12 chemistry impq_ch08_d_and_f_block_elements_01

76 XII – Chemistry
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d – AND f – BLOCK ELEMENTS
Electronic Configuration of Transition Metal/Ions
The d-block element is called transition metal if it has partly filled
d-orbitals in the ground state as well as in its oxidised state.
The general electronic configuration of transition metal is (n–1) d1–10ns1–2.
Exceptions in electronic configuration are due to (a) very little engery difference
between (n–1) d and ns orbitals and (b) extra stability of half filled and completely
filled orbitals in case of Cr and Cu in 3d series.
Cr : Is2 2s2 2p6, 3s2 3p6 4s1 3d5
Cu : Is2 2s2 2p6, 3s2 3p6 4s1 3d10
To write the electronic configuration of Mn+, the electrons are first removed
from ns orbital and then from (n - 1) d orbitals of neutral, atom (if required).
For example, the electronic configuration of Cu+, Cu2+ and Cr3+ are respectively
3d10 4s°, 3d9 4s° and 3d3 4s°.
The following questions can be answered with the help of above.
(i) Scandium (Z = 21) is a transition element but zinc (Z = 30) is not.
(ii) Copper (Z = 29) and silver (Z = 47) both have fully filled d-orbitals
i.e., (n - 1) d10. why are these elements are regarded as transition
elements?
(iii) Which of the d-block elements are not regarded as transition
elements?
UNDERSTANDING fus H vap H AND a H
In transition metals unpaired (n - l)d electrons as well as ns electrons take
part in interatomic bonding. Larger the number of unpaired (n - 1) d electrons, the
stronger is the interatomic bonding and large amount of energy is required to
overcome the interatomic interaction.
M(s) H
fus
θ∆
→ M(1)
M(1)
H
vap
θ∆
→ M(vapour)
M(s)
H
a
θ∆
→ M(g)
These enthalpies are related as
fus H < Dvap H < a H
UNIT-8
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The following questions can be explained using the above reasoning.
(i) Which has higher m.p.? V (Z = 23) or Cr (Z = 24) ?
(ii) Explain why Fe (Z = 26) has higher m.p. than cobalt (Z = 27).
Metals of second (4d) and third (5d) transition series have greater
enthalpies of atomisation than corresponding elements of first transition series
on account of more frequent metal metal bonding due to greater spatial
extension of 4d and 5d orbitals than 3d orbitals.
LANTHANOID CONTRACTION AND ITS CONSEQUENCE
The 4f orbitals screen the nuclear .charge less effectively because they
are large and diffused. The filling of 4f orbitals before 5d orbitals results in the
gradual increase in effective nuclear charge resulting in a regular decrease in
atomic and ionic radii. This phenomenon is called lanthanoid contraction. The
corresponding members of second and third transition series have similar radii
because the normal size increase down the group of d-block elements almost
exactly balanced by the lanthanoid contraction.
This reasoning is applied in answering the following questions.
(i) Elements in the’following pairs have identical (similar) radii : Zr/Hf, Nb/Ta
and Mo/W. Explain why?
(ii) Why do Zr and Hf have very similar physical and chemical properties and
occur together in the same mineral?
VARIATION IN IONISATION ENTHALPY
With the filling of (n - 1) d orbitals effective nuclear charge increases
resulting in the increase in first ionisation enthalpy. There are some irregular
variations.
The first ionisation enthalpy of chromium is lower because the removal
of one electron produces extra stable d5 configuration and that of zinc is higher
because the removal of electron takes place from fully filled 4s orbital.
Second ionization enthalpy of Zn (i H2 = 1734 kj)/mol) is lower than
second ionization enthalpy of Cu (1958 kj/mol). This is because removal of
second electron in Zn produces stable d10 configuration while the removal of
second electron in Cu disrupts the d10 configuration with a considerable loss
in exchange energy to from less stable d9 configuration.
Cr = 3d5 4s1
Zn = 3d10 4s2; Zn2+ = 3d10
Cu = 3d10 4s1; Cu2+ = 3d9

