Download 1. 1. Write down the electronic configuration of: 2. Why are Mn 2+

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1. 1. Write down the electronic configuration of:
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(i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+
(ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+
•
Solution:
(a) Cr3+
(b) Cu+
(c) Co2+
(d) Mn2+
(e) Pm2+
(f) Ce4+
(g) Lu2+
(h) Th4+
1S2 2S2 2P6 3S23P6 3d3
1S2 2S2 2P6 3S23P6 3d10
1S2 2S2 2P6 3S23P6 3d7
1S2 2S2 2P6 3S23P6 3d5
(Xe) 4F5
(Xe) 4Fo
(Xe) 4f14 5d1
(Rh) 5Fo
2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to
their +3 state?
•
Solution:
In Mn2+ compounds the ion possess stable half electronic configuration of 3d5 whereas in
Fe2+ compounds the ion has configuration 3d6. If it loses one electron it can attain the configuration
3d5 which is very stable. Therefore Fe2+ ions tend to be oxidised easily to Fe3+ to attain half filled
electronic configuration.
3. Explain briefly how +2 state becomes more and more stable in the
first half of the first row transition elements with increasing atomic
number.
•
Solution:
in
With Scandium, having 3d1 configuration, d-orbitals are not stabilised. That is why it exhibits stable
+3 o.s, loosing both two 4S electrons and one 3d electron. After addition of each electron in 3d, the
3d - orbitals are more and more stabilised and hence they exhibits more stable +2 o.s due to loss of
two 4s electrons only. Also the sum of IE1 and IE2 decides the stability of the +2 state. Thus upto
Chromium +2 o.s. is less stable and from Chromium onwards +2 o.s. is more stable. Thus Mn2+ is
very stable because of stability of d – orbitals (half filled orbitals).
4. To what extent do the electronic configurations decide the stability of
oxidation states in the first series of the transition elements? Illustrate
your answer with examples.
•
Solution:
Zero, half filled and completely filled electronic configuration makes the oxidation state of the first
series of transition elements more stable. For e.g. in Scandium the +3 (O) state is very stable as the
1
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Sc3+ ion possesses 3d0 configuration. Similarly Manganese exhibits +2 and +7 oxidation states in
most of its compounds as the Mn2+ and Mn+7 ions possesses the electronic configuration of 3d5 and
3dO respectively. For Iron the +3 oxidation state is more stable as the electronic configuration is 3d5.
For Zinc, Zn2+ ion is very stable as it possess 3d10 structure.
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5. What may be the stable oxidation state of the transition element with
the following d electron configurations in the ground state of their atoms :
3d3,3d5, 3d8 and 3d4?
•
Solution:
Ground State
Electronic configuration
3d3
3d5
3d8
3d4
The stable (o) state
+5
+2
+2
does not exist
6. Name the oxo metal anions of the first series of the transition metals
in which the metal exhibits the oxidation state equal to its group number.
•
Solution:
Dichromate ion, Cr2O72-
Permanganate ion, MnO
Chromate ion, CrO42Vanadium dioxide ion, VO2+
Scandium oxide ion
Chromic oxide, CrO3
Vanadium tetroxide ion, VO43-
State of the metal
+6
+7
+6
+5
+3
+6
+5
7. What is lanthanoid contraction? What are the consequences of
lanthanoid contraction?
•
in
Solution:
The overall decrease in atomic and Ionic radii from Lanthanum to Luticium is a unique feature in the
chemistry of Lantanides. This is due to the imperfect shielding of one electron by another in the
same sub-shell. However the shielding of one 4f electron by another is less than one d electron by
another and with the increase in nuclear charge along the series, there is fairly regular decrease in
the sizes with increasing atomic numbers.
Consequences: The atomic radii of Zr (160pm) and Hf (159 p.m) are almost similar. In general the
atomic radii of pre Lanthanide elements and post Lanthanide elements are almost the same.
Lanthanides occur together due to their similarities in Atomic radii and the difficulty is faced in their
separation.
8. What are the characteristics of the transition elements and why are
they called transition elements? Which of the d-block elements may not be
regarded as the transition elements?
