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

Ans: As an effect of lanthanoid 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. 

Ans: ${\text{Ce  =  4}}{{\text{f}}^{\text{1}}}{\text{ 5}}{{\text{d}}^{\text{1}}}{\text{ 6}}{{\text{s}}^{\text{2}}}$

${\text{C}}{{\text{e}}^{{\text{3 + }}}}{\text{  =  4}}{{\text{f}}^1}{\text{ 5}}{{\text{d}}^0}{\text{ 6}}{{\text{s}}^0}$ 

As in \[{\text{C}}{{\text{e}}^{{\text{3 + }}}}\],

4f has only single electron. Cerium holds the capacity to lose this electron also and attain \[{\text{4}}{{\text{f}}^0}\] configuration which is more stable. 

\[{\text{C}}{{\text{e}}^{{\text{4 + }}}}{\text{  =  4}}{{\text{f}}^0}{\text{ 5}}{{\text{d}}^0}{\text{ 6}}{{\text{s}}^0}\]

Hence it shows 4+ oxidation states.

Ans: \[{\text{2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{  +  16}}{{\text{H}}^ + }{\text{  +  5}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ }} \to {\text{ 2M}}{{\text{n}}^{{\text{2 + }}}}{\text{  +  8}}{{\text{H}}_{\text{2}}}{\text{O  +  10C}}{{\text{O}}_{\text{2}}}\]

\[{\text{KMn}}{{\text{O}}_{\text{4}}}\] oxidises the oxalic acid to CO2 and reduces itself to \[{\text{M}}{{\text{n}}^{{\text{2 + }}}}\] state. \[{\text{M}}{{\text{n}}^{{\text{2 + }}}}\]is colourless that’s why it seems that \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] has been disappeared. 

Ans: This is due to the inter conversion of dichromate (orange) to chromate ion (yellow). 

${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}{\text{   }}\underset{{{{\text{H}}^{\text{ + }}}}}{\overset{{{\text{O}}{{\text{H}}^{\text{ - }}}}}{\longleftrightarrow}}{\text{   Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ }}$

${\text{(orange)                   (yellow)}}$ 

Ans: \[{\text{KMn}}{{\text{O}}_{\text{4}}}\] act as an oxidising agent however it’s activity as oxidising agent is influenced by the pH of the solution i.e. acidic, basic or neutral solution. 

${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{    +    2KOH      }} \to {\text{   2}}{{\text{K}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{                      (Orange)                (yellow)}}$

${\text{ 2}}{{\text{K}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{      }} \to {\text{    }}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{  +  }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O}}$

${\text{(yellow)                                    (orange)}}$ 

Ans: As an effect of lanthanide contraction, the second and third rows of transition elements resemble each other. For example, zirconium and hafnium have similar radius i.e. 160pm and 159pm 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.

Ans: The sum of sublimation energy and ionisation enthalpy to oxidise \[{\text{cu}}\left( {\text{s}} \right){\text{ to c}}{{\text{u}}^{{\text{2 + }}}}\]is so highly that it is not compensated by the hydration enthalpy of \[{\text{Cu}}\]. Due to this, the \[{\text{E}}\]of \[{\text{Cu}}\]is positive. 

While in case if \[{\text{Zn}}\], the \[{\text{E}}\] value is negative or more negative than the expected value because when the electrons are removed from the 4s-orbital. \[{\text{Zn}}\] acquires a stable \[{\text{3}}{{\text{d}}^{{\text{10}}}}\] configuration state.

Ans: As the positive charge of the ion increases or we can say that oxidation state of a transition element increases, its size decreases and as per Fajan’s rule, more the charge on the metal ion, more is its tendency to form covalent compounds because positively charged cation attracts the electron cloud strongly towards itself. 

Ans: As per \[{\text{(n + l)}}\] rule, 4s has lower energy than 3d-orbital. 

${{{3d  -  n  +  l  =  3 + 2  =  5}}}$

${{\text{4s  -  n  +  l  =  4 + 0  =  4}}}$ 

So, 4s are filled first.

After filling of electrons, 4s-orbital moves beyond 3d-orbital and 4s electrons are loosely held by the nucleus. Hence, electrons are removed first during the process of ionisation.

Ans: Ionisation enthalpies are the main factor that influence the reactivity of transition elements. Higher the ionisation enthalpy, lesser is the reacting of the transition element. 

When we move along the period from \[{\text{Sc to Cu}}\], a regular increase in the ionisation enthalpy is observed which results in the almost regular decrease in the reactivity of elements.