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Now the following questions can be accounted for :
(i) Why is second ionization enthalphy of Cr (Z = 24) more than that
Mn (Z = 25) (Hint. Cr+ (d5  d4), Mn
+
(3d5 4s1  3d5).
(ii) Which has more second ionisation enthalpy?
Cu (Z = 29) or Zn (Z = 30) (Hint. Cu+ (d10  d9), Zn+ (3d10 4s1
 3d10).
[iii) Second ionization enthalpy of Mn (Z = 25) is less than that of
Fe (Z=26) but third ionisation enthalpy of Mn is more than that of Fe.
Why?
Hint : Mn+ (3d5 4s1  3d5) Fe+ (3d6 4s1  3d6)
Mn2+ (3d5  3d4) Fe2+ (3d6  3d5)
Relationship between E
red and stability of various oxidation states.
Transition metals show variable oxidation states due to incompletely filled
d-orbitals. These variable oxidation states differ from each other by unity, e.g.,
Mn (II), Mn (III), Mn (IV), Mn (V), Mn (VI) and M (VII). Scandium is the only
transition element which exclusively shows the oxidation state of +3.
Standard Electrode potential E
M
2+/M
can be calculated from the follwoing
parameters :
The reducing property of a transition metal will be higher if rH has a
large negative value which is possible if hyd H more than compensates
(a H + a H1 + i H1 + i H2).

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More negative the r H, the more positive will be standard oxidation
potential and hence, more negative will be standrd reduction potential. 2+Cu
Cu
E∅
positive because (i H1 + i H2) i.e., energy required to produce
Cu2+ is not balanced by hyd H of Cu2+.
Since the sum of i H1 and i H2 generally increases with the increase
in the atomic number of the transition metal, therefore 2+
M M
E∅
value becomes
less and less negative.
2+
M M
E∅
values for Mn, Zn and Ni are more negative than expected trend. This is
because i H2 for Mn and Zn produces stable d5 configuration (Mn2+) and d10
configuration (Zn2+) are produced and therefore requirement of energy is less.
whereas 2+Ni
Ni
E∅
is more negative due to highest negative hyd H which
is – 2121 kj/mol for Ni2+.
Example : Why is E value for Mn3+ / Mn2+ couple much more
positive than for Cr3+ / Cr2+ or Fe3+ / Fe2+.
Solution : Mn2+ (d5) → Mn3+ (d4)
has much larger third ionisation energy due to disruption of extrastability
of half filled d5 configuration.
Cr2+ (d4) → Cr3+ (d3)
Cr3+ has half-filled t2g level. Hence Cr2+ is oxidised easily to stable Cr3+
ion. Hence E value is compartively less.
Fe2+ (d6) → Fe3+ (d5)
Comparatively low value of E is also due to extra stability of d5
configuration of Fe3+.
Example : Which is stronger reducing agent Cr2+ or Fe2+ and why?
Solution : Cr2+ (d4) → Cr3+ (d3 or half-filled t
3
2g
In water medium [Cr (H2O)6]3+ has more CFSE than [Fe (H2O)6]3+. Hence
Cr2+ in a stronger reducing agent.

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Stability of Higher Oxidation States :
Higher oxidation states are shown by transition metals in fluorides, oxides,
oxocations and oxoanions. The ability of fluorine to stablise the highest oxidation
state is due to either higher lattice energy as in case of CoF3 or higher bond
enthalpy terms for higher covalent compounds, e.g., VF5 and CrF6.
Transition metals show highest oxidation state in oxides and oxocations
and oxoanions, e.g., VO4
+ and VO4
–. The ability of oxygen to stabilise these
high oxidation states exceeds that of fluorine due to its ability to form multiple
bonds to metals.
The following questions can be explained using the above concepts.
(i) All Cu (II) halides are known except the iodides. Why?
[Hint : 2Cu2+ + 4I– → Cu2I2 + I2]
(ii) Why do Cu (I) compounds undergo disproportionation in water?
[Hint : hyd H of Cu2+ more than compensates the 1 H2 of copper]
(iii) Highest fluoride of Mn is MnF4 but the highest oxide is Mn2O7.
(iv) E values of 3d series varies irregularly.
(v) Why is Cr2+ is reducing and Mn3+ oxidising when both have d4
configuration?
[Hint : Cr2+ (d4) → Cr2+ (t3
2g; half filled t2g level)
Mn3+ (d4) + e– → Mn2+ (d5; half filled d-level)
(vi) Why is highest oxidation state shown in oxocations and oxoanions?
Properties of Transition Elements
Transition matals with partly filled d-orbitals exhibit certain characterstic
properties. For example they display a variety of oxidtion states, form coloured
ions and enter into complexe formation. Transition metals and their compounds
exhibit catalytic properties and are generally paramagnetic in nature.

