L-8
d&f-BLOCK
ELEMENTS
BASIC CONCEPTS/IMPORTANT FORMULA/EQUATIONS
“The elements which have partially filled
d-sub orbit or the elements in which the last electron enters in (n-1)
d-orbitals are called transition elements.”
The d-block
elements are called transition elements also because they exhibit transitional behaviour between
highly reactive ionic-compound-forming s-block elements (electropositive
elements) on one side, and mainly the covalent-compound-forming p-block
elements (electronegative elements) on the other side.
Electronic Configuration of Transition
Elements
From the point of
view of electronic configuration, the elements which have partially filled d-orbitals
in their neutral atoms or in their common ions are called transition
elements. Thus, the outer electronic configuration of the transition elements
is (n – 1)d1–10
ns1–2, where n
is the outermost shell, and (n –
1) stands for the penultimate shell.
Ques:-Why
are Zinc, Cadmium and Mercury not considered as the Transition Elements?
Ans:
- In zinc
cadmium and mercury the last electron enters in s-orbital not in the (n-1)
d-orbital, so these elements are not called transition elements. Their
electronic configurations are (n – 1) d10 ns2.
Since, in these metals d-orbitals are completely filled, hence these do
not exhibit the general characteristic properties of the transition elements.
Therefore, these metals are not considered as transition elements.
General Trends in the Chemistry of First
Row Transition Elements (3d-series)
1. Electronic
Configuration
All d-block
elements exhibit 3d1–10 4s1–2 electronic
configuration. Some characteristic features of the electronic configurations of
the transition elements are, Atoms of all transition elements consist of an
inner core of electrons having noble gas configuration. For example,
Sc
: [Ar] 3d1 4s2 Y : [Kr] 4d1
5s2
La : [Xe] 5d1 6s2
2. Atomic
Radii
The atomic radii
of 3d-series of elements are compared with those of the neighbouring s-
and p-block elements.
164 147 135 129 137 126 125 125 128 137
in pm
The atomic radii of transition elements
show the following characteristics.
Ques.:-The
atomic radii and atomic volumes of d-block
elements in any series decrease with increase in the atomic number. The
decrease however, is not regular. The atomic radii tend to reach minimum near
at the middle of the series, and increase slightly towards the end of the
series, why?
Ans:
- When we go in any transition series from left
to right, the nuclear charge increases gradually by one unit at each element.
The added electrons enter the same penultimate shell, (inner d-shell).
These added electrons shield the outermost electrons from the attraction of the
nuclear charge. The increased nuclear charge tries to reduce the atomic radii,
while the added electron tries to increase the atomic radii. At the beginning
of the series, due to smaller number of electrons in the d-orbitals, the
effect of increased nuclear charge predominates, and the atomic radii decrease.
In the middle of the series, the atomic radii tend to have a minimum value as
observed Later in the series, when the number of d-electrons increases,
the increased shielding effect and the increased repulsion between the
electrons tend to increase the atomic radii.
Ques.:-The
atomic radii increase while going down in each group. However, in the third
transition series (5d series) from hafnium (Hf) and onwards, the elements have
atomic radii nearly equal to those of the second transition series elements,
why?
Ans:
- The atomic radii increase while going down the
group. This is due to the introduction of an additional shell at each new
element down the group. A nearly equal radius of second (4-d series) and third
transition series (5d series) elements is due to a special effect called lanthanide
contraction. In the 5d- series of transitions elements, after
lanthanum (La), the added 14 electrons go to the inner most 4f orbitals
(antepenultimate orbitals). The 4f electrons have poor shielding effect.
But due to addition of 14 extra protons in the nucleus the outermost electrons
experience greater nuclear attraction. So size of elements of 5-d series
becomes smaller then 4-d series.
3.
Ionic Radii
For ions having
identical charges, the ionic radii decrease slowly with the increase in the
atomic number across a given series of the transition elements.
EXPLANATION.
The gradual decrease in the values of ionic radius across the series of
transition elements is due to the increase in the effective nuclear charge.
