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

Ans: 

${\text{CuC}}{{\text{O}}_{\text{3}}}{\text{ }}\xrightarrow{{{\text{heat}}}}{\text{ CuO  +  C}}{{\text{O}}_{\text{2}}}$

${\text{                                          (D)}}$ 

${\text{Ca(OH}}{{\text{)}}_{\text{2}}}{\text{  +  C}}{{\text{O}}_{\text{2}}}{\text{ }} \to {\text{ CaC}}{{\text{O}}_3}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{                                     (E)}}$ 

${\text{CaC}}{{\text{O}}_3}{\text{  +  C}}{{\text{O}}_{\text{2}}}{\text{ +  }}{{\text{H}}_{\text{2}}}{\text{O  }} \to {\text{    Ca(HC}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$

${\text{                                                clear sol}}{\text{.}}$ 

${\text{CuO   +  CuS }}\xrightarrow{{{\text{heat}}}}{\text{3Cu  +  S}}{{\text{O}}_{\text{2}}}{\text{ }}$

${\text{                                    (A) }}$ 

${\text{ Cu }} + 4{\text{HN}}{{\text{O}}_3}{\text{ }} \to {\text{ Cu(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{  + 2N}}{{\text{O}}_{\text{2}}}{\text{  +  2}}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{           (conc}}{\text{.)           (B)}}$ 

${\text{Cu(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}{\text{  +  4N}}{{\text{H}}_{\text{3}}}{\text{ }} \to {\text{ }}\left[ {{\text{Cu(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{4}}}} \right]{({\text{N}}{{\text{O}}_{\text{3}}}{\text{)}}_{\text{2}}}$

${\text{                                         (C) blue solution}}$ 

Ans: ${\text{(A)  =  FeC}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{4}}}$

${\text{(B)  =  N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$

${\text{(C)  =  N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_7}$

${\text{(D)  =  }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ 

${\text{4FeCr}}{{\text{O}}_4}{\text{  +  8NaC}}{{\text{O}}_{\text{3}}}{\text{  +  7}}{{\text{O}}_{\text{2}}}{\text{ }}\xrightarrow{{}}{\text{ 8N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{  +  2F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}{\text{  +  8C}}{{\text{O}}_{\text{2}}}$

${\text{ (A)                                                     (B)}}$

${\text{2NaCr}}{{\text{O}}_{\text{4}}}{\text{  +  2}}{{\text{H}}^{\text{ + }}}{\text{ }} \to {\text{ N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_7}{\text{  +  2N}}{{\text{a}}^{\text{ + }}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_7}{\text{   +  KCl}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_7}{\text{ +  2NaCl }}$

${\text{    (C)                                    (D)  }}$ 

Ans: ${\text{(A)  =  Mn}}{{\text{O}}_{\text{2}}}$

${\text{(B)  =  }}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}$

${\text{(C)  =  KMn}}{{\text{O}}_{\text{4}}}{\text{ }}$

${\text{(D)  =  KI}}{{\text{O}}_{\text{3}}}$ 

${\text{2Mn}}{{\text{O}}_{\text{2}}}{\text{  +  4KOH  +  }}{{\text{O}}_{\text{2}}}{\text{ }}\xrightarrow{{}}{\text{ 2}}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}{\text{  +  2}}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{ (A)                                              (B)}}$

${\text{3Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{  +  4}}{{\text{H}}^{\text{ + }}}{\text{ }} \to {\text{ 2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{  +  Mn}}{{\text{O}}_{\text{2}}}{\text{  +  2}}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{                                    (C)}}$

${\text{2Mn}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{   +  2}}{{\text{H}}_{\text{2}}}{\text{O  +  KI}} \to {\text{ 2Mn}}{{\text{O}}_{\text{2}}}{\text{  +  2S}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{  +  KI}}{{\text{O}}_{\text{3}}}$

${\text{                                              (A)                         (D)  }}$ 

Ans: As per fajan’s rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from \[{\text{La to Lu}}\] in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, \[{\text{L}}{{\text{a}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]is ionic and \[{\text{L}}{{\text{u}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]is covalent.

Ans: The stability of Oxo salts is directly proportional to the size of atom, as we move from \[{\text{La to Lu}}\], the size of the atom decreases and hence the stability of oxo salts also decreases.

Ans: As we move along the lanthanide series, the atomic size decreases. As a result, the charge/size ratio increases and the stability of the complexes also increases.

Ans: As an effect of lanthanide contraction, zirconium and hafnium have similar radius of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii.

(e) Trends in acidic character of lanthanide oxides.