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Crystal Field Theory :
Calculation of CFSE : Each electron occupying t2g orbital results in the
lowering of energy by – 0.40 0 and each electron occupying the eg orbital
increases the energy by + 0.60 0. If x is the no. of electrons occupying t2g
orbitals and ‘y’ is the no. of electrons occupying the eg orbitals, then CFSE
is given by
CFSE = (–0.40 0 x + 0.60 0y)
= (–0.40 x + 0.60 y) 0
Formation of Coloured Ions : Degeneracy of d-orbitals is lifted in presence
of the field of ligands approaching the central metal ion. For example, in the
octahedral crystal field of ligands, the d-orbitals are split into two set of d-
orbitals (i) t2g orbitals of lower energy : these are dxy dyz, dxz and (ii) eg orbitals
of higher energy i.e., dx2–y2 and dz2.
When visible light is incident on the octahedral transition metal complex, an
electron is excited from t2g level to eg level. During this d-d transition, a
characteristic wave length of visible light is absorbed and therefore transmitted
light appears coloured. The colour of complex is complementry to the colour
absorbed by the transition metal complex.

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No d-d transition occurs if d-orbitals are empty or fully filled and therefore, such
ions may be colourless.
Exceptions : AgBr, Agl, have fully filled d-orbitals but are coloured due to
transference of electron cloud from Br– or I– to Ag+ (d10) when white light is
incident on AgBr / Agl. During this process also characteristic wave length of
visible light is absorbed. Similarly MnO4
– (purple), CrO4
2– (yellow) and Cr2O7
2–
(orange) are coloured due to charge transfer from oxide ions to the central
metal ions although they have no d-electrons.
Comparison of oxidising powers of KMnO4 and K2Cr2O7
MnO2
– + 2H2O + 3e– → Mn2 + 4OH– E = + 1.69 V
MnO4
– + 8H+ + 5e– → Mn2+ + 4H2O E = + 1.52 V
Cr2O7
2– + 14H+ + 6e– → 2Cr3+ + 7H2O E = + 1.33 V
Electrode potential values shows that acidified KMnO4 is a stronger
oxidising agent than acidified K2Cr2O7. But KMnO4 in faint alkaline medium is
a stronger oxidising agent than acidified KMnO4.
For example, KMnO4 oxidises KI to I2 in acidic medium but to KIO3 in
alkaline medium.
(a) MnO4
– + 8H+ + 5e– → Mn2+ + 4H2O]  2
2I– → I2 + 2e–]  2
Overall : 2Mn4
– + 10 I– + 16 H– → 2Mn2+ + 8H2O + 5I2
(b) MnO4
– + 2H2O + 3e– → MnO2 + 4OH–]  2
I– + 6 OH– → IO3
– + 3H2O + 6e–
Overall : 2Mn4
– + H2O + I– → 2MnO2 + IO3
– + 2OH–
Following questions involving the oxidising actions of KMnO4 and K2Cr2O7
may be answered
(i) How do acidified KMnO4 and acidified K2Cr2O7 reacts separately with (a)
SO2, (b) H2S (c) FeSO4?
(ii) Write the ionic equations of KMnO4 (acidified) with (a) oxalate ion (b) Mohr
salt (c) NO2
– and (d) Iron (II) oxalate.
[Hint : (d) Both Fe2+ and C2O4
2– are oxidised to Fe3+ and CO2 respectively.]