2
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•
Solution:
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The transition elements are the elements whose ‘d’ atomic orbitals are incomplete and possess the
general electronic configuration (n-1) d1-10 ns1-2. As a result of progressive filling of the electrons in
the ‘d’ orbital they show properties slightly different from the normal elements. As there is transition
in chemical properties on moving from left to right of the periodic table they are known as Transition
Elements. They possess certain general characteristic properties (i) variation in atomic and ionic
sizes (ii) variable oxidation states (iii) Magnetic properties (iv) complex formation (v) Alloy formation
(vi) catalytic properties (vii) Formation Interstitial compounds .
Zn, Cd, Hg possess completely filled ‘d’ orbitals in their atoms or important ions and do not satisfy
the general electronic configuration. They may be considered as ‘non’ Transition Elements, but they
are regarded as transition elements because of their common properties with others.
9. In what way is the electronic configuration of the transition elements
different from that of the non transition elements?
•
Solution:
In non transition elements ,the incoming electron enters the nth electronic level or outermost level.
But in transition elements, the last electron enters the penultimate or (n-1) level.
10. What are the different oxidation states exhibited by the lanthanoids?
•
Solution:
In Lanthanides La3+ and Lu3+ compounds are predominant species. However occasionally +2 and + 4
ions in solution or in solid compounds are also obtained. This irregularity arises mainly from the extra
stability of empty, half filled or filled sub shell. Thus the formation of Ce4+ is favoured by its noble
gas configuration. Pr, Nd, Tb and Dy also exhibit +4 state in the oxides or the type MO2. Eu2+ is
formed by losing two ‘s’ electrons and its f7 configuration accounts for the formation of this ion.
Eu2+ is a strong reducing agent changing to common +3 state.
11. Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iv) Transition metals and their many compounds act as good catalyst.
•
in
(iii) The transition metals generally form coloured compounds.
Solution:
(i) Para magnetism arises from the presence of un paired electrons, the magnetic moment is
determined by the number of unpaired electrons and is calculated by using the formula, taking spin
into consideration.
µ=
3
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The magnetic moment increases with increasing number of unpaired electrons. Thus the observed
magnetic moment gives a useful indication about the number of unpaired electrons present in the
atom or molecule or ion.
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(ii) Enthalpies of Atomisation: Strong metallic bonds between the atoms of the Transition elements
are responsible for the high melting and boiling points. This is clearly seen in their high enthalpies of
atomisation. The strength of the metallic bond depends upon the number of unpaired 'd' electrons
(half filled d orbitals). Greater is the number of unpaird electrons stronger is the metallic bonding.
Hence they possess high enthalpies of atomisation.
(iii) Formation of coloured compounds: Most of the transition metal compounds (ionic as well as
covalent) are coloured both in the solid state and in aqueous solution in contrast to compounds of S
and P block elements. This property is due to presence of unpaired electron(s) in the d-orbital and its
transition within the d-orbtials. The energy of excitation corresponds to the frequency of light
absorbed. This frequency generally belongs to the visible region, as less energy is involved in the
transition of electron. The colour observed corresponds to the complementary colour of the light
absorbed. The frequency of the light absorbed is determined by the nature of the legand.
(iv) The transition metals and their compounds are known for their catalytic activity. This activity is
ascribed to their ability to adopt multiple oxidation states and form complexes. Vanadium (v) oxide,
finely divided Iron oxide and nickel are some examples of catalysts. The solid surface involve the
formation of bonds between reactant molecule s and atoms of the surface of the catalyst. This has
the effect of increasing the concentration of the reactants at the catalyst surface and also weakening
of the bonds in the reacting molecules. Also because the transition metal ions can change their
oxidation states, they become more effective as catalysts.
12. What are interstitial compounds? Why are such compounds well
known for transition metals?
•
Solution:
Interstitial compounds are those which are formed when small atoms like H N or C are trapped inside
the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor
covalent. Many of the transition metals form interstitial compounds, particularly with small non-metal
atoms such as hydrogen, boron, carbon and nitrogen. These small atoms enter into the void sites
between the packed atoms of the crystalline metal e.g TiC, Mn4N, TiH2, etc. The formulae
corresponds to non-stoichiometric compounds. The characteristics of these compounds are:
(i) They have high melting points, higher than those of pure metals.
(ii) They are very hard as compared to diamond
(iv) They are chemically inert.
in
(iii) They retain metallic conductivity.
