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Published on: 28/09/2019
The d- and f- Block Elements
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Questions + Answers key
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1.
(a) How would you account for the following?
(i) Highest fluoride of Mn is MnF4 whereas the highest oxide is Mn2O7.
(ii) Transition metals and their compounds show catalytic properties.
(b) Complete the following equation:
3MnO42- + 4H+ ⟶
2.
Write the complete chemical equation for each the following:
(i) An alkaline solution of \(KMnO_{ 4 }\) reacts with an iodide.
(ii) An excess of \(SnCI_{ 2 }\) solution is added to a solution of mercury (II) chloride.
(iii) Potassium chromate is acidified with sulphuric acid.
3.
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
4.
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
5.
Which is stronger reducing agent \({ Cr }^{ 2+ }\) or \({ Fe }^{ 2+ }\) and why?
6.
In the series (Z = 21) to Zn (Z = 30), the enthalpy of zinc is the lowest 126 kJ mol-1 Why?
7.
A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What difference stages of the reducing do these represent and how are they carried out?
8.
Complete the following reactions:
(a) Cr2O72-+14H++6e-\(\longrightarrow \).......... +7H2O
(b) CrO42-+..........\(\rightleftharpoons \)........\(\rightleftharpoons \).......+H2O
(c) MnO4-+2H2O+3e- \(\overrightarrow { medium } \)........ +4OH-
9.
Identify the first row transition metal ions which have outer electronic configurations of 3d4 and 3d6 and describe their oxidation states.
10.
Give reasons for each of the following:
(i) Transition metal fluorides are ionic in nature, whereas bromides and chlorides are usually covalent in nature.
(ii) Size of trivalent lanthanoid cations decreases with increase in the atomic number.
(iii) Chemistry of all the lanthanoids is quite similar.
11.
Explain the following:
(i) The transition elements have great tendency for complex formation.
(ii) There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers.
(iii) Lanthanum and Lutetium do not show colouration in solutions.
(At. No : La = 57, Lu = 71)
12.
How would you account for the following?
(I) The Eo value for the Mn3+ /Mn2+ couple is much more positive than that for the Cr3+/Cr2+ couple or Fe 3+/Fe 2+ couple.
(ii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
(iii) The atomic radii ofthe metals ofthe third (Sd) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
13.
Compare the stability of +2 oxidation state for the elements of the first transition series.
14.
When a chromite ore(A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify (A) (B) (C) and write the reactions.
1.
(a) (i) Transition metals show variable oxidation states, therefore, they and their compounds act as catalyst.
(ii) Oxygen can form double bond, therefore, it can form Mn2O7 , whereas 'F' cannot form double bonds, so, it can form MnF4 .
(b) \(3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{MnO}_{2}+2 \mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O}\)
2.
(i) The black powder acts as a catalyst. The catalysts accelerate the speed of reactions without itself undergoing any permanent changes.
(ii) Black powder is manganese dioxide (MnO2).
(iii) Black powder cannot be used for all decomposition reactions because catalysts are highly specific in nature. A catalyst which can catalyze one reaction may have no effect on another reaction even, if that reaction is very similar.
(iv) Glycerol slows down the decomposition of H2O2 and it is called negative.catalyst.
3.
This is because of ability to form multiple bonds.
4.
The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., Fe+ and Fe3+, Cu+ and Cu2+ etc. In the case of non-transition elements, the oxidation states differ by units of two, e.g., Pb2+ and Pb4+, S2+ and S4+ etc. Moreover, in transition elements, the higher oxidation states are more stable for heavier elements in a group. For example, in group 6, Mo (VI) and W (VI) are more stable than Cr (VI). In P block elements, the lower oxidation states are more stable for heavier members due to inert pair effect, e.g., in group 16, Pb (II) is more stable than Pb (IV).
5.
\({ Cr }^{ 2+ }\) is a stronger reducing agent than \({ Fe }^{ 2+ }\). This is because the configuration of \({ Cr }^{ 2+ }\) changes from \({ d }^{ 4 }\) to \({ d }^{ 3 }\) configuration is stable \(({ t }_{ 2g }^{ 3 })\) being half-filled \({ t }_{ 2g }\) level.