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Oxidation States of Lanthanoids and Actinoids
The most common and most stable oxidation state of lanthanoids is +3. They
also show oxidation states of +2, and +4 if the corresponding lanthanoid ions
have extra stable 4f0, 4f7 and 4f14 configuration.
Ce4+ + e– → Ce3+, + 3 more stable O.S. than + 4
Tb2+ + e– → Tb3+, + 3 more stable O.S. than + 4
Hence Ce4+ and Tb4+ are strong oxidants.
Eu2+ → Eu3+ + e–, +3 more stable O.S. than +2
Yb2+ → Yb3+ + e–, + 3 more stable O.S. than + 3
Hence, Eu2+ and Yb2+ are strong reducing agents.
Actinoids also show most common O.S. of + 3 but it is not always
most stable. Actinoids also show higher oxidation states, e.g., Th (+4),
Pa (+5), U (+ 6) and Np (+ 7).
Example : La (5d16s2), Gd (4f75d16s2) and Lu (4f14 5d16s2) have
abnormally low third ionisation enthalpies. Why?
Solution : La3+, Gd3+ and Lu3+ have stable configurations 4f0, 4f7 and 4f14
respectively.
Answer the following question –
(i) Name the two members of lanthanoid series which show + 2 oxidation
state
(ii) Name the lanthanoid element which shows +4 O.S.
Participation of 5f electrons of actinoids in bonding.
5f orbitals in actinoids are not as burried as 4f orbials in lanthanoids and hence
5f electrons can participate in bonding to a far greater extent.
There in a gradual decrease in the size of atoms or M3+ ions across the
actionoid series. Since 5f orbitals are larger and more diffuse than 4f orbitals,
their penetration towards the inner core of electrons is less than the penetration
of 4f elecrons. Hence 5f electrons screen the nuclear charge less effectively
than 4f electrons in lanthanoids. Consequently effective nuclear charge in
actinoids increases at faster rate as compared with lanthanoids. Hence actinoid
contraction from element to element is more than the lanthanoid contraction.
The following question can explained with the above reasoning :
Explain why Actinoid contraction from element to element is greater than
lanthanoid contraction.

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VERY SHORT ANSWER TYPE QUESTIONS
(1 - MARK QUESTIONS)
1. Write the electronic configuration of Cr3+ ion (atomic number of Cr = 24)?
3. Explain CuSO4. 5H2O is blue while ZnSO4 and CuSO4 are colourless?
4. Why is the third ionisation energy of Manganese (Z = 25) is unexpectedly
high?
[Hint : The third electron is to be removed from stable configuration Mn2+
(3d5). It requires higher energy.]
5. Which element among 3d– transition elements, exhibit the highest oxidation
state?
[Hint : Mn (+7)]
6. Silver (Ag) has completely filled d-orbitals (4d10) in its ground state. How
can you say that it is a transition element.
7. In 3d series (Sc → Zn), the enthalpy of atomisation of Zn is low. Why?
[Hint : Poor interatomic bonding in zinc.]
8. Out of the following elements, identify the element which does not exhibit
variable oxidation state?
Cr, Co, Zn.
9. The +3 oxidation state of lanthanum (Z = 57), gadolinium (Z = 64) and
lutetium (Z = 71) are especially stable. Why?
10. Mention one consequence of Lanthanoid Contraction?
11. The first ionization enthalpies of 5d– series elements is higher than those
of 3d and 4d series elements why?
[Hint : Increasing value of effective nuclear charge due to lanthanoid
contraction.]
12. Why Mn2+ compounds are more stable than Fe2+ compounds towards
oxidation to their +3 state?

