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Published on: 05/09/2019
The d- and f- Block Elements
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Questions + Answers key
Take MCQ Chemistry Test

1.
Explain why mercury (I) ion exists as \({ Hg }_{ 2 }^{ 2+ }\) ion while copper (I) ion exists as Cu+ ion.
2.
Give reasons for the following : Variations in the radii of transition elements are not as pronounced as those of representative elements.
3.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
4.
In what way is the electronic configuration of the transition elements different from that of the non transition elements?
5.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
6.
Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.
7.
Name a transition element which does not exhibit variable oxidation states.
8.
Compare the chemistry of the actinoids with that of lanthanoids with reference to the following :
(i) Electronic configuration
(ii) Oxidation states
(iii) Chemical reactivity
9.
Describe the preparation of potassium permanganate from pyrolusite ore. Write the ionic equation for the reaction that takes place between acidified KMnO4 solution and iron (II) ions
10.
Why do the transition elements exhibit higher enthalpies of atomisation?
11.
(a) Comlete the following chemical equations for reactions in aqueous media:
(i) Cr2O72-+H++Fe2+ \(\longrightarrow \)
(ii) MnO4-+I-+H+\(\longrightarrow \)
(b) How many unpaired electrons are present in Mn2+ ion (At. no. of Mn =25)? How does it influence magnetic behaviour of Mn2+ ions?
12.
Account for the following :
(a) Transition metals show variable oxidation states.
(b) Zn, Cd and Hg are soft metals
(c) Eo value for the Mn3+ /Mn2+ couple is highly positive (+ 1.57 V) as compared to Cr3+ /Cr2+.
(ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.
13.
(a) Write the electronic configuration of Ce3+ ion, and calculate the magnetic moment on the basis of 'spin-only' formula. [Atomic No. of Ce = 58]
(b) Account for the following
(i) The enthalpies of atomisation of the transition metals are high.
(ii) The lowest oxide of a transition metal is basic, the highest is amphoteric/ acidic.
(iii) Cobalt (II) is stable in aqueous solution but in the presence of complexing agents, it is easily oxidised.
14.
KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is
\(\frac { 2 }{ 5 } \)
\(\frac { 3 }{ 5 } \)
\(\frac { 4 }{ 5 } \)
\(\frac { 1 }{ 5 } \)
15.
Which is the strongest base among the following ?
La(OH)3
Lu(OH)3
Ce(OH)3
Yb(OH)3
16.
In the reaction : \(NaCI+{ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+conc.{ H }_{ 2 }{ SO }_{ 4 }\underrightarrow { heat } X+{ Na }_{ 2 }{ SO }_{ 4 }+{ K }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 },X\) is a raddish brown gas which gives a yellow solution on passing through water and a yellow precipitate on treating the solution with lead acetate solution X could be
CI2
CrO3
H2CrO4
CrO2CI2
17.
Which of the following ion is colourless in aqueous solution ?
Fe2+
Mn2+
Ti3+
Sc3+
18.
In first transition series which of the following has lowest enthalpy of atomisation ?
Sc
Cu
Yi
Zn
19.
General electronic configuration of d - block elements is
(n - 1) d1-10ns1-2
ns2np1-6
(n - 2) f0-14(n - 1)1-2ns2
(n-1) d1-5ns1-2
20.
Transition metals in which vacant spaces are occupied by small atoms such as hydrogen, carbon etc. are called _____________________.
1.
The electronic configuration of Hg(I), i.e., Hg+ is [Xe] 4f145d10 6s1 and thus has one electron in the valence 6s-orbital. If this were so, all Hg (I) compounds should be paramagnetic but actually they are diamagnetic. This behaviour can be explained if we assume that the singly filled 6s-orbitals of the two Hg+ ions overlap to form a Hg-Hg covalent bond. Thus, Hg+ions exist as dimeric species, i.e., Hg22+. In contrast, the electronic configuration of Cu(l) ion, i.e., Cu+ is [Ar] 3d10.Therefore, it has no unpaired b electrons to form dimeric species, i.e., Cu22+ and hence it always exists as Cu+ ion.
2.
As we proceed along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the d-subshell increases the screening effect which tends to counterbalance the effect of the increased nuclear charge
3.
Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
4.
In transition elements penultimate d-orbitals are progressively filled whereas in non-transition elements outermost s or p-orbitals are progressively filled.
