D and F Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
Learn D and F Block Elements MCQ questions & answers in Inorganic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
101.
Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified $$KMn{O_4}$$ for complete oxidation?
A
$$FeS{O_4}$$
B
$$FeS{O_3}$$
C
$$Fe{C_2}{O_4}$$
D
$$Fe{\left( {N{O_2}} \right)_2}$$
Answer :
$$FeS{O_4}$$
$$FeS{O_4}$$ will require the least amount of acidified $$KMn{O_4}$$ for complete oxidation.
102.
Amount of oxalic acid present in a solution can be determined by its titration with $$KMn{O_4}$$ solution in the presence of $${H_2}S{O_4}.$$ The titration gives unsatisfactory result when carried out in the presence of $$HCl,$$ because $$HCl$$
A
gets oxidised by oxalic acid to chlorine
B
furnishes $${H^ + }$$ ions in addition to those from oxalic acd
C
reduces permanganate to $$M{n^{2 + }}$$
D
Oxidises oxalic acid to carbon doxide and water
Answer :
reduces permanganate to $$M{n^{2 + }}$$
The titration of oxalic acid with $$KMn{O_4}$$ in presence of
$$HCl$$ gives unsatisfactory result because of the fact that $$KMn{O_4}$$ can also oxidise $$HCl$$ along with oxalic acid.
$$HCl$$ on oxidation gives $$C{l_2}$$ and $$HCl$$ reduces $$KMn{O_4}$$ to $$M{n^{2 + }}$$ thus the correct answer is (C).
103.
Which of the following is correct representation of reaction of acidified permanganate solution with sulphurous acid?
No explanation is given for this question. Let's discuss the answer together.
105.
Although Zirconium belongs to $$4d$$ transition series and Hafnium to $$5d$$ transition series even then they show similar physical and chemical properties because
A
both belong to $$d$$ - block
B
both have same number of electrons
C
both have similar atomic radius
D
both belong to the same group of the periodic table
Answer :
both have similar atomic radius
Due to lanthanoid contraction $$Zr$$ and $$Hf$$ have nearly equal size.
106.
A reduction in atomic size with increase in atomic number is a characteristic of elements of
A
$$d$$ - block
B
$$f$$ - block
C
radioactive series
D
high atomic masses
Answer :
$$f$$ - block
$$f$$ - block elements show a regular decrease in atomic size due to lanthanide/actinide contraction.
107.
Match the catalysts (Column I) with products (Column II).
A
A - iii; B - iv; C - i; D - ii
B
A - ii; b - iii; C - i; D - iv
C
A - iii; B - i; C - ii; D - iv
D
A - iv; B - iii; C - ii; D - i
Answer :
A - iii; B - i; C - ii; D - iv
$$\left( A \right)\,\,{V_2}{O_5} \to $$ Preparation of $${H_2}S{O_4}$$ in contact process
$$\left( B \right)\,\,TiC{l_4} + Al{\left( {Me} \right)_3} \to $$ Polyethylene ( Ziegler-Natta catalyst )
$$\left( C \right)\,\,PdC{l_2} \to $$ Ethanal ( Wacker’s process )
$$\left( D \right)\,\,$$ Iron oxide $$ \to N{H_3}$$ in ( Haber’s process )
108.
Which pair of compounds is expected to show similar colour in aqueous medium?
A
$$FeC{l_2}\,{\text{and}}\,CuC{l_2}$$
B
$$VOC{l_2}\,{\text{and}}\,CuC{l_2}$$
C
$$VOC{l_2}\,{\text{and}}\,FeC{l_2}$$
D
$$FeC{I_2}\,{\text{and}}\,MnC{l_2}$$
Answer :
$$VOC{l_2}\,{\text{and}}\,CuC{l_2}$$
Colour of transition metal ion salt is due to $$d{\text{ - }}d$$ transition of unpaired elctrons of $$d$$ - orbital. Metal ion salt having similar number of unpaired electrons in $$d$$ - orbitals
shows similar colour in aqueous medium.
$${{\text{V}}^{4 + }}:\left[ {Ar} \right]3{d^1}$$
$$C{u^{2 + }}:\left[ {Ar} \right]3{d^9}$$
Number of unpaired electrons = $$1$$
109.
The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows
highest magnetic moment.
A
$$3{d^7}$$
B
$$3{d^5}$$
C
$$3{d^8}$$
D
$$3{d^2}$$
Answer :
$$3{d^5}$$
No explanation is given for this question. Let's discuss the answer together.
110.
The correct order of $$E_{\frac{{{M^{2 + }}}}{M}}^ \circ $$ values with negative sign for the four successive elements $$Cr, Mn, Fe$$ and $$Co$$ is
A
$$Fe > Mn > Cr > Co$$
B
$$Cr > Mn > Fe > Co$$
C
$$Mn > Cr > Fe > Co$$
D
$$Cr> Fe > Mn > Co$$
Answer :
$$Mn > Cr > Fe > Co$$
$${E^ \circ }$$ values for $${\frac{{{M^{2 + }}}}{M}}$$ with negative signs are $$Cr = - 0.90\,V,Mn = - 1.18\,V,$$ $$Fe = - 0.44\,V,Co = - 0.28\,V$$
Thus, the order is $$Mn > Cr > Fe > Co.$$