4.
Ionisation Energies
The ionisation
energies (now called ionisation enthalpies, ΔIH) of the
elements of first transition series are given below:
The following generalizations can be
obtained from the ionisation energy values given above.
Ques.:-The
ionisation energies of these elements are high, and in most cases lie between
those of s- and p-block elements. This indicates that
the transition elements are less electropositive than s-block elements.
Ans:
- Transition metals have smaller atomic radii
and higher nuclear charge as compared to the alkali metals. Both these factors
tend to increase the ionisation energy, as observed. The ionisation energy in
any transition series increases with atomic number; the increase however is not
smooth and as sharp as seen in the case of s- and p-block
elements.
EXPLANATION.
The ionisation energy increases due to the increase in the nuclear charge with
atomic number at the beginning of the series. Gradually, the shielding effect
of the added electrons also increases. This shielding effect tends to decrease
the attraction due to the nuclear charge.
These two opposing
factors lead to a rather gradual increase in the ionisation energies in any
transition series.
Ques.:-The
first ionisation energies of 5d-series
of elements are much higher than those of the 3d- and 4d-series
elements, why?.
Ans:
- In the 5d- series of transitions
elements, after lanthanum (La), the added 14 electrons go to the inner most 4f
orbitals (antepenultimate orbitals). The 4f electrons have poor
shielding effect. But due to addition of 14 extra protons in the nucleus the
outermost electrons experience greater nuclear attraction. So size of elements
of 5-d series becomes smaller then 4-d series. This leads to higher ionisation
energies for the 5d-series of transition elements.
5. Metallic
Character
All transition
elements are metals. These are hard, and good conductor of heat and
electricity. All these metals are malleable, ductile and form alloys with other
metals. These elements occur in three types, e.g., face-centered cubic (fcc),
hexagonal closepacked (hcp) and body-centred cubic (bcc),
structures.
EXPLANATION.
The ionisation energies of the transition elements are not very high. The
outermost shell in their atoms have many vacant/partially filled orbitals.
These characteristics make these elements metallic in character.
The
hardness of these metals, suggests the presence of covalent bonding in these
metals. The presence of unfilled d-orbitals favours covalent bonding.
Metallic bonding in these metals is indicated by the conducting nature of these
metals. Therefore, it appears that there exists covalent and metallic
bonding in transition elements. The strength of inter atomic interactions
becomes stronger as the number of unpaired electrons increases. Cr, Mo and W
have maximum number of unpaired electrons so these metals are very hard.
Ques.:- Why
is the energy of atomization is very high for d- block elements?
6. Melting
and Boiling Points
EXPLANATION.
Atoms of the transition elements are closely packed and held together by strong
metallic bonds which have appreciable covalent character. This leads to high
melting and boiling points of the transition elements.
The strength of
the metallic bonds depends upon the number of unpaired electrons in the
outermost shell of the atom. Thus, greater is the number of unpaired electrons
stronger is the metallic bonding. In any transition element series, the number
of unpaired electrons first increases from 1 to 5 and then decreases back to
zero. The maximum five unpaired electrons occur at Cr (3d series). As a
result, the melting and boiling points first increase and then decrease showing
maxima around the middle of the series.
The low melting points of
Zn, Cd, and Hg may be due to the absence of unpaired d-electrons in their atoms.
7.
Oxidation States
Most of the
transition elements exhibit several oxidation states, i.e., they show
variable valency in their compounds. Some common oxidation states of the first transition
series elements are given below .
Ques.:- Why
do d-block elements show variable oxidation states?
Ans.:- The
outermost electronic configuration of the transition elements is (n – 1) d1–10 ns2.
The energy of (n – 1) d and ns-
orbitals are nearly same, so along with the ns-electrons (n –
1) d-electrons also involved in
oxidation state so these elements shows variable oxidation states. Also it
arises due to partially filled d-orbital.