13. How is the variability in oxidation states of transition metals
different from that of the non transition metals? Illustrate with examples.
•
Solution:
In transition metals there exists less energy gap between (n-1) d and ns atomic orbitals. Therefore,
electrons are lost from ns atomic orbital as well as and (n-1)d orbitals in order to attain empty d
orbitals, half filled ‘d’ orbitals and completely filled ‘d’ orbital configuration which are extra stable
4
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where as in non-transition metals the electrons are lost only from the same energy level. Transition
metals show variable oxidation states, involving ns electrons and (n-1) d electrons.(eg) Mn exhibits
+2 and +7 states.
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Non transition elements also show more than one valency, using ns and np electrons only. They
show o.s. either equal to the valance electrons or (8-valance electrons). They exhibit variable
valencies with a difference of one. (Eg) Iodine exhibits valencies ---------------1,+1,+3,+5,+7.Also in
p block as you go down, lower oxidation states become more stable due to inert pair effect.
14. Describe the preparation of potassium dichromate from iron
chromite ore. What is the effect of increasing pH on a solution of
potassium dichromate?
•
Solution:
Potassium dichromate is prepared by fusing the chromite ore Fe Cr2 O4 with sodium or potassium
carbonate in free excess of air.
4FeCr2O4+8Na2CO3+7O2 →€€8Na2CrO4+2Fe2O3+8CO2
The fused mass is dissolved in water. The yellow solution of sodium Chromate is filtered and acidified
with sulphuric acid when sodium dichromate Na2 Cr2 O7 2H2O is formed.
2Na2CrO4 + 2H+ →€Na2 Cr2O7+2Na+ + H2O
Sodium dichromate is more soluble than potassium dichromate. The latter is prepared by treating the
solution Sodium dichromate with potassium chloride.
Na2 CrO7 + 2Kcl K2 →€Cr2O7 + 2 Nacl
Orange crystals of potassium dichromate crystallise out on fractional crystallisation.
Potassium dichromate solution changes to potassium chromate by increasing the PH of the solution.
Cr2O
+ 2OH- →€2CrO
+ H 2O
On decreasing the PH of the chromate solution it charges dichromate solution.
+ 2H+ →€CrO
+ H 2O
in
2CrO
15. Describe the oxidising action of potassium dichromate and write the
ionic equations for its reaction with:(a) iodide (b) iron(II) solution and (c)
H 2S
•
Solution:
In acidic solution the oxidising action of potassium dichromate can be represented as follows
5
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+ 14H+ + 6e- → 2Cr3+ + 7H2O
Cr2O
(a) It oxidises I- to I2
+ 14H+ + 6I- → 2Cr3+ + 7H2O + 3I2
Cr2O
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(b) It oxidises Fe2+ to Fe3+ ion
+ 14H+ + 6Fe2+ → 2Cr3+ +7H2O + 6Fe3+
Cr2O
2H+ + S2-
(c) H2S
Cr2O
+ 14H+ + 6e- →€2Cr3+ + 7H2O
3[S2- →€S + 2e-]
____________________________________
Cr2O
+ 14H+ + 3S2- →€2Cr3+ + 7H2O + 3S
16. Describe the preparation of potassium permanganate. How does the
acidified permanganate solution react with (a) iron(II) ions (b) SO2and (c)
oxalic acid? Write the ionic equations for the reactions.
•
Solution:
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an
oxidising agent like KNO3. This produces dark green K2MnO4 which disproportionates in neutral or
acidic solution to give permanganate.
2MnO2 + 4KOH + O2
3 MnO
2K2 MnO4 + 2H2O
+ 4H+
2MnO
+ MnO2 + 2H2O
Acidic solution MnO
+ 8H+ + 5e-
(a) MnO + 8H+ + 5Fe2+
(b) 4MnO
+ 12H+ + 10SO2
Mn2+ +4H2O E° = 1.55V
in
Potassium permanganate is a strong oxidising agent in acidic medium. The relevant half reactions
are:
5Fe3+ + Mn2+ + 4H2O
4Mn2+ + 6H2O + 10SO3
6
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(c) MnO
+ 8H+ +5
Mn2+ + 4H2O + 10CO2
17. For M2+/M and M3+/M2+ systems the EV values for some metals are
as follows:
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Cr2+/Cr -0.9V Cr3+/Cr2+ -0.4 V
Mn2+/Mn -1.2V Mn3+/Mn2+ +1.5 V
Fe2+/Fe -0.4V Fe3+/Fe2+ +0.8 V
Use this data to comment upon:
(a) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+and
(b) the ease with which iron can be oxidised as compared to a similar process for either
chromium or manganese metal.
