Chapter 8: The D-block and F-block Elements

Ans: (A) \[\frac{2}{5}\]

\[{\text{2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{  +  5}}{{\text{S}}^{{\text{2 - }}}}{\text{  + 16}}{{\text{H}}^{\text{ + }}}{\text{ }} \to {\text{ 2M}}{{\text{n}}^{{\text{2 + }}}}{\text{  +  5S  +  }}{{\text{H}}_{\text{2}}}{\text{O}}\]

From the reaction,

For 5 moles of \[{\text{S}}\] , two moles of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] were required.

Therefore, for one  mole of \[{\text{S}}\] , the number of moles of \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] required will be = \[\frac{2}{5}\].

Ans: (A) ${V_2}{O_5}, C{r_2}{O_3} $

Oxides in the lower oxidation state are basic and in the higher oxidation state they are acidic. They are amphoteric in the intermediate oxidation state. Therefore, \[{{\text{V}}_{\text{2}}}{{\text{O}}_{\text{5}}}{\text{, C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{ }}\]are amphoteric in nature.

Ans: (A) $\left[ {Xe} \right] 4{f^7} 5{d^1} 6{s^2} $

Gadolinium belongs to the 4f series; it has atomic no.= 64. It has extra stability due to a half-filled 4f subshell.

\[{\text{Gd }}\left( {Z = 64} \right){\text{ }} \to {\text{ }}\left[ {{\text{Xe}}} \right]{\text{ 4}}{{\text{f}}^{\text{6}}}{\text{ 5}}{{\text{d}}^{\text{2}}}{\text{ 6}}{{\text{s}}^{\text{2}}}{\text{ }}\]

Ans: (D) They are chemically very reactive.

Interstitial compounds/alloys are substances that are formed when a small atom like carbon, hydrogen, boron, nitrogen can occupy space in their lattices. All the properties mentioned above are true for interstitial compounds except (D), they are chemically inert.

Ans: (B) 3.87 B.M. 

${\text{Cr }} \to {\text{ (Z = 24)}}$

${\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ }} \to {\text{ (Z = 21)}}$

${\text{1}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{s}}^{\text{2}}}{\text{ 2}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{s}}^{\text{2}}}{\text{ 3}}{{\text{p}}^{\text{6}}}{\text{ 3}}{{\text{d}}^{\text{3}}}{\text{ 4}}{{\text{s}}^{\text{0}}}$

${\text{n  =  3 ( 3 unpaired electrons )}}$

${{\mu   =  }}\sqrt {{\text{n ( n  +  2 )}}} $

${{\mu   =  }}\sqrt {{\text{3 ( 3  +  2 )}}} $

${{\mu   =  }}\sqrt {{\text{3( 5 )}}} $

${{\mu   =  }}\sqrt {15{\text{ }}} {\text{  =  3}}{\text{.87 B}}{\text{.M}}{\text{.}}$ 

Ans: (C) \[{\text{I}}{{\text{O}}_{\text{3}}}^{\text{ - }}\]

Iodide is oxidised to iodate.  

${\text{2KMn}}{{\text{O}}_{\text{4}}}{\text{  +  KI  +  }}{{\text{H}}_{\text{2}}}{\text{O }} \to {\text{ 2KOH  +  2Mn}}{{\text{O}}_{\text{2}}}{\text{  +  KI}}{{\text{O}}_{\text{3}}}$

${\text{                  iodide                                               iodate}}$ 

Ans: (A) is incorrect.

Copper lies below hydrogen in electrochemical series and hence cannot liberate hydrogen from acids. 

${\text{Cu  +  2}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{  }} \to {\text{ CuS}}{{\text{O}}_{\text{4}}}{\text{  +  S}}{{\text{O}}_{\text{2}}}{\text{  +  2}}{{\text{H}}_{\text{2}}}{\text{O}}$

Ans:  (C) $S{n^{4 + }} $

\[{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}{\text{  +  14}}{{\text{H}}^{\text{ + }}}{\text{  +  3S}}{{\text{n}}^{{\text{2 + }}}}{\text{  }} \to {\text{ 2C}}{{\text{r}}^{{\text{3 + }}}}{\text{  +  3S}}{{\text{n}}^{{\text{4 + }}}}{\text{  +  7}}{{\text{H}}_{\text{2}}}{\text{O}}\]

In the above redox reaction, oxidation of \[{\text{S}}{{\text{n}}^{{\text{2 + }}}}\] is taking place. It gets converted to \[{\text{S}}{{\text{n}}^{{\text{4 + }}}}\] whereas, reduction of chromium is occurring from +6 oxidation state to +3. 

Ans: (D) in covalent compounds fluorine can form single bond only while oxygen forms double bond

Oxygen has the capacity to form multiple bonds which enables it to form a variety of covalent compounds. 

In \[\left( {{\text{M}}{{\text{n}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)\] also, 6 oxygen are doubly bonded to two manganese atoms and one oxygen is forming bridge between two. 

While in \[\left( {{\text{Mn}}{{\text{F}}_{\text{4}}}} \right)\], four fluorine atoms are singly bonded to manganese atom giving it a +4 oxidation state.

Therefore, due to capability of oxygen to have multiple bonds in covalent compounds, manganese is having higher oxidation state of +7 in \[\left( {{\text{M}}{{\text{n}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)\] . 

Ans:  (C)  both have similar atomic radius

As an effect of lanthanide contraction, zirconium and hafnium have similar radii of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanide contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield well, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and results in the decrease in their size and atomic radii.