Therefore, the
number of oxidation states shown by these elements depends upon the number of d-electrons
it has. For example, Sc having a configuration 3d1 4s2 may show
an oxidation state of + 2 (only s-electrons are lost) and + 3 (when d-electron
is also lost). The highest oxidation state which an element of this group
might show is given by the total number of ns- and (n – 1) d-electrons.
The relative
stability of the different oxidation states depends upon the factors such as,
electronic configuration, nature of bonding, stereochemistry, lattice energies
and solvation energies.
Ques:-Why
highest oxidation states are shown by oxide and fluorides?
The highest
oxidation states are found in fluorides and oxides because fluorine and oxygen
are the most electronegative elements.
The highest
oxidation state shown by any transition metal is eight. The oxidation state of
eight is shown by Ru and Os.
An
examination of the common oxidation states reveals the following conclusions:
(a)
The variable oxidation states shown
by the transition elements are due to the participation of outer ns- and
inner (n – 1) d-electrons in bonding.
(b)
Except scandium, the most common
oxidation state shown by the elements of first transition series is + 2. This
oxidation state arises from the loss of two 4s electrons. This means
that after scandium, d-orbitals become more stable than the s-orbital.
(c)
The greatest number of oxidation
states is observed near middle of the series. Eg:- Mn show +2 to +7 O.S. The
highest oxidation states are observed in fluorides and oxides. The highest
oxidation state shown by any transition element (by Ru and Os) is +8.
(d)
The transition elements in the + 2
and + 3 oxidation states mostly form ionic bonds. In compounds of the higher
oxidation states (compounds formed with fluorine or oxygen), the bonds are
essentially covalent. For example, in permanganate ion MnO4–,
all bonds formed between manganese and oxygen are covalent.
(e)
Within a group, the maximum oxidation
state increases with atomic number. For example, Iron shows the common oxidation
state of + 2 and + 3, but ruthenium and osmium in the same group form compounds
in the + 4, + 6 and + 8 oxidation states.
(f)
Transition metals also form compounds
in low oxidation states such as + 1 and 0. For example, nickel in nickel
tetracarbonyl, Ni(CO)4 has zero oxidation state. Fe(CO)5
The bonding in the compounds of transition
metals in low oxidation states is not always very simple.
8. Electrode
Potentials (E°)
Standard electrode
potentials of half-cells involving 3d-series of transition elements are
negative except Cu.The negative values of E° for the first series of
transition elements (except for Cu2+/Cu) indicate that:
These metals
should liberate hydrogen from dilute acids, ,
M +
2H+ → M2+ +
H2(g)
2M +
6H+ → 2M3+ + 3H2(g)
i.e., the reactions
are favourable in the forward direction. In actual practice however, most of
these metals react with dilute acids very slowly. Some of these metals get
coated with a thin protective layer of oxide. Such an oxide layer prevents the
metal to react further.
These metals
should act as good reducing agents. There is no regular trend in the E°
values. This is due to irregular variation in the ionisation and sublimation
energies across the series. Relative stabilities of transition metal ions in
different oxidation states in aqueous medium can be predicted from the
electrode potential data. To illustrate this, let us consider the following:
M(s)
→ M(g)
ΔH1 = Enthalpy
of sublimation, ΔsubH
M(g)
→ M+(g)
+ e– ΔH2
= Ionisation
energy, IE
M+(g)
→ M+(aq)
ΔH3 = Enthalpy
of hydration, ΔhydH
Adding these equations one gets,
M(s)
→ M+
(aq) + e– ΔH
= ΔH1 + ΔH2 + ΔH3
= ΔsubH + IE + ΔhydH
The ΔH represents
the enthalpy change required to bring the solid metal M to the monovalent ion
in aqueous medium, M+(aq).
The reaction, M(s)
→ M+(aq) + e–, will be favourable only if ΔH is negative.
More negative is the value of ΔH, more favourable will be the formation
of that cation from the metal. Thus, the oxidation state for which ΔH
value is more negative will be more stable in the solution.