5.
Mn exhibits largest number of oxidation states because it has 7 electrons in 's' as well as 'd' orbitals which can take part in bond formation.
6.
It is due to increase in ionisation enthalpy, tendency to lose electron decreases, therefore, reactivity decreases.
7.
Scandium (Z = 21) does not exhibit variable oxidation states.
8.
(i) Electronic configuration
Lanthanoids = [Xe] 4f0-14 5d0-1 6s2
Actinoids = [Rn] 5f0-14 6d0-1 7s2
(ii) Oxidation states In lanthanoids, +3 oxidation state is most common along with + 2 and + 4. While in actinoids, there is a greater range of oxidation states because 5f, 6d and 7s levels are of comparable energies.They show + 2, + 3, + 4, + 5, + 6 and + 7 oxidation states. Common oxidation state in actinoids is + 3.
(iii) Chemical reactivity Lanthanoids are less reactive than actinoids. Actually, earlier members of lanthanoids are quite reactive similar to calcium but with increasing atomic number, they behave more like aluminium. Lanthanoids react with dilute acids to liberate H2 gas while actinoids react with boiling water and gives a mixture of oxide and hydride.
9.
Reaction between acidified KMnO4 and iron (II) ions KMnO4 oxidises ferrous salts to ferric salt.
5Fe2+ + Mn\(O_4^-\) + 8H+ ⟶ Mn2+ + 4H2O + 5Fe3+
10.
Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.
11.
(a)\((i) Cr_2O_7^{2-} + 14 H^+ + 6Fe^{2+}\longrightarrow\ 2Cr^{3+} + 6Fe^{3+} + 7H_2O\)
\((ii) 2 MnO_4^- + 10 I^- + 16 H^+\longrightarrow 2Mn^{2+} + 8H_2O + 5I_2\)
(b) Mn2+: 3d54s0 has S unpaired electrons.It is highly paramagnetic and it is attracted by magnet.
12.
(i) (a) Due to the comparatively smaller size of the metal ions, their high ionic charges and the availability of vacant d-orbitals for bond formation, transition metals form a large number of complex compounds.
(b) As oxidation number (or oxidation state) of an element increases ionic character decreases. In general, the oxides in lower oxidation states of metals are basic and in their higher oxidation state, the oxides are amphoteric.
In lower oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. In higher oxidation state, valence electrons are involved in bonding and hence, electrons are not available for donation. Instead, their effective nuclear charge is high and hence they behave as acids.
(c) Mn3+(3d4) is less stable than Mn2+(3d5) because Mn2+ has stable half-filled configuration. Cr3+ has stable 3d3(t32g) configuration, therefore, Cr3+ cannot be reduced to Cr2+. That's why, EO value for the Mn3+ / Mn2+ couple is much more positive than Cr3+ /Cr2+. In other words, Mn3+ is a strong oxidising agent.
(ii) Similarity Both lanthanoids and actinoids exhibit +3 oxidation state predominantly.Difference Lanthanoids have less tendency towards complex formation while actinoids have greater tendency towards complex formation.
13.
\((a) Ce(58)-[Xe]{ 4f }^{ 1 }{ 5d }^{ 1 }{ 6s }^{ 2 }\)
\({ Ce }^{ 3+ }(58)-[Xe]{ 4f }^{ 1 }\)
Spin only formula:
Magnetic moment
\(=\sqrt { 4S(S+1) } =\sqrt { 4\times \frac { 1 }{ 2 } (\frac { 1 }{ 2 } +1) }\)
\( \\ =\sqrt { 2\times \frac { 3 }{ 2 } } =\sqrt { 3 } =1.732B.M.\)
(i) It is because of strong metallic bonds due to a large number of unpaired electrons in d-orbitals.
(ii) It is because of transition metal i.e. 1 lowest oxidation state are more metallic and in higher oxidation state are least metallic, therefore oxides in lower oxidation state are basic whereas in higher oxidation state are amphoteric acidic.
(iii) Strong oxidizing agents provide energy for loss of one more electron from CO2+
14.
(a)
\(\frac { 2 }{ 5 } \)
15.
(a)
La(OH)3
16.
(d)
CrO2CI2
17.
(d)
Sc3+
18.
(d)
Zn
19.
(a)
(n - 1) d1-10ns1-2
20.
( )
interstitial compounds
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