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14. Calculate the magnetic moment of Cu2+ (Z = 29) on the basis of “spin-only”
formula.
[Hint : µ = n (n + 2) B.M.]
15. What is the shape of chromate ions?
[Hint : Tetrahedral]
16. Why does vanadium pentoxide act a catalyst?
[Hint : In V2O5, Vanadium shows variable oxidation sates.]
17. What are interstitial compounds?
18. The transition metals and their compounds are known for their catalytic
activity. Give two specific reasons to justify the statement.
19. Write the chemical equation for the reaction of thiosulphate ions and
alkaline potassium permanganate.
[Hint : 8MnO4
– + 3S2O3
2– + H2O → 8MnO2 + 2OH– + 6SO4
2–].
20. Mention the name and formula of the ore from which potassium dichromate
is prepared.
[Hint : FeCr2O4 (Chromite)].
21. Write the electronic configuration of Lu3+ (At. No. = 71).
22. What is the most common oxidation state of actinoids?
23. Write the names of the catalyst used in the :
(a) Manufacture of sulphuric acid by contact process.
(b) Manufacture of polythene.
24. Mention the name of the element among lanthanoids known to exhibit +4
oxidation state.
25. Name one ore each of manganese and chromium.
26. Why is Cd2+ ion white?
*27. Draw the structure of dichromate anion.
*28. Arrange the following monoxides of transition metals on the basis of
decreasing basic character TiO, VO, CrO, FeO.[Hint : TiO > VO > CrO >
FeO]

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SHORT ANSWER TYPE QUESTIONS
(2 - MARK QUESTIONS)
1. Write the chemical equation, when the yellow colour of aqueous solution
of Na2CrO4 changes to orange on passing CO2 gas?
2. The stability of Cu2+ (aq) is more than that of Cu+ (aq). Why?
3. Indicate the steps in the preparation of
(a) K2Cr2O7 from Chromite ore.
(b) KMnO4 from Pyrolusite ore.
4. Give reason for : –
(a) In permanganate ions, all bonds formed between manganese and
oxygen are covalent.
(b) Permanganate titrations in presence of hydrochloric acid are
unsatisfactory.
5. Write complete chemical equations for
(a) oxidation of Fe2+ by Cr2O7
2– in acidic medium
(b) oxidation of Mn2+ by MnO4
– in neutral or faintly alkaline medium.
6. (a) Why do transition metals show high melting points?
(b) Out of Fe and Cu, which one would exhibit higher melting point?
[Hint. (i) Strong interatomic bonding arising from the participation of ns
and unpaired (n – 1) d-electrons.
(ii) Fe has higher melting point due to presence of more unpaired electrons
3d-orbitals.
7. Describe giving reason which one of the following pairs has the property
indicated :
(a) Cr2+ or Fe2+ (stronger reducing agent).
(b) Co2+ or Ni2+ (lower magnetic moments).
8. Of the ions Co2+, Sc3+, Cr3+ which one will give colourless aqueous solution
and how will each of them respond to magnetic field and why?
[Hint : Co2+ (3d7); Cr3+ (3d4); Sc3+ (3d°)]
9. Complete the following equations :
(a) MnO2 + KOH + O2
→ (b) Na2Cr2O7 + KCl →

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10. Transition metals show low oxidation states with carbon monoxide.
[Hind : CO is a  acceptor ligand capable of forming a bond by accepting
 electrons from the filled d-orbitals of transition metal and CO also form
 bond by donating  electrons to transition metal orbital.
11. For the first row transition metals the enthalpy of atomisation value are :
Sc Ti V Cr Mn Fe Co Ni Cu Zn
aH/kJ mol–1 326 473 515 397 281 416 425 430 339 26
Assign reason for the following :
(a) Transition elements have higher values of enthalpies of atomisation.
(b) The enthalpy of atomisation of zinc is the lowest in 3d - series.
12. Account for the following :
(a) Copper shows its inability to liberate hydrogen gas from the dilute
acids.
(b) Scandium (Z = 21) does not exhibit variable oxidation states.
13. Copper (I) compounds undergo disproportionation. Write the chemical
equation for the reaction involved and give reason.
14. Iron (III) catalyses the reaction :
15. Complete the equations :
(a) MnO4
– + NO2
– + H+ →
(b) KMnO4
513 k
→
16. The following two reactions of MNO3 with Zn are given.
(a) Zn + conc. HNO3
→ Zn(NO3)2 + X + H2O
(b) Zn + dil.HNO3
→ Zn(NO3)2 + Y + H2O
Identify X and Y and write balanced equations.
[Hint : X is NO2 and Y is N2O].

