D and F Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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161.
Select the correct option, among $$Sc\left( {{\text{III}}} \right),Ti\left( {{\text{IV}}} \right),Pd\left( {{\text{II}}} \right)$$ and $$Cu\left( {{\text{II}}} \right)$$ ions
A
All are paramagnetic.
B
All are diamagnetic.
C
$$Sc\left( {{\text{III}}} \right),Ti\left( {{\text{IV}}} \right)$$ are paramagnetic and $$Pd\left( {{\text{III}}} \right),$$ $$Cu\left( {{\text{II}}} \right)$$ are diamagnetic.
D
$$Sc\left( {{\text{III}}} \right),Ti\left( {{\text{IV}}} \right)$$ are diamagnetic and $$Pd\left( {{\text{II}}} \right),$$ $$Cu\left( {{\text{II}}} \right)$$ are paramagnetic.
Answer :
$$Sc\left( {{\text{III}}} \right),Ti\left( {{\text{IV}}} \right)$$ are diamagnetic and $$Pd\left( {{\text{II}}} \right),$$ $$Cu\left( {{\text{II}}} \right)$$ are paramagnetic.
No explanation is given for this question. Let's discuss the answer together.
162.
Arrange the oxides of manganese according to increasing acidic strength.
A
$$MnO < M{n_3}{O_4} < M{n_2}{O_3} < Mn{O_2} < M{n_2}{O_7}$$
B
$$M{n_2}{O_7} < Mn{O_2} < M{n_2}{O_3} < M{n_3}{O_4} < MnO$$
C
$$Mn{O_2} < M{n_2}{O_7} < M{n_3}{O_4} < M{n_2}{O_3} < MnO$$
D
$$M{n_3}{O_4} < M{n_2}{O_3} < M{n_2}{O_7} < Mn{O_2} < MnO$$
Acidic strength of oxides of transition metals increases with increase in oxidation number.
$$\mathop M\limits^{ + 2} nO\mathop {,M}\limits^{ + \,\frac{8}{3}} {n_3}{O_4},\mathop M\limits^{ + 3} {n_2}{O_3},$$ $$\mathop M\limits^{ + 4} n{O_2},\mathop M\limits^{ + 7} {n_2}{O_7}$$
Hence acidic strength is of the order of
$$\mathop {MnO}\limits_{{\text{Basic}}} < M{n_3}{O_4} < \mathop {M{n_2}{O_3}}\limits_{{\text{Amphoteric}}} < $$ $$Mn{O_2} < \mathop {M{n_2}{O_7}}\limits_{{\text{Acidic}}} $$
163.
Iron is rendered passive by treatment with concentrated
A
$${H_2}S{O_4}$$
B
$${H_3}P{O_4}$$
C
$$HCl$$
D
$$HN{O_3}$$
Answer :
$$HN{O_3}$$
Conc. $$HN{O_3}$$ renders iron passive by forming a thin protective film of $$F{e_3}{O_4}$$ on its surface.
164.
Identify the incorrect statement among the following.
A
There is a decrease in the radii of the atoms or ions as one proceeds from $$La$$ or $$Lu$$
B
Lanthanide contraction is the accumulation of successive shrinkages
C
As a result of lanthanide contraction, the properties of $$4d$$ series of the transition elements have no similarities with the $$5d$$ series of elements
D
Shielding power of $$4f$$ electrons is quite weak
Answer :
As a result of lanthanide contraction, the properties of $$4d$$ series of the transition elements have no similarities with the $$5d$$ series of elements
The regular decrease in the radii of lanthanide ions from $$L{a^{3 + }}$$ to $$L{u^{3 + }}$$ is known as lanthanides contraction.
It is due to the greater effect of the increased nuclear charge than that of screening effect ( shielding effect ).
As a result of lanthanide contraction, the atomic radii of element of $$4d$$ and $$5d$$ come closer, so the properties of $$4d$$ and $$5d$$ - transition element shows the similarities.
165.
What happens when potassium iodide reacts with acidic solution of potassium dichromate?
A
It liberates iodine.
B
Potassium sulphate is formed.
C
Chromium sulphate is formed.
D
All the above products are formed.
Answer :
All the above products are formed.
No explanation is given for this question. Let's discuss the answer together.
166.
In the silver plating of copper, $$K\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$ is used instead of $$AgN{O_3}.$$ The reason is
A
a thin layer of $$Ag$$ is formed on $$Cu$$
B
more voltage is required
C
$$A{g^ + }$$ $$ions$$ are completely removed from solution
D
less availability of $$A{g^ + }$$ $$ions,$$ as $$Cu$$ cannot displace $$Ag$$ from $${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^ - }$$ $$ion$$
Answer :
less availability of $$A{g^ + }$$ $$ions,$$ as $$Cu$$ cannot displace $$Ag$$ from $${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^ - }$$ $$ion$$
In the silver plating of copper, $$K\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$ is used instead of $$AgN{O_3}.$$ The reason is less availability of$$A{g^ + }$$ $$ions,$$ as $$Cu$$ cannot displace $$Ag$$ from $${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^ - }$$ $$ion.$$
167.
$$CuS{O_4}$$ reacts with $$KCN$$ solution and forms
A
$$Cu\left( {CN} \right)$$
B
$$Cu{\left( {CN} \right)_2}$$
C
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
D
$${K_4}\left[ {Cu{{\left( {CN} \right)}_6}} \right]$$
168.
The chemical processes in the production of steel from haematite ore involve
A
reduction
B
oxidation
C
reduction followed by oxidation
D
oxidation followed by reduction
Answer :
reduction followed by oxidation
Haematite ore $$\left( {F{e_2}{O_3}} \right)$$ is first reduced to cast iron which is then oxidised for removing carbon (impurity) as $$C{O_2}.$$
169.
The lanthanide contraction is responsible for the fact that
A
$$Zr\,{\text{and}}\,Zn$$ have the same oxidation state
B
$$Zr\,{\text{and}}\,Hf$$ have about the same radius
C
$$Zr\,{\text{and}}\,Nb$$ have similar oxidation state
D
$$Zr\,{\text{and}}\,Y$$ have about the same radius
Answer :
$$Zr\,{\text{and}}\,Hf$$ have about the same radius
NOTE: In vertical columns of transition elements, there is an increase in size from first member to second member as expected but from second member to third member, there is very small chang in size and some times sizes
are same. This is due to lanthanide contraction this is the reason for $$Zr\,{\text{and}}\,Hf$$ to have same radius.
170.
The $$Ce (Z = 58)$$ belongs to IIIrd group of periodic table. If it furnish one $$\alpha $$ particle to form an element $$'X',$$ then $$X$$ belongs to
A
IIIrd group
B
IInd group
C
Ist group
D
zero group
Answer :
IInd group
$$_{58}Ce$$ is lanthanoid. Lanthanoid are from 57 to 71 all present in IIIrd group. Hence $$_{58}Ce$$ forms $$_{56}X$$ element on emission of one $$\alpha $$ particle with belongs to IInd group.