•
Solution:
(a) Comparing E° values, they are in the increasing order
Cr3+/Cr2+ < Fe3+/Fe2+<Mn3+/Mn2+.
Hence the tendency to get reduced increases. It means Cr3+ is more stable than Fe3+ which is more
stable than Mn3+.
(b) The oxidation electrode potentials of the metals are for Fe (+0.4V) ,Cr(+0.9V) and for.
Mn(+1.2V).The order in which the metal gets oxidized is Mn>Cr>Fe.
18. Predict which of the following will be coloured in aqueous solution?
Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
•
Solution:
The transition metal ion exhibits colour in aqueous solution due to the presence of an unpaired
in
electron in the ‘d’ orbital. Ti3+, V3+, Mn2+, Fe3+, Co2+ and MnO
unpaired electrons.
ions exhibit colour as they possess
19. Compare the stability of +2 oxidation state for the elements of the
first transition series.
•
Solution:
Mn2+ and Zn2+ ions are more stable as contain half filled and fulfilled d- electrons. The rest of the
M2+ ions are less stable, with variation. Fe2+ ion is less stable and easily oxidised to Fe3+ which
contains half filled d-orbital electrons.
7
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20. Compare the chemistry of actinoids with that of the lanthanoids with
special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity.
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•
Solution:
Lanthanides
Electronic Configuration
In Lanthanides 4f level gets Progressively filled
Oxidation States
In Lanthanides the +3(o.s) State is more stable
Occasionally they also exhibit +2 and +4 o.s
Actinides
In actinides 5f level gets progressively filled
In Actinides the 5f level get progressively filled. In
actinides +4 oxidation state is very common. The
other Oxidation states which are commonly
exhibited are +5, +6 and +7 respectively.
Atomic and Ionic Sizes
There is over all decrease in Atomic and ionic radii The atomic and ionic sizes for actinides also
from Lanthanium to Lutitium.
decreases progressively.
Chemical Reactivity
They are radioactive elements. Actinides are
All lanthanides show similar chemical properties. highly reactive metals, especially when finely
They tarnish readily on exposure to air. They form divided. With boiling water they give a mixture of
Ln2O3 except Ce which forms CeO2. All form
oxide and hydroxide they combine with nonLn(OH)3 which are stronger bases than Al(OH)3. metals at moderate temperatures. They do not
They combine with non metals like O2, S and
react with alkalies. They are slightly affected by
hologens. They act as reducing agents. They have Nitric acid owing to the formation of Protective
less tendency to from complexes.
oxide layer. They have greater tendency to form
complexes.
21. How would you account for the following:
(a) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.
(b) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is
easily oxidised.
(c) The d1 configuration is very unstable in ions.
Solution:
The E0 value of Cr3+/Cr2+ is negative while that of Mn3+/Mn2+ is positive.So Cr
oxidized to Cr3+ while Mn has greater tendency to remain in Mn2+state.
2+
in
•
can easily get
22. What is meant by ‘disproportionation’? Give two examples of
disproportionation reaction in aqueous solution.
•
Solution:
When a particular oxidation state becomes less stable relative to other oxidation states, one lower,
8
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one higher, it is said to undergo disproportionation, for e.g. Manganese (vi) becomes unstable
relative to Mn (vii) and manganese (iv) in acidic solution
3MnVI
+ 4H+ = 2MnVIIO
+ MnIVO2+2H2O
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23. Which metal in the first series of transition metals exhibits +1
oxidation state most frequently and why?
•
Solution:
Cu exhibits +1 oxidation most frequently in first series of transition metals. This is due to ‘d’ orbitals
consisting of 10 electrons giving rise to a stable configuration.
24. Calculate the number of unpaired electrons in the following gaseous
ions: Mn3+,Cr3+, V3+ and Ti3+. Which one of these is the most stable in
aqueous solution?
