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The d-And-f-Block Elements

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Question
CBSEENCH12011206
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Acidified K2Cr2O7 solution turns green when Na2SO3 is added to it. This is due to the formation of 

  • CrO42-

  • Cr2(SO3)3

  • CrSO4

  • Cr2(SO4)3

Solution

D.

Cr2(SO4)3

K2Cr2O7 +3Na2SO3 + 4H2SO4 → 3Na2SO4 +K2SO4 + Cr(SO4)3 +4H2O
Chromium(III) sulfate gives green colour.

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Question
CBSEENCH12011221
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Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation?

  • FeSO4

  • FeSO3

  • FeC2O4

  • Fe(NO2)2

Solution

A.

FeSO4

FeSO4 will require the least amount of acidified KMnO4 for complete oxidation.

Question
CBSEENCH12011061
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Because of lanthanoid contraction, of elements which of the following elements have nearly same atomic radii ? (Number in the parenthesis are atomic number).

  • Ti (22) and Zr(40)

  • Zr(40) and Nb (41)

  • Zr (40) and Hf (72)

  • Zr (40) and Ta(73)

Solution

C.

Zr (40) and Hf (72)

Because of the lanthanoid contraction Zr (atomic radii 160 pm) and Hf (atomic radii 158 pm) have nearly same atomic radii.
Lanthanoids include the elements from lanthanum La (Z=57) to lutetium Lu(Z =71).Zirconium  Zr (40) belong to the second transition series (4d) and Hf (72) belongs to third transition series (5d). Lanthanoid contraction is associated with the intervention of the 4f orbitals which are filled before the 5d- series of elements starts. The filling of 4f if orbitals before 5d orbitals result in a regular decrease in atomic radii which compensates the expected increases in atomic size with increasing atomic number. As a result of this lanthanoid contraction, the elements of second and third transition series have almost similar atomic radii.

Question
CBSEENCH12011195
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For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?

(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)

  • Fe > Mn > Co> Cr 

  • Co > Mn > Fe > Cr

  • Cr > Mn > Co > Fe

  • Mn > Fe > Cr > Co

Solution

D.

Mn > Fe > Cr > Co

This can be understood on the basis of Eo values for M2+/M.

Eo/V Cr Mn Fe Co
M2+/M -0.90 -1.18 -0.44 -0.28

Eo value of Mn is more negative than expected from general trend due to the extra stability of half- filled Mn2+ ion. Thus, the correct order should be
Mn > Cr > Fe > Co
An examination of Eo values for redox couple M3/M2+ shows that Cr2+ is a strong reducing agent and left parenthesis straight E subscript straight M to the power of 3 plus end exponent divided by straight M to the power of 2 plus end exponent end subscript superscript straight o space equals space 0.41 space straight V right parenthesis liberates H2 from dilute acids.
2Cr2+ (aq) + 2H+ (aq)  --> 2 Cr3+ (aq) + H2 (g)
Therefore, the correct order is Mn > Fe > Cr > Co.

Question
CBSEENCH12011177
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Four successive members of the first series of the transition metals are listed below. For which one of them, the standard potential value left parenthesis straight E subscript straight M to the power of 2 plus end exponent divided by straight M end subscript superscript straight o right parenthesis has a positive sign?

  • Co (Z=27)

  • Ni (Z=28)

  • Cu (Z=29)

  • Fe (Z=26)

Solution

C.

Cu (Z=29)

In electrochemical series metals with positive standard potential are placed below hydrogen. Out of the given transition metals, only Cu is placed below hydrogen in the electrochemical series. Therefore, it must have a positive sign for standard potential.

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