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NCERT Exemplar · Class 12 · Chemistry

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

Practice each question, then open the answer or worked solution.

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Multiple Choice Questions (Type-II)

10 questions
Q22Written answer

Generally, transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

(A) $KMn{O_4} $

(B)  $Ce{\left( {S{O_4}} \right)_2} $

(C) $TiC{l_4} $

(D) $C{u_2}C{l_2}$ 

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Worked solution

Ans: (A) & (B)

Due to charge transfer phenomenon, which can be understood, for an instant, an electron is transferred from oxygen to the central metal atom, where oxidation state of metal is reduced by one and oxygen gets converted from \[{{\text{O}}^{\text{ - }}}\]to \[{{\text{O}}^{{\text{2 - }}}}\]. 

This transfer imparts Colour to the compound. 


Q23Written answer

Transition elements show magnetic moments due to spin and orbital motion of electrons. Which of the following metallic ions have almost the same spin only magnetic moment?

(A) $C{o^{2 + }} $

(B) $C{r^{2 + }} $

(C) $M{n^{2 + }} $

(D) $C{r^{3 + }}$ 

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Worked solution

Ans: (A) & (D) 

The magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of the magnetic moment.

\[{\text{C}}{{\text{o}}^{{\text{2 + }}}}{\text{ }}\]and \[{\text{C}}{{\text{r}}^{{\text{3 + }}}}\]have same unpaired electrons i.e. 3 unpaired electrons. Hence they have the same spin only magnetic moment. 


Q24Written answer

In the form of dichromate, Cr(VI) is a strong oxidizing agent in acidic medium

but \[Mo\] (VI) in \[Mo{O_3}\] and \[W\] (VI) in \[W{O_3}\] are not because

(A) \[Cr\] (VI) is more stable than \[Mo\] (VI) and \[W\] (VI)

(B) \[Mo\] (VI) and \[{\text{W}}\] (VI) are more stable than \[Cr\] (VI)

(C) higher oxidation states of heavier members of group-6 of transition series are more stable

(D) lower oxidation states of heavier members of group-6 of transition series are more stable.

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Ans: (B) and (C) 

For the heavier metals of d-block, higher oxidation state is more favourably stable than the lower oxidation state. For example: \[{\text{Mo}}\] (VI) and \[{\text{W}}\] (VI) are more stable than \[{\text{Cr}}\] (VI). For lighter d-block elements like chromium, +3 oxidation state is more stable. So, \[{\text{Cr}}\] (VI) in dichromate act as a strong oxidising agent by reducing itself from +6 to +3 oxidation states While \[{\text{Mo}}\] (VI) and \[{\text{W}}\] (VI) does not. 


Q25Written answer

Which of the following actinoids show oxidation states up to +7?

(A) $Am $

(B) $Pu $

(C) $U $

(D) $Np$ 

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Ans:  (B) & (D) 

Actinides show a variety of oxidation states like +3, +5, +6 and +7. 

From the given option, plutonium and neptunium show all oxidation states up to +7. 


Q26Written answer

General electronic configuration of actinides is \[\left( {n -- 2} \right){f^{1 - 14}}\left( {n -- 1 } \right){d^{0 - 2}}n{s^2}\]. Which of the following actinoids have one electron in 6d orbital?

(A) U (Atomic no. 92) 

(B) Np (Atomic no. 93) 

(C) Pu (Atomic no. 94)

(D)  Am (Atomic no. 95) 

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Ans: (A) and (B)

Electronic configuration:

${}_{{\text{92}}}{\text{U -  5}}{{\text{f}}^{\text{3}}}{\text{ 6}}{{\text{d}}^{\text{1}}}{\text{ 7}}{{\text{s}}^{\text{2}}}$

${}_{{\text{93}}}{\text{Np -  5}}{{\text{f}}^4}{\text{ 6}}{{\text{d}}^{\text{1}}}{\text{ 7}}{{\text{s}}^{\text{2}}}$ 


Q27Written answer

Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanides?