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17. Titanium shows magnetic moment of 1.73 BM in its compound. What is the
oxidation number of Ti in the compound?
[Hint : O.N. of Ti = +3].
(a) Transition metals and majority of their compounds act as good
catalysts.
(b) From element to element, actionoid contraction is greater than
lanthanoid contraction
19. Calculate the number of electrons transferred in each case when KMnO4
acts as an oxidising agent to give
(i) MnO2 (ii) Mn2+ (iii) Mn(OH)3 (iv) MnO4
2– respectively.
[Hint : 3, 5, 4, 1].]
20. Calculate the number of moles of KMnO4 that is needed to react completely
with one mole of sulphite ion in acidic medium.
[Hint : 2/5 moles].
SHORT ANSWER TYPE QUESTIONS (3 - MARK QUESTIONS)
(a) La(OH)3 is more basic than Lu(OH)3
(b) Zn2+ salts are white.
(c) Cu(I) compounds are unstable in aqueous solution and undergo
disproportination.
2. Describe the oxidising action of potassium dichromate with following. Write
ionic equations for its reaction with.
(a) Iodide ion (b) Iron (II) (c) H2S.
3. (a) Deduce the number of 3d electrons in the following ions :
Fe3+, Cu2+ and Sc3+.
(b) Why do transition metals form alloys.
(c) Write any two characteristics of interstitial compounds.
*4. In the following reaction, Mn(VI) changes to Mn(VII) and Mn(IV) in acidic
solution.
3MnVIO4
2– + 4H+ → 2MnVIIO4
– + MnIVO2 + 2H2O

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(a) Explain why Mn(VI) changes to Mn(VII) and Mn(IV).
(b) What special name is given to such type of reactions?
5. What happens when
(a) thiosulphate ions react with alkaline KMnO4.
(b) ferrous oxalate reacts with acidified KMnO4.
(c) sulphurous acid reacts with acidified KMnO4
Write the chemical equations for the reactions involved.
7. Name the catalysts used in the
(a) manufacture of ammonia by Haber’s Process
(b) oxidation of ethyne to ethanol
(c) photographic industry.
*8. Among TiCl4, VCl3 and FeCl2 which one will be drawn more strongly into
a magnetic field and why?
[Hint : Among these halides the transition metal ion having maximum
number of unpaired electrons will be drawn strongly into the magnetic
field.
Ti4+ = 3d0 no. of unpaired e– = 0 µ = 0
V3+ = 3d2 no. of unpaired e– = 2 µ = 2.76 BM
Fe2+ = 3d6 no. of unpaired e– = 4 µ = 4.9 BM]
9. Complete the following equations
(a) Mn04
2– + H+ → ......... + ...........+ .............
(b) KMnO4
Heat
→
(c) H+ + MnO4
– + Fe2+ + C2O4
2– →
10. How do you account for the following?
(a) With the same d-orbital configuration (d4), Cr2+ is a reducing agent
while Mn3+ is an oxidiising agent.
(b) The actinoids exhibit a larger number of oxidation states than the
corresponding members in the lanthanoid series.
(c) Most of transition metal ions exhibit characteristic colours in aqueous
solutions.