•
Solution:
Ions
Mn3+
Cr3+
V3+
Ti3+
•
Number of unpaired electrons
4
3
2
1
Of these Cr3+ is most stable as it has half filled t2g set of d orbitals.
25. Give examples and suggest reasons for the following features of the
transition metal chemistry:
(a) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(b) A transition metal exhibits highest oxidation state in oxides and fluorides.
(c) The highest oxidation state is exhibited in oxoanions of a metal.
Solution:
(a) The lowest oxide of transition metal is basic due to its large size of central metal atom. It readily
furnish electrons and behave as Lewis base whereas in higher oxidation state the central metal atom
is smaller in size and possess more polarisability and readily accept the electrons. Thus the acidic
nature is observed in these oxides.
in
•
(b) The oxygen and fluorine are the most electronegative atoms hence when they combine with
transition metals they readily oxidise the transition metal to a higher oxidation state.
(c) The oxo anions of transition metals are generally formed from the reaction of metal with oxygen
at high temperatures. As oxygen is more electronegative the transition metal atom loses the
electrons to exist in higher oxidation states.
9
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26. Indicate the steps in the preparation of:(a) K2Cr2O7 from chromite
ore. (b) KMnO4 from pyrolusite ore.
•
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Solution:
(a) The chromite ore is fused with Sodium Carbonate in excess of air or in the presence of oxidising
agent.
4Fe.Cr2O4 + 8 Na2CO3
8Na2CrO4 + 2Fe2O3 + 8CO2
It is filtered. The filterate is yellow solution of Sodium Chromate. The yellow solution is now acidified
with sulphuric acid.
2Na2CrO4 + 2H
Na2Cr2O7+2Na+ + H2O
The Sodium dichromate is converted to K2Cr2O7 by addition of potassium chloride
Na2Cr2O7 + 2KCl
K2Cr2O7 + 2 NaCl
The K2Cr2O7 is seperated from NaCl by fractional crystallisation.
(b) Potassium permanganate is obtained by the fusion of MnO2 with alkali metal hydroxide and
an oxidizing agent like KNO3.
2MnO2 + 4KOH+O2
2K2 MnO4+2H2O
Potassium manganate is treated with Chlorine when KmnO4 is formed.
2K2MnO4 + Cl2
2KCl + 2KMnO4
27. What are alloys? Name an important alloy which contains some of
the lanthanoid metals. Mention its uses.
•
in
Solution:
An alloy is a blend of metals prepared by mixing components. Alloys are homogeneous solid
solutions in which atoms of one metal are randomly distributed among the atoms of the other.
Lanthanides are used for the production of alloy steels for plates and pipes. A well known alloy is
Misch metal which consists of a lanthanides metal on 95% and iron 5% and tracer of S, C, Ca and Al.
A good deal of Misch metal is used in Mg based alloy to produce bullets, shell and lighter flint. Mixed
oxides of Lanthanides are used as catalysts in petroleum cracking.
28. What are inner transition elements? Decide which of the following
atomic numbers are the atomic numbers of the inner transition elements :
29, 59,74, 95, 102, 104.
•
Solution:
Inner Transition elements are the elements in which f-orbitals are filled. They possess the general
electronic configuration (n-1)d0-1. The neutral atom of ion should possess the following electronic
configuration.
10
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Atomic No.
e- Configuration
29
1S2, 2S2, 2P6, 3S2, 3P6, 4S1, 3d10
59
1S2 2S2 2P6 3S2 3P6 4S2 3d10 4P65S2
4d10 Sp6 6S2 4 f3
74
[Xe] 4f14 Sd5 6S1
[Rn] 5f7 7S2
102
[Rn] 5 f14 7S2
103
[Rn] 5f14 6 d2 7S2
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95
Except element with atomic number 29,all are inner transitional elements.
29. 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.
•
Solution:
The lanthanoid elements show only +3,+2 and +4 oxidation states. But actinoids exhibit +6 and+7
also in addition to this.
Most of the Actinides elements are Radioactive in nature which makes the study of these elements
rather difficult. When compared with lanthanides there is much variation in chemical properties in the
first half of actinide series. 5f electrons are more effectively shielded from the nuclear charge than all
the 4f electrons of the corresponding Lanthanides. The oxidation states of Actinides increases from
+4 to +5 to +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements. The ion
in +3 and +4 oxidation states however tend to hydrolyse to form oxo ions due to the uneven
distribution of oxidation states.
