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

The outer electronic configuration of $$C{e^{3 + }}$$  is $$4{f^1}.$$
$$\eqalign{ & {\text{Hence, }}\mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1\left( {1 + 2} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.732\,B.M. \cr} $$

Cuprous compound contains $$C{u^ + }$$  $$ion$$  which has small size, so the hydration is maximum and hence, the system has lower energy which result in stability of the compounds.

Green colour of $$CuC{r_2}{O_7}$$   is due to blue colour of $$C{u^{2 + }}\,ions$$   and yellow colour of $$C{r_2}O_7^{2 - }\,ions.$$

Basic strength decreases from $$La{\left( {OH} \right)_3}$$  to $$Lu{\left( {OH} \right)_3}.$$
Hence, (1) is incorrect.
$$Ce = \left[ {Xe} \right]4{f^1}5{d^1}6{s^2};C{e^{4 + }}$$       $$ = \left[ {Xe} \right]$$
$$Yb = \left[ {Xe} \right]4{f^{14}}6{s^2};Y{b^{2 + }}$$      $$ = \left[ {Xe} \right]4{f^{14}}$$
$$Lu = \left[ {Xe} \right]4{f^{14}}5{d^1}6{s^2};L{u^{3 + }}$$      $$ = \left[ {Xe} \right]4{f^{14}}$$

Electronic configuration of $${}_{63}Eu = {\left[ {Xe} \right]_{54}}4{f^7}6{s^2}$$     Electronic configuration of $${}_{64}Gd = {\left[ {Xe} \right]_{54}}4{f^7}5{d^1}6{s^2}$$
Electronic configuration of $${}_{65}Tb = {\left[ {Xe} \right]_{54}}4{f^9}6{s^2}$$

$$CaC{O_3}\mathop \to \limits^{Heat} CaO + C{O_2};$$
$$\mathop {CaO}\limits_{{\text{Flux}}} + \mathop {Si{O_2}}\limits_{{\text{impurity}}} \to CaSi{O_3}\left( {{\text{slag}}} \right).$$

Oxidation state of $$Fe$$  in $$F{e_3}{O_4}$$  is calculated as
$$\eqalign{ & \mathop F\limits^x {e_3}\mathop {{O_4}}\limits^{ - 2} \cr & 3x + \left( { - 2 \times 4} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x = + 8 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + \frac{8}{3} \cr} $$

When $$S{O_2}$$  is passed through acidified $${K_2}C{r_2}{O_7}$$   solution, green chromium sulphate is formed. In this reaction, oxidation state of $$Cr$$  changes from + 6 to + 3.
$$\mathop {{K_2}C{r_2}{O_7}}\limits_{OS\,\,{\text{of}}\,\,Cr = + 6} + {H_2}S{O_4} + 3S{O_2} \to $$      \[{{K}_{2}}S{{O}_{4}}+\underset{\begin{smallmatrix} OS\,\,\text{of}\,Cr=+3 \\ \text{Green} \end{smallmatrix}}{\mathop{C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}}}\,+{{H}_{2}}O\]
The appearance of green colour is due to the reduction of chromium metal.

The compounds with lower oxidation number and which cannot reduced by one another can exist together. Thus, $$FeC{l_2}$$  and $$SnC{l_2}$$  can exist together as $$F{e^{2 + }}$$  cannot be reduced by $$S{n^{2 + }}.$$

Nitriding is the process of heating of steel in the presence of ammonia. In this process the layer of iron nitride is formed which prevent the rusting of iron.