Electrode
potential for a Mn+/M half-cell is a measure of
the tendency for the reaction,
Mn+(aq) + n
e– → M(s)
Thus, this
reduction reaction will take place if the electrode potential for Mn+/M
half-cell is positive. The reverse reaction,
M(s) → Mn+(aq)
+ n e–
involving the formation of Mn+(aq)
will occur if the electrode potential is negative, i.e., the tendency
for the formation of Mn+(aq) from the metal
M will be more if the corresponding E° value is more negative. In other
words, the oxidation state for which E° value is more negative (or
less positive) will be more stable in the solution.
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When an element exists in more than one oxidation states, the standard
electrode potential (E°) values can be used in predicting the relative
stabilities of different oxidation states in aqueous solutions. The following
rule is found useful.
The
oxidation state of a cation for which ΔH(= ΔsubH + IE + ΔhydH)
or E° is more negative (or less positive) will be more stable.
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Trends
in the M+2 / M+ Standard Electrode Potentials
The
observed values of Eo of the solid metal atoms M to M+2
ions in solution and their standard electrode potentials compared in Fig.
Ques:-Why is Cr+2 reducing and
Mn+3 oxidizing when both have d4 configuration?
Ques:-Which is a stronger reducing agent
Cr+2 or Fe +2 and why?
9. Formation
of Coloured Ions: - Most of the compounds of
the transition elements are coloured in the solid state and/or in the solution
phase. The compounds of transition metals are coloured due to the presence of
unpaired electrons in their d-orbitals. This occurs as follows.
Typical splitting
for octahedral and tetrahedral geometries are shown in Fig. 9.4.
Relationship between the colour of
the absorbed radiation and that of the transmitted light is given in Table 9.4.
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Colour
of the
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Colour
of the
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||
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absorbed
light
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transmitted
light
|
absorbed
light
|
transmitted
light
|
|
IR
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White
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green
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Red
|
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Red
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Blue-green
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Blue
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Orange
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Orange
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Blue
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Indigo
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Yellow
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Yellow
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Indigo
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Violet
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Yellow-green
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Yellow-green
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Violet
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UV
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White
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Green
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Purple
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|
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10. Magnetic
Properties:
- Most of the transition elements and
their compounds show paramagnetism.
The paramagnetism first increases in any transition element series, and then
decreases. The maximum paramagnetism is seen around the middle of the series.
The paramagnetism is described in Bohr Magneton (BM) units. The
paramagnetic moments of some common ions of first transition series are given
below in Table 9.5 on the next page.
EXPLANATION:
A substance which is attracted by magnetic field is called paramagnetic substance.
The substances which are repelled by magnetic field are called diamagnetic
substances. Paramagnetism is due to the presence of unpaired electrons in
atoms, ions or molecules.
The magnetic
moment of any transition element or its compound/ion is given by (assuming no
contribution from the orbital magnetic moment),
where, S is the total spin (n ×
s) : n is the number of unpaired electrons and s is equal
to 1/2 (representing the spin of an unpaired electron).
11. Formation
of Complex Ions
Transition metals
and their ions show strong tendency for complex formation. The cations of
transition elements (d-block elements) form complex ions with certain
molecules containing one or more lone-pairs of electrons, viz., CO, NO,
NH3 etc., or with anions such as, F–, Cl–, CN–
etc. A few typical complex ions are,
[Fe(CN)6]4–,
[Cu(NH3)4]2+, [Y(H2O)6]2+, [Ni(CO)4], [Co(NH3)6]3+,
[FeF6]3–
EXPLANATION.
This complex formation tendency is due to,
(a)
Small size of the transition metal
cations.
(b)
High positive charge density
(c)
The availability of vacant inner d-orbitals
of suitable energy to accept lone pair of electrons.
12. Formation
of Interstitial Compounds
Transition
elements form a few interstitial compounds with elements having small atomic
radii, such as hydrogen, boron, carbon and nitrogen. The small atoms of these
elements get entrapped in between the void spaces (called interstices)
of the metal lattice. Some characteristics of the interstitial compounds are,
(a)
These are non-stoichiometric
compounds and cannot be given definite formulae.