30. Which is the last element in the series of the actinoids? Write the
electronic configuration of this element. Comment on the possible
oxidation state of this element.
•
Solution:
Last element in
the actinoid
series
Lawrencium (Lr)
Outer electronic
configuration
103
5f14 6d1 7s2
Outer electronic
configuration of
M3+ ion
5f14
Outer electronic
configuration of
M4+ ion
5f13
in
•
Atomic
number
The actinoids show in general +3 oxidation state. The first few elements of the actinoid series exhibit
higher oxidation states. But, Lawrencium exhibit +3 oxidation state more commonly than other
oxidation states because by doing so it gets a completely filled configuration.
31. Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and
calculate its magnetic moment on the basis of ‘spin-only’ formula.
•
Solution:
The electronic configuration of Ce3+ ion [Xe] 4f'
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Magnetic moment, µ =
(where n is number of unpaired electrons)
µ=
µ = 1.7BM.
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32. Name the members of the lanthanoid series which exhibit +4
oxidation states and those which exhibit +2 oxidation states. Try to
correlate this type of behaviour with the electronic configurations of these
elements.
•
Solution:
Cerium exhibits Ce4+ o.s as it as f0 configuration and stable noble gas structure. Pr,Nd,Tb and Dy also
exhibits o.s. IV in their oxides. Eu2+ is due to f7. Yb2+ is given f14 configuration. The existence of ions
in +2 and +4 o.s. in solution or in solid the state is due to extra stability of empty, half filled and full
filled f-orbitals.
33. Write the electronic configurations of the elements with the atomic
numbers 61, 91, 101, and 109.
•
Solution:
Atomic No.
61
91
101
109
Electronic Configuration
[Xe] 4f5 6s2
[Rn] 5f2 6d1 7s2
[Rn] 5f13 7S2
[Rn] 5f14 7S2 6d7
34. Compare the general characteristics of the first series of the
transition metals with those of the second and third series metals in the
respective vertical columns. Give special emphasis on the following
points:(i) electronic configurations (ii) oxidation states (iii) ionisation
enthalpies and (iv) atomic sizes.
•
Solution:
(i) The elements of first transition series involve progressive filling of 3d orbitals whereas that of
second and third series involve filling of 4d and 5d subshells respectively.
in
(ii) For the elements if first transition series +2 and +3 oxidation states are common and these
elements form many stable complexes in these oxidation states. For the elements of second and
third series higher oxidation states are more important. These elements form more stable
compounds in higher oxidation states. For example, [CrCl6]3- is very stable whereas no equivalent
complexes of Mo and W are known. On the contrary OsO4 and PtF6 are quite stable and no
corresponding compounds for first transition series are known.
(iii) The metals of second and third transition series have higher Ionisation enthalpies than the
elements of first series. The elements of second and third series form many compounds with M-M
bond.
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(iv) The atomic radii of elements of second and third transition series are larger than those of the
elements of first series. Because of lanthanoid contraction, the radii of third series are almost equal
to those of second row.
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35. Write down the number of 3d electrons in each of the following ions:
Ti2+, V2+,Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you
expect the five 3d orbitals to be occupied for these hydrated ions
(octahedral).
•
Solution:
(i) Ti2+: [Ar] 3d2
Under the influence of an octahedral legand field the d-orbitals split into two groups of different
energies. These are t2g orbitals (dxy, dxz and dyz) and eg orbitals (dz2 and dx2-y2),t2g orbitals are of
lower energy than eg orbitals. Thus, the two d- electrons in Ti2+ would be present in two of the three
t2g orbitals. Thus there would be two unpaired electrons.
(ii) V2+: [Ar] 3d3
The three d-electrons in V2+ would be present in three different t2g orbitals. Thus, there would be
three unpaired electrons.