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LONG ANSWER TYPE QUESTIONS (5 - MARK QUESTIONS)
1. A green compound ‘A’ on fusion with NaOH in presence of air forms yellow
compound ‘B’ which on acidification with dilute acid, gives orange solution
of compound ‘C’. The orange solution when reacted with equimolar
ammonuim salt gives compound ‘D’ which when heated liberates nitrogen
gas and compound ‘A’. Identify compounds A to D and write the chemical
equation of the reactions involved.
[Hint : ‘A’ = CrO3; ‘B’ = Na2CrO4; ‘C’ = Na2Cr2O7 ‘D’ = (NH4)2 Cr2O7
2. Assign reasons for the following :
(a) There is no regular trends in E° values of M2+/M systems in 3d
series.
(b) There is gradual decrease in the ionic radii of M2+ ion in 3d series.
(c) Majority of transition metals form complexes.
(d) Ce3+ can be easily oxidised to Ce4+
(e) Tantalum and palladium metals are used to electroplate coinage
metals.
(a) Actinoids display a variety of oxidation states.
(b) Yb2+ behaves as a good reductant.
(c) Cerium (iv) is a good analytical reagent.
(d) Transition metal fluorides are ionic in nature while chlorides and
bromides are covalent in nature.
(e) Hydrochloric acid attacks all the actinoids.
*4. Explain by giving suitable reason :
(a) Co(II) is stable in aqueous solution but in the presence of complexing
agent it is readily oxidised.
(b) Eu2+, Yb2+ are good reductants whereas Tb4+ is an oxidant.
(c) AgCl dissolves in ammonia solution
(d) Out of Cr2+ or Fe2+, which one is a stronger reducing agent?
(e) The highest oxidation state is exhibited in oxoanions of a transition
metal.

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5. When a white crystalline compound A is heated with K2Cr2O7 and conc.
H2SO4, a reddish brown gas B is evolved, which gives a yellow coloured
solution C when passed through NaOH. On adding CH3COOH and
(CH3COO)2 Pb to solution C, a yellow coloured ppt. D is obtained. Also on
heating A with NaOH and passing the evolved gas through K2HgI4 solution,
a reddish brown precipitate E is formed.
Identify A, B, C, D and E and write the chemical equations for the reactions
involved.
[Hint : (A) NH4Cl, (B) CrO2Cl2 (g), (C) Na2CrO4
*6. (a) Describe the preparation of potassium dichromate (K2Cr2O7). Write
the chemical equations of the reactions involved.
(b) “The chromates and dichromates are interconvertible by the change
in pH of medium.” Why? Give chemical equations in favour of your
answer.
7. Explain giving reasons :
(a) Transition metals are less reactive than the alkali metals and alkaline
earth metals.
(b) 2+Cu
Cu
E∅
has positive value
(c) Elements in the middle of transition series have higher melting points.
(d) The decrease in atomic size of transition elements in a series is very
small.
8. (a) Compare the chemistry of the actinoids with that of lanthanoids with
reference to—
(i) electronic configuration
(ii) oxidation states
(iii) chemical reactivity.

AK
(b) How would you account for the following :
(i) of the d4 species, Cr2+ is strongly reducing while Mn3+ is
strongly oxidising.
(ii) the lowest oxide of a transition metal is basic whereas highest
is amphoteric or acidic.
9. (a) What is meant by disproportionation of an oxidation state. Give one
example.
(b) Explain why europium (II) is more stable than Ce(II)?
[Hint : (a) When particular state becomes less stable relative to
other oxidation states, one lower and one higher, it is said to undergo
disproportionation, for example,
(b) Eu (II)= [Xe] 4f7 5d0 (4f subshell is half filled)
Ce (II)= [Xe] 4f1 5d0 (5d Subshell is empty and 4f subshell has only
one electron which can be easily lost.)]
10. (a) For M2+/M and M3+/M2+ systems, the E values for some metals are
as follows :
Cr2+/Cr = – 0.9V and Cr3+/Cr2+ = – 0.4V
Mn2+/Mn = – 1.2 V and Mn3+/Mn2+ = + 1.5V
Fe2+/Fe = – 0.4V and Fe3+/Fe2+ = + 0.8V
Use this data to comment upon :
(i) the stability of Fe3+ in acid solution as compared to that of
Cr3+ and Mn3+
(ii) the ease with which iron can be oxidised as compared to a
similar process for either chromium or manganese.
(b) How is the variability in oxidation states of transition metals different
from that of the non-transition metals? Illustrate with examples.
For more important question's visit :
www.4ono.com

12 chemistry impq_ch08_d_and_f_block_elements_01

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