(b)
These compounds show essentially the
same chemical properties as the parent metals, but differ in physical
properties such as density and hardness.
Steel and cast iron are hard due to the
formation of interstitial compound with carbon. Some non-stoichiometric
compounds are, VSe 0.98
(Vanadium selenide), Fe0.94O,
and titanium hydride TiH1.7.
Some properties
1. Interstitial
compounds are hard and dense. This is because; the smaller atoms of lighter
elements occupy the interstices in the lattice, leading to a more closely
packed structure.
2. Mp are higher
and
3. They are
chemically inert. Due to greater electronic interactions, the strength of the
metallic bonds also increases.
13. Catalytic
Properties
Most of the
transition metals and their compounds particularly oxides have good catalytic
properties. Platinum, iron, vanadium pentoxide, nickel, etc., are important
catalysts. Platinum is a general catalyst. Nickel powder is a good catalyst for
hydrogenation of unsaturated organic compounds such as, hydrogenation of oils.
Some typical industrial catalysts are:
(a)
Vanadium pentoxide (V2O5)
is used in the Contact process for the manufacture of sulphuric acid,
(b)
Finely divided iron is used in the
Haber’s process for the synthesis of ammonia.
EXPLANATION.
Most transition elements act as good catalyst because of,
(a)
The presence of vacant d-orbitals.
(b)
The tendency to exhibit variable
oxidation states.
(c)
The tendency to form reaction
intermediates with reactants. The presence of defects in their crystal
lattices.
14. Alloy
Formation
Transition metals
form alloys among themselves. The alloys of transition metals are hard and high
melting as compared to the host metal. Various steels are the alloys of iron
with metals such as chromium, vanadium, molybdenum, tungsten, manganese etc.
EXPLANATION.
The atomic radii of the transition elements in any series are not much
different from each other. As a result, they can very easily replace each other
in the lattice and form solid solutions over an appreciable composition range.
Such solid solutions are called alloys.
15. Chemical
Reactivity
The d-block
elements (transition elements) have lesser tendency to react, i.e.,
these are less reactive as compared to s-block elements.
EXPLANATION.
Low reactivity of transition elements is due to,
(i)
their high ionisation energies,
(ii)
low heats of hydration of their ions,
(iii)
Their high heats of sublimation.
General Chemical Properties of First Row
Transition Metal Compounds
The transition
metals form a number of binary compounds with non-metals, e.g., carbon,
nitrogen, phosphorus, oxygen, sulphur and halogens. The chemical reactivity of
transition elements may be seen through the study of their oxides, sulphides
and halides.
Oxides of First Row Transition Elements
Transition metals
of first row (3d-series) generally react with oxygen at higher
temperatures. Because of the tendency to exhibit variable oxidation states,
these metals form a number of oxides of different varieties.
Some general
characteristics of the oxides of 3d-transition series are given below.
(i) Formulae.
The general formulae of the oxides of first row transition metals (3d-series)
are, MO, M2O3, M3O4, MO2,
M2O5 and MO3.
(ii) Acidic,
basic or amphoteric character. The character of an oxide depends upon the
oxidation state of the metal in it. For example,
The oxides of
metals in low oxidation states are basic. For example, TiO,
VO, MnO, Cu2O etc., are basic in nature.
The oxides of
metals in high oxidation states are acidic. For example, V2O5,
CrO3, Mn2O7 are acidic.
The oxides of
metals in the intermediate oxidation states, are generally amphoteric.
For example, CuO, Cr2O3,
MnO2 etc., are amphoteric.
Important oxides
of the first transition series elements are given in Table 9.6.
SOME IMPORTANT COMPOUNDS OF
TRANSITION ELEMENTS
Potassium Permanganate,
(KMnO4)
Potassium
permanganate is a salt of an unstable acid HMnO4 (permanganic acid).
