D and F Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry

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In equation $$\left( i \right)\,\,F{e_2}{\left( {S{O_4}} \right)_3}$$   and in equation $$\left( {ii} \right)F{e_2}{\left( {S{O_4}} \right)_3}$$   on decomposing will form oxide instead of $$Fe$$ . The correct sequence of reactions is
$$\eqalign{ & Fe\mathop \to \limits^{{O_2}.heat} F{e_3}{O_4}\mathop \to \limits^{CO,{{600}^{ \circ C}}} FeO\mathop \to \limits^{CO,{{700}^{ \circ C}}} Fe \cr & \cr} $$

The reaction of $$MnO_4^ - \,\,{\text{and}}\,\,SO_3^{2 - }$$    in acidic medium is derived as follows:
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4\left. {{H_2}O} \right] \times 2$$
$$\frac{{SO_3^{2 - } + {H_2}O \to SO_4^{2 - } + 2{H^ + } + \left. {2{e^ - }} \right] \times 5}}{{2MnO_4^ - + 5SO_3^{2 - } + 6{H^ + } \to 2M{n^{2 + }} + 5SO_4^{2 - } + 3{H_2}O}}$$
Hence, $$2$$ mole $$2MnO_4^ - \equiv 5mol\,\,SO_3^{2 - }$$
i.e., $$\frac{2}{5}mol\,\,MnO_4^ - \equiv 1\,\,mol\,\,SO_3^{2 - }$$

$$M{n^{2 + }}\,{\text{in}}\,\,MnS{O_4}.4{H_2}O\,\,{\text{has}}\,\,{d^5}$$       (five unpaired electrons); $$C{u^{2 + }}\,{\text{in}}\,CuS{O_4}.5{H_2}O\,\,{\text{has}}\,\,{d^9}$$      configuration (one unpaired electron); $$F{e^{2 + }}\,{\text{in}}\,FeS{O_4}.6{H_2}O\,\,{\text{has}}\,\,{d^6}$$      configuration (four unpaired electron); and $$N{i^{2 + }}\,{\text{in}}\,NiS{O_4}.6{H_2}O\,\,{\text{has}}\,\,{d^8}$$     configuration (two unpaired electron). Thus $$CuS{O_4}.5{H_2}O$$    has lowest degree of paramagnetism.

$$P = Mn{O_2},Q = {K_2}Mn{O_4},$$      $$R = KMn{O_4},S = KI{O_3}$$
$$\mathop {2Mn{O_2}}\limits_{\left( P \right)} + 4KOH + {O_2} \to $$      $$\mathop {2{K_2}Mn{O_4}}\limits_{\left( Q \right)} + 2{H_2}O$$
$$3MnO_4^{2 - } + 4{H^ + } \to $$     $$\mathop {2MnO_4^ - }\limits_{\left( R \right)} + Mn{O_2} + 2{H_2}O$$
$$2MnO_4^ - + {H_2}O + KI \to $$      $$\mathop {2Mn{O_2}}\limits_{\left( P \right)} + 2O{H^ - } + \mathop {KI{O_3}}\limits_{\left( S \right)} $$

In case $$HCl$$  is used, it is oxidised to $$C{l_2}.$$
$$2KMn{O_4} + 16HCl \to $$     $$2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2}$$

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

\[CuS{{O}_{4}}\left( aq. \right)\xrightarrow{{{H}_{2}}S}\underset{\begin{smallmatrix} \text{Black ppt}\text{.} \\ \left( M \right) \end{smallmatrix}}{\mathop{2CuS}}\,\xrightarrow[\text{Excess}]{KCN}{{\left[ \underset{\left( N \right)}{\mathop{Cu}}\,{{\left( CN \right)}_{4}} \right]}^{3-}}+\underset{\left( O \right)}{\mathop{{{\left( CN \right)}_{2}}}}\,\uparrow \]

$${K_2}C{r_2}{O_7}$$  contains $$C{r^{6 + }}\left( {3{d^0}} \right)$$   which is diamagnetic but coloured due to charge transfer spectra.
$${\left( {N{H_4}} \right)_2}\left[ {TiC{l_6}} \right]$$    contains $$T{i^{4 + }}\left( {3{d^0}} \right),$$   which is diamagnetic and colourless.
$$VOS{O_4}$$  contains $${V^{4 + }}\left( {3{d^1}} \right),$$   which is paramagnetic and coloured.
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$    contains $$C{u^ + }\left( {3{d^{10}}} \right),$$   which is diamagnetic and colourless.

$$2Mn{O_2} + 4KOH + {O_2} \to $$      $$\mathop {2{K_2}Mn{O_4}}\limits_{{\text{dark green}}} + 2{H_2}O$$