\left( {{\text{Ionic}}\,{\text{radii}}\,{\text{of}}\,L{u^{3 + }}} \right)}} = \frac{{71}}{{57}} \cr & \Rightarrow \,{\text{Ionic}}\,{\text{radii}}\,{\text{of}}\,L{u^{3 + }} = 0.85\,\mathop {\text{A}}\limits^o \cr} $$

$$\eqalign{ & A{g_2}S{O_4} \to A{g^ + }\left( {4{d^{10}}} \right) - {\text{colourless}} \cr & Cu{F_2} \to C{u^{2 + }}\left( {3{d^9}} \right) - {\text{coloured}} \cr & Zn{F_2} \to Z{n^{2 + }}\left( {3{d^{10}}} \right) - {\text{colourless}} \cr & C{u_2}C{l_2} \to C{u^ + }\left( {3{d^{10}}} \right) - {\text{colourless}} \cr} $$

Common oxidation states of $$Ce$$ (Cerium) are + 3 and + 4