The Mn is in + 7 state in this compound.
Preparation. Potassium
permanganate is obtained from pyrolusite as follows.
(i)
Conversion of pyrolusite to potassium
manganate. When manganese dioxide is fused with
potassium hydroxide in the presence of air or an oxidising agent such as
potassium nitrate or chlorate, potassium manganate is formed, possibly via
potassium manganite.
(ii)
Oxidation of potassium manganate to
potassium permanganate. The potassium manganate so obtained
is oxidised to potassium permanganate by either of the following methods.
(a) By chemical method: The
fused dark-green mass is extracted with a small quantity of water.
The filtrate is
warmed and treated with a current of ozone, chlorine or carbon dioxide.
Potassium manganate gets oxidised to potassium permanganate and the hydrated
manganese dioxide precipitates out. The reactions taking place are,
When CO2 is passed:
When chlorine or ozone is passed:
The purple
solution so obtained is concentrated and dark purple, needle-like crystals
having metallic lustre are obtained.
(b) Electrolytic method:
Presently, potassium manganate (K2MnO4) is oxidised
electrolytically.
The electrode reactions are,
The purple
solution containing KMnO4 is evaporated under controlled conditions
to get crystalline sample of potassium permanganate.
Physical properties.
(i)
KMnO4 crystallizes as dark
purple crystals with greenish luster (m.p. 523 K).
(ii)
It is soluble in water to an extent
of 6.5 g per 100 g at room temperature. The aqueous solution of KMnO4
has a purple colour.
Chemical properties. Some
important chemical reactions of KMnO4 are given below:
(i)
Action of heat. KMnO4
is stable at room temperature, but decomposes to give oxygen at higher
temperature
(ii)
Oxidising action. KMnO4
is a powerful oxidising agent in neutral, acidic and alkaline media. The nature
of reaction is different in each medium. The oxidising character of KMnO4
(to be more specific, of MnO4–) is indicated by high
positive reduction potentials for the following reactions.
There are a large
number of oxidation-reduction reactions involved in the chemistry of manganese
compounds. Some typical reactions are
(a) In the
presence of excess of reducing agent in acidic solutions permanganate ion gets
reduced to manganous ion, e.g.,
(b) An
excess of reducing agent in an alkaline solution reduces permanganate ion only
to manganese dioxide, e.g.,
(c) In faintly acidic and neutral
solutions, manganous ion is oxidised to manganese dioxide
by permanganate.
(d) In strongly basic solutions,
permanganate oxidises manganese dioxide to manganate ion.
(e) In acidic medium, KMnO4
oxidises,
(i) Ferrous salts to ferric
salts
This reaction
forms the basis of volumetric estimation of Fe2+ in any solution by
KMnO4.
(ii) Oxalic acid to carbon
dioxide
(iii) Sulphites to sulphates
(iv) Iodides to iodine in acidic
medium
Potassium Dichromate (K2Cr2O7)
Potassium
dichromate is one of the most important compound of chromium, and also among
dichromates. In this compound Cr is in the hexavalent (+ 6) state.
Preparation. It
can be prepared by any of the following methods:
(i)
From potassium chromate:
Potassium
dichromate can be obtained by adding a calculated amount of sulphuric acid to a
saturated solution of potassium chromate.
K2Cr2O7
crystals can be obtained by concentrating the solution and crystallisation.
(ii)
Manufacture from chromite
ore: K2Cr2O7 is generally manufactured
from chromite ore (FeCr2O4). The process involves the following steps.
(a)
Preparation of sodium chromate.
Finely powdered chromite ore is mixed with soda ash and quicklime. The mixture
is then roasted in a reverberatory furnace in the presence of air. Yellow mass
due to the formation of sodium chromate is obtained.
(b)
Conversion of chromate into
dichromate. Sodium chromate solution obtained in step (a)
is treated with concentrated sulphuric acid when it is converted into sodium
dichromate.
