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

$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \to $$  
no of unpaired electron $$= 0$$
$${\left[ {MnC{l_4}} \right]^{2 - }} \to $$  
no of unpaired electron $$= 5$$
$${\left[ {CoC{l_4}} \right]^{2 - }} \to $$  
no of unpaired electron $$= 3$$
The greater the number of unpaired electrons, greater the magnitude of magnetic moment. Hence the correct order will be $${\left[ {MnC{l_4}} \right]^{2 - }} > {\left[ {CoC{l_4}} \right]^{2 - }} > {\left[ {Fe{{\left( {Cn} \right)}_6}} \right]^{4 - }}$$

The catalytic activity of transitional metals is due to their variable oxidation states.

In lanthanides ( at. no. of elements 57 to 71 ) the electronic configuration of outermost shells are $$\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){s^2}{p^6}{d^{0\,\,{\text{to}}\,\,1}}n{s^2}.$$

$$N{a_2}C{r_2}{O_7}$$   is used as primary standard in volumetry. Hydration energy of $$N{a^ + }$$  is greater than $${K^ + }$$  because of smaller size of $$N{a^ + },N{a_2}C{r_2}{O_7}$$    is more soluble than $${K_2}C{r_2}{O_7}.$$

$$Z{n^{2 + }}\left( {3{d^{10}}} \right)$$   has zero unpaired electron (colourless).
$$N{i^{2 + }}\left( {3{d^8}} \right)$$   has 2 unpaired electrons (coloured).
$$C{u^{2 + }}\left( {3{d^9}} \right)$$   has 1unpaired electron (coloured).
$$C{r^{2 + }}\left( {3{d^4}} \right)$$   has 4 unpaired electrons (coloured).

$$\left( {Fe\,{\text{becomes}}\,{\text{passive}}\,{\text{on}}\,{\text{reaction}}\,{\text{with}}\,{\text{concentrated}}\,HN{O_3}} \right).\,\,{\text{However,}}\,{\text{cold}}\,{\text{relatively}}\,{\text{conc}}{\text{.}}\,HN{O_3}\,{\text{reacts}}\,{\text{with}}\,Fe\,{\text{as}}\,{\text{below}}{\text{.}}$$
$$\eqalign{ & Fe + 6HN{O_3} \to Fe{\left( {N{O_3}} \right)_3} + 3N{O_2} + 3{H_2}O \cr & Cu + 4HN{O_3}\,\left( {conc.} \right) \to Cu{\left( {N{O_3}} \right)_2} + 2N{O_2} + 2{H_2}O \cr & 4Au + 8NaCN + {O_2} + 2{H_2}O \to 4Na\left[ {Au{{\left( {CN} \right)}_2}} \right] + 4NaOH \cr & Zn + 2\,NaOH \to N{a_2}Zn{O_2} + {H_2} \cr} $$

\[\begin{align} & {{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\text{heat}}{{N}_{2}}+C{{r}_{2}}{{O}_{3}}+4{{H}_{2}}O \\ & N{{H}_{4}}N{{O}_{2}}\to {{N}_{2}}+2{{H}_{2}}O\,\,\,\,\,\left( \text{Same}\,\,\text{gas}\,\,\text{i}\text{.e}\text{.,}\,\,{{N}_{2}} \right) \\ & N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O \\ & M{{g}_{3}}{{N}_{2}}+3{{H}_{2}}O\to 3Mg{{\left( OH \right)}_{2}}+2N{{H}_{3}} \\ & 2NaN{{O}_{2}}+{{H}_{2}}{{O}_{2}}\to N{{a}_{2}}{{O}_{2}}+2HN{{O}_{2}} \\ \end{align}\]

$$HgC{l_2}$$  and $${I_2}$$  both when dissolved in water containing $${I^ - }$$ $$ions,$$  the pair of species formed is $$HgI_4^{2 - }$$  and $$I_3^ - .$$
In aqueous solution, $${I_2}$$  reacts with $${I^ - }$$  and maintains the following equilibrium.
$${I_2} + {I^ - } \rightleftharpoons I_3^ - $$
$$H{g^{2 + }}$$  gives ppt. of $$Hg{I_2}$$  on reaction with $${I^ - }.$$
But $$Hg{I_2}$$  is soluble in excess of $${I^ - }$$
$$H{g^{2 + }} + 2{I^ - } \to \mathop {Hg{I_2}}\limits_{{\text{Red}}\,{\text{ppt}}.} \downarrow 2C{l^ - }$$
$$Hg{I_2} + 2{I^ - } \rightleftharpoons {\left[ {Hg{I_4}} \right]^{2 - }}$$

$$Zr\left( {40} \right),Hf\left( {72} \right)$$    have similar properties as they belong to same group and have same valence shell electronic configuration.

$$\eqalign{ & CrO_4^{2 - }\left( {aq} \right) + 4{H_2}O + 3{e^ - } \to Cr{\left( {OH} \right)_3} + 5O{H^ - }, \cr & {E^ \circ } = - 0.13\,V \cr & C{r_2}O_7^{2 - }\left( {aq.} \right) + 14{H^ + }\left( {aq.} \right) + 6{e^ - } \to 2C{r^{3 + }}\left( {aq.} \right) + 7{H_2}O, \cr & {E^ \circ } = 1.33\,V \cr} $$