6.
The high enthalpies of atomization of transition elements are due to the participation of electrons (n-1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are S involved in the formation of metallic bonds.
7.
Oxidising behavior of KMnO 4 depends on the pH of the solution.
In acidic medium, it gives colorless \({ Mn }^{ 2+ }\) ions
\({ MnO }_{ 4 }^{ - }+{ 8H }^{ + }+{ 5e }^{ - }\)\(\rightarrow \) \({ Mn }^{ 2+ }\)\(+\) \({ 4H }_{ 2 }{ O }\)
In alkaline medium, green colored \({ MnO }_{ 4 }^{ 2- }\) ions are formed
\({ MnO }_{ 4 }^{ 2- }\)\(+\)\({ e }^{ - }\rightarrow { MnO }_{ 4 }^{ 2- }\)
In neutral medium, brown colored \({ MnO }_{ 2 }\) are formed
\({ MnO }_{ 2 }^{ - }+{ 2H }_{ 2 }{ O }+{ 3e }^{ - }\rightarrow { MnO }_{ 2 }+{ 4OH }^{ - }\)
8.
(a) Cr2O72-+14H++6e-\(\longrightarrow \)2Cr3+ +7H2O
(b) CrO42-+2H+\(\rightleftharpoons \)2HCrO4-\(\rightleftharpoons \)Cr2O72-+H2O
(c) MnO4-+2H2O+3e- \(\overrightarrow { medium } \)MnO2 +4OH-
9.
Cr2+ has electronic configuration 3d4 Chromium also shows +3 and +6 oxidation states. Fe2+ has electronic configuration 3d6. Iron has oxidation states +2 and +3.
10.
(i) F is more electronegative than CI and Br, therefore, fluorides are ionic; whereas chlorides and bromides are covalent.
(ii) It is due to poor shielding effect off-electrons, effective nuclear charge increases, so, ionic size decreases.
(iii) It is due to similar ionic size which is due to lanthanoid contraction, they resemble in their properties.
11.
(i) It is due to the presence of vacant d-orbitals of suitable energy, the smaller size of cautions and higher charge.
(ii) It is due to increase in effective nuclear charge gradually because unpaired electrons increase in the beginning and then decreases. There is repulsion between paired electrons.
(iii) It is due to absence of unpaired electrons, they do not absorb light from visible region and do not radiate color.
12.
(i) It is because Mn2+ is more stable than Mn3+ due to stable half filled 3d5 configuration, whereas Cr3+(t2g 3) and Fe3 (3d5) are more stable than Cr2+ sg and Fe2+ respectively.
(ii) It is because oxygen and fluorine are strong oxidising agents, highly electronegative, small size and can provide energy for formation of transition metal ion in higher oxidation state.
(iii) It is due to lanthanoid contraction which is due to poor shielding effect off-electrons.
13.
Elements (+ 2 state ) 21Sc2+'22Ti2+ 23V2+ 24 Cr2+ and 25Mn2+have more stable +2 oxidation state and the outer electronic configuration are 3d1 , 3d2 , 3d3 , 3d4 and 3d5, respectively. In all the elements listed, the removal of two 4s electrons (in Cr2+, 1e- from 4s and 1e- from 3d), the 3d-orbitals get gradually occupied. Since, the number of empty d-orbitals decreases or the number of unpaired electrons in 3d-orbitals increases with increase in atomic number of cations, so the stability of the cations (M 2+) increases from Sc2+ to Mn.
14.
'A' is iron chromite (FeCr2 04 ), '8' is sodium chromate (Na2CrO4 ), 'C' is sodium dichromate (Na2CrO7) and 'D' is potassium dichromate (K2Cr2O7).
\(4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+7 \mathrm{O}_{2} \longrightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}\) + 2Fe2O3 + 8CO2 (g)
'A' 'B'
(Chromite ore) (Yellow solution)
\(2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \longrightarrow 2 \mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)
'C' 'C' 'D'
(Orange crystals)
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