On concentration,
the less soluble sodium sulphate, Na2SO4.10H2O
crystallizes out. This is filtered hot and allowed to cool when sodium
dichromate, Na2Cr2O7.2H2O,
separates out on standing.
(c)
Conversion of sodium dichromate to
potassium dichromate. Hot concentrated solution of sodium
dichromate is treated with a calculated amount of potassium chloride, when potassium
dichromate being less soluble crystallizes out on cooling.
Chemical properties.
(i) Action of alkalies. With
alkalies, it gives chromates. For example, with KOH,
On acidifying, the
colour again changes to orange-red owing to the formation of dichromate.
Actually, in
dichromate solution, the Cr2O7 2– ions are in
equilibrium with CrO4 2– ions.
(iv) Oxidising
nature. In neutral or in acidic solution, potassium dichromate acts as an
excellent oxidising agent, and Cr2O7 2– gets
reduced to Cr3+. The standard electrode potential for the reaction,
is + 1.31 V. This
indicates that dichromate ion is a fairly strong oxidising agent, especially in
strongly acidic solutions. That is why potassium dichromate is widely used as
an oxidising agent, for quantitative estimation of the reducing agents such as,
Fe2+.
It oxidises,
(a) Ferrous
salts to ferric salts
Ionic equation:
(b) Sulphites to sulphates and arsenites to
arsenates.
Ionic equation:
Similarly, arsenites are oxidised to
arsenates.
(c) Hydrogen halides to halogens.
Ionic equation:
(d) Iodides to iodine
Ionic equation:
Thus, when KI is
added to an acidified solution of K2Cr2O7 iodine gets liberated.
(e) It oxidises H2S to S.
Ionic equation:
(i)
Formation of insoluble chromates. With
soluble salts of lead, barium etc., potassium dichromate gives insoluble
chromates. Lead chromate is an important yellow pigment.
(ii)
Chromyl chloride test. When
potassium dichromate is heated with conc. H2SO4 in the presence of a soluble
chloride salt, the orange-red vapour of chromyl chloride (CrO2Cl2)
is formed.
Chromyl chloride
vapour when passed through water give yellow-coloured solution containing
chromic acid.
Structure of Chromate and Dichromate Ions
f-block elements
Inner-Transition Elements: Lanthanoides
and Actinoids
The elements which
in their elemental or ionic form have partly filled f-orbitals are
called f-block elements. As the f-orbitals lie inner to the
penultimate (second outermost) shell i.e. antepenultimate orbitals,
therefore these elements having partially filled f-orbitals are also
known as inner-transition elements.
There general electronic
configuration is
(n-2)f1-14(n-1)d0
or 1ns2
There are two
series of inner-transition elements, each having 14 elements. The elements in
which 4f orbitals are progressively filled are called lanthanides.
The elements in which 5f orbitals are progressively filled are termed actinides.
Lanthanides thus
belong to the first inner-transition series, while actinides belong to the
second inner-transition series.
Lanthanoides
The fourteen
elements (atomic no. 58 – 71) after lanthanum are known as lanthanides or
lanthanons. All these elements closely resemble one another in their
properties. Because of their limited availability, these are also known as the rare
earth elements.
Names and the
outer-electronic configurations of the lanthanides are given in Table 9.10.
General Characteristics of Lanthanides
General
physical characteristics of lanthanides are described below:
(1) Electronic configuration.
The
outer-electronic configurations of lanthanides are given in Table 11.9. There
is however, some uncertainty about the correctness of these configurations. The
5d and 4f energy levels are very close-by. It is not always
possible to decide with certainty whether the electron has entered 5d or
4f level. Due to the extra-stability of half-filled and completely
filled orbitals, there is a tendency to acquire f7 and f14
configurations wherever possible. The general electronic configuration of lanthanides
may be described as 4f1–14 5d 0–1 6s2.
(2) Oxidation states.