(A) $Ce $

(B) $Eu $

(C) $Yb $

(D) $Ho$ 

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Ans: (B) and (C) 

${\text{Ce (Z = 57)  =  }}\left[ {{\text{Xe}}} \right]{\text{ 4}}{{\text{f}}^{\text{5}}}{\text{ 5}}{{\text{d}}^{\text{0}}}{\text{ 6}}{{\text{s}}^{\text{2}}}{\text{ , O}}{\text{.S}}{\text{.  =   + 3, + 4}}$

${\text{Eu (Z = 63)  =  }}\left[ {{\text{Xe}}} \right]{\text{ 4}}{{\text{f}}^{\text{7}}}{\text{ 5}}{{\text{d}}^{\text{0}}}{\text{ 6}}{{\text{s}}^{\text{2}}}{\text{, O}}{\text{.S}}{\text{.  =   + 2, + 3}}$

${\text{Yb (Z = 70)  =  }}\left[ {{\text{Xe}}} \right]{\text{ 4}}{{\text{f}}^{{\text{14}}}}{\text{ 5}}{{\text{d}}^{\text{0}}}{\text{ 6}}{{\text{s}}^{\text{2}}}{\text{, O}}{\text{.S}}{\text{.  =   + 2, + 3}}$

${\text{Ho (Z = 67)  =  }}\left[ {{\text{Xe}}} \right]{\text{ 4}}{{\text{f}}^{{\text{11}}}}{\text{ 5}}{{\text{d}}^{\text{0}}}{\text{ 6}}{{\text{s}}^{\text{2}}}{\text{, O}}{\text{.S}}{\text{.  =   + 3}}$ 


Q28Written answer

Which of the following ions show higher spin only magnetic moment value?

(A) $T{i^{3 + }} $

(B) $M{n^{2 + }} $

(C)  $F{e^{2 + }} $

(D) $C{o^{3 + }}$ 

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Ans: (B) and (C) 

Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.

Electronic configuration:

${\text{T}}{{\text{i}}^{{\text{3 + }}}}{\text{    =  3}}{{\text{d}}^{\text{1}}}{\text{   =  1 unpaired electron}}$

${\text{M}}{{\text{n}}^{{\text{2 + }}}}{\text{  =  3}}{{\text{d}}^{\text{5}}}{\text{  =  5 unpaired electrons}}$

${\text{F}}{{\text{e}}^{{\text{2 + }}}}{\text{   =  3}}{{\text{d}}^{\text{6}}}{\text{  =  4 unpaired electrons}}$

${\text{C}}{{\text{o}}^{{\text{2 + }}}}{\text{  =  3}}{{\text{d}}^{\text{7}}}{\text{  =  3 unpaired electrons}}$ 

Out of all given options, \[{\text{M}}{{\text{n}}^{{\text{2 + }}}}\]and \[{\text{F}}{{\text{e}}^{{\text{2 + }}}}\] have unpaired electrons in number of five and four respectively. Hence, have higher spin only magnetic moment. 


Q29Written answer

Transition elements form binary compounds with halogens. Which of the following elements will form \[M{F_3}\] type compounds?

(A) $Cr $

(B) $Co $

(C) $Cu $

(D) $Ni$ 

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Ans: (A) and (B) 

\[{\text{Cr}}\] and \[{\text{Co}}\]  form tri halide compounds.


Q30Written answer

Which of the following will not act as oxidizing agents?

(A) $Cr{O_3}$

(B) $Mo{O_3}$

(C) $ W{O_3} $

(D) $Cr{O_4}^{2 - }$ 

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Ans: (B) and (C) 

A compound act as oxidising agent when it’s central atom is reduced to its lower oxidation state. This occurs only when the lower oxidation state of metal is more stable than the higher oxidation states. 

In metal \[{\text{W}}{{\text{O}}_{\text{3}}}\] and \[{\text{Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}\], both \[{\text{W}}\] and \[{\text{Cr}}\] are stable in their higher oxidation state and won’t get reduced to their lower oxidation state. Therefore, will not act as oxidising agents. 


Q31Written answer

Although +3 is the characteristic oxidation state for lanthanide but cerium

also shows +4 oxidation state because 

(A) it has variable ionization enthalpy

(B) it has a tendency to attain noble gas configuration

(C) it has a tendency to attain \[{{\text{f}}^{{^\circ }}}\] configuration

(D) it resembles \[P{b^{4 + }}\]

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Ans: (B) and (C) 

Electronic configuration of cerium:

${\text{Ce  -  4}}{{\text{f}}^{\text{1}}}{\text{5}}{{\text{d}}^{\text{1}}}{\text{6}}{{\text{s}}^{\text{2}}}$

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

Lead has the tendency to lose its four outer electrons to attain Noble gas configuration and this removal of electrons corresponds to the \[{{\text{f}}^{{^\circ }}}\] configuration and cause it to be stable in +4 oxidation state.