Ans: Acidic or basic character of oxides is related to the nature of oxides. As we move along the lanthanide series from La to Lu, the covalent character increases. 

Therefore, their basic character decreases or acidic character increases. 

Ans: Out of all the elements of the first transition series copper has the highest second ionisation enthalpy.

Electronic configuration of Copper is: \[{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^1}\]

After the Loss of first electron from the 4s copper acquires \[{\text{3d}}{{\text{ }}^{{\text{10}}}}\] configuration which is stable. Therefore, removal of second electron from the field 3-D orbital is very difficult and requires high amount of energy.

Ans: Among the elements of first transition series zinc has the highest third ionisation enthalpy. Electronic configuration of zinc is: \[{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{2}}}\]

After the loss of two electrons from 4s orbital, Z and +2 Ion acquires \[{\text{3d}}{{\text{ }}^{{\text{10}}}}\] fully filled configuration which is highly stable therefore removal of third electron from \[{\text{3d}}{{\text{ }}^{{\text{10}}}}\]orbital will be more difficult and requires a large amount of energy.

Ans: Zinc is the element of the first transition series that has the lowest enthalpy of atomisation. The electronic configuration of zinc is: \[{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{2}}}\].This is because it has filled 3d-subshell and filled 4s-subshell so no unpaid electron is available for metallic bonding.

(b) Identify the metal and justify your answer:

\[\left( I \right) Carbonyl M{\left( {CO} \right)_5}\]

Ans: \[{\text{Fe}}{\left( {{\text{CO}}} \right)_{\text{5}}}\]as per EAN Rule. 

EAN Rule: Effective atomic number (EAN), number that represents the total number of electrons surrounding the nucleus of a metal atom in a metal complex. 

\[\left( {II} \right) M{O_3}F\]

Ans: The M in the given complex is in +7 O.S.

Manganese is the only element in the first transition series that shows +7 oxidation state, therefore this compound is \[{\text{Mn}}{{\text{O}}_{\text{3}}}{\text{F}}\].

Ans: Interstitial compounds are formed when small atoms like \[{\text{H, C and N}}\] get trapped inside the crystal lattice of transition metals.

Characteristics:

They are hard and rigid in nature having high melting points. 

Like pure metals only, they are conductors. They can gain chemical inertness. 

Ans: ${\text{2}}{{\text{I}}^{\text{ - }}}{\text{  +  }}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{8}}}^{{\text{2 - }}}{\text{ }}\xrightarrow{{{\text{Fe(III)}}}}{\text{ }}{{\text{I}}_{\text{2}}}{\text{  +  2S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$

${\text{ role of fe ions:}}$

${\text{2F}}{{\text{e}}^{{\text{3 + }}}}{\text{  +  2I -  }} \to {\text{ 2F}}{{\text{e}}^{{\text{2 + }}}}{\text{  +  }}{{\text{I}}_{\text{2}}}$

${\text{2F}}{{\text{e}}^{{\text{2 + }}}}{\text{  +  }}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{8}}}^{{\text{2 - }}} \to {\text{ 2F}}{{\text{e}}^{{\text{3 + }}}}{\text{ +  2S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$ 

Ans: 1. Fine powdered state of Nickel is used in the hydrogenation of oils into fats. 

2. Iron is used in the formation of ammonia in Haber’s process. 

3. In contact process in manufacture of \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\], vanadium is used in its oxide form- \[{{\text{V}}_{\text{2}}}{{\text{O}}_{\text{5}}}\].

Ans: ${\text{KMn}}{{\text{O}}_{\text{4}}}{\text{ }}\xrightarrow{\vartriangle }{\text{ }}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}{\text{    +    Mn}}{{\text{O}}_{\text{2}}}{\text{  +  }}{{\text{O}}_{\text{2}}}$

${\text{     (A)                           (B)          (C)}}$

${\text{Mn}}{{\text{O}}_{\text{2}}}{\text{  +  KOH }} \to {\text{ 2}}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}{\text{  +  2}}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{Mn}}{{\text{O}}_{\text{2}}}{\text{  +  4NaCl  +  4}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\xrightarrow{{}}{\text{ MnC}}{{\text{l}}_{\text{2}}}{\text{  +  2NaHS}}{{\text{O}}_{\text{4}}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O  +  C}}{{\text{l}}_{\text{2}}}$

${\text{                                                         (D)}}$ 

${\text{(A)  =  KMn}}{{\text{O}}_{\text{4}}}$

${\text{(B)  =  }}{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}$

${\text{(C)  =  Mn}}{{\text{O}}_{\text{2}}}$

${\text{(D)  =  MnC}}{{\text{l}}_{\text{2}}}$