All Lanthanoides exhibit a common stable oxidation
state of +3. in addition some lanthaniodes shows +2 and +4 oxidation state
also. These are shown by those elements which by doing so attain the stable f0, f7 and f14
configurations. For example:
(i) Ce and Tb
exhibit +4 oxidation states.
Cerium (Ce) and terbium (Tb) attain f0 and f7
configuration respectively when they get +4 oxidation state, as shown below:
Ce4+
: [Xe]4f0
Tb4+ : [Xe]4f7
(ii) Eu and Yb
exhibit +2 oxidation states.
Europium and yetterbium get f7 and f14
configuration in +2 oxidation state, as shown below:
Eu2+
: [Xe]4f7
Yb2+ : [Xe]4f14
(iii) La, Gd, and
Lu exhibit only +3 oxidation states due to empty, half filled and fulfilled
4f-sub orbit.
The stability of different oxidation state has strong effect on the
properties of those elements. For example, Ce(IV) is favoured because of its
noble gas configuration. But it is strong oxidant changing to common +3
oxidation state.
Similarly, Eu2+ is stable because of its half filled 4f7 configuration.
However, it is a strong reducing agent changing to Eu3+ (common
oxidation state.) Similarly, Yb2+ having the configuration 4f14
is a reductant. Samarium also behaves like europium exhibiting both +2 and +3
oxidation states.
Important
note: - Irrespective of noble gas
configuration 4f0 Ce+4 is strong oxidizing agent and it
changes to +3 state. It is because the Eo value for Ce+4|
Ce+3 is +1.74 V which suggests that it can oxidise water. But its
reaction rate is very slow so Ce+4 is good analytical reagent.
Similarly
Eu+2
is stable with half filled 4f7 configuration but Eu+2 is strong reducing agent and
it changes to +3 state. It is because the Eo value for Eu+3
| Eu+2 is negative.
(3) Magnetic properties.
La3+
and Lu3+ are diamagnetic, while the trivalent ions of the rest of
the lanthanides are paramagnetic in nature. The paramagnetic moment values of
the lanthanide ions are higher than those expected on the basis of the number
of unpaired electrons. This occurs due to an appreciable contribution from
orbital angular momentum.
(4) Reduction potentials and metallic
character. The standard electrode (reduction) potentials of the
lanthanide ions become less negative across the series. Thus, their reducing
power decreases in going from Ce to Lu. The highly negative E° values
indicate these elements to be highly electropositive metals capable of
displacing hydrogen from water.
The
M(OH)3 are ionic and basic in character. These hydroxides are
stronger than Al(OH)3 and weaker than Ca(OH)2. The
basic strength decreases in going from La to Lu.
(5) Atomic and ionic size: Lantha-nide
contraction. The
atomic and ionic sizes decrease steadily in going from Ce to Lu. This decrease
can be explained as follows.
EXPLANATION. In
the atoms of lanthanides, the nuclear charge increases with
atomic number, and the added electrons go
to the inner 4f orbitals. The shielding effect of 4f electrons
from the increased nuclear charge, is poor. Thus, as the atomic number
increases, the effective nuclear charge experienced by each 4f electron
increases. This causes a slight reduction in the entire 4f shell. The
successive contractions accumulate and the total effect for all the lanthanides
is called lanthanide contraction.
The variation of
ionic radii of lanthanide ions is shown in Fig. 9.16.
The 4f electrons
also shield the valence shell from contracting appreciably. In lanthanides, the
decrease of radius for fourteen elements (Ce to Lu) is 15 pm. This may be
compared with the second period decrease of 81 pm in the radii for 7 elements
(Li to F) and with that of the third period elements (Na to Cl), 86 pm. Consequences
of lanthanide contraction. The lanthanide contraction has a highly
significant
effect on the relative properties of the elements which precede and follow
lanthanides in the periodic table. Some important consequences of lanthanide
contraction are:
(i)
The
radius of La3+ ion, for example, is 22 pm larger than that of Y3+
ion which lies immediately above it in the periodic table. On this basis, if
the fourteen lanthanides had not intervened, the radius of Hf 4+