(iii) Cr3+ : [Ar] 3d3 - (t2g)3 --- Three unpaired electrons
(iv) Mn2+ : [Ar] 3d5 - (t2g)3(eg)2 --- Five unpaired electrons
(v) Fe2+ : [Ar] 3d6 - (t2g)4(eg)2 --- Four unpaired electrons
(vi) Fe3+ : [Ar] 3d5 - (t2g)3(eg)2 --- Five unpaired electrons
(vii) Co2+: [Ar] 3d7 - (t2g)5(eg)2 --- Three unpaired electrons
(viii) Ni2+ : [Ar] 3d8 - (t2g)6(eg)2 --- Two unpaired electrons
(ix) Cu2+ : [Ar] 3d9 - (t2g)6(eg)3 --- One unpaired electron
•
in
36. Comment on the statement that elements of the first transition
series possess many properties different from those of heavier transition
elements.
Solution:
Some points of difference between properties of elements of the first transition series and those of
heavier transition elements are given below:
(i) For elements of first series +2 and +3 oxidation states are more common while for the heavier
transition elements higher oxidation states are more common.
(ii) M-M bonding is rare in the elements of first series but is quite common in heavier transition
elements.
1
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(iii) The elements of first transition series do not form complexes with coordination no. 7 or 8
whereas heavier transition elements do so.
(iv) The elements of first transition series form low spin or high spin complexes depending upon
strength of the ligand field. On the other hand heavier transition elements form low spin complexes
irrespective of the strength of the ligand field.
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37. What can be inferred from the magnetic moment values of the
following complex species ?
Example Magnetic Moment (BM)
(i)K4[Mn(CN)6] 2.2
(ii)[Fe(H2O)6]2+ 5.3
(iii)K2[MnCl4] 5.9
•
Solution:
Calculate the magnetic moment by applying formula
BM
The magnetic moment of 2.2BM corresponds to n=1. Thus, in
K4[Mn(CN)6] there is only one unpaired electron in 3d subshell.
Thus, the distribution of five 3d electrons in Mn(II) is (t2g)5.
This indicates that there are four unpaired electrons in the complex. Thus, the six 3d electrons in
Fe(II) are distributed as (t2g)4(eg)2 .
This indicates that there are five unpaired electrons in the complex. Hence, the five 3d-electrons in
Mn(II) are distributed as (t2g)3(eg)2 in the given complex.
38. Silver atom has completely filled d orbitals (4d10) in its ground
state.How can you say that it is a transition element?
Solution:
Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d),
hence a transition element.
in
•
39. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation
of zinc is the lowest, i.e., 126 kJ mol–1. Why?
•
Solution:
In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in
all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of
metallic bonds.
14
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40. The E(-) (M2+/M) value for copper is positive (+0.34V). What is
possibly the reason for this? (Hint: consider its high ∆aH(-) and low ∆hydH(-))
•
Solution:
The positive value of E(-) (M2+/M) for copper is due to its high enthalphy of atomization and high
values of enthalpies of ionization (∆iHI + ∆iHII).
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41. How would you account for the irregular variation of ionisation
enthalpies (first and second) in the first series of the transition elements?
•
Solution:
Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of
different 3d-configurations (e.g., d0, d5, d10 are exceptionally stable).
42. Why is the highest oxidation state of a metal exhibited in its oxide or
fluoride only?
•
Solution:
Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its
highest oxidation state.
43. Which is a stronger reducing agent Cr2+ or Fe2+ and why ?
•
Solution:
Cr2+ is stronger reducing agent than Fe2+
Reason: d4
But d6
d3 occurs in case of Cr2+ to Cr3+
d5 occurs in case of Fe2+ to Fe3+
In a medium (like water) d3 is more stable as compared to d5.
44. Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).
•
Solution:
The electronic configuration of the M2+(aq) ion (Z = 27) would be [Ar] 3d7 It would contain three
=
BM = 3.87 BM
in
unpaired electrons. The ‘spin only’ magnetic moment is given by the relation,
BM
45. Explain why Cu+ ion is not stable in aqueous solutions?
•
Solution:
Cu+ in aqueous solution undergoes disproportionation, i.e.,
2Cu+(aq)
Cu2+(aq) + Cu(s)
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The E0 value for this is favourable.
46. Actinoid contraction is greater from element to element than
lanthanoid contraction. Why?
Solution:
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•
This is due to poor shielding effect of 5f electrons in actinoids compared to poor shielding effect by 4f
electrons in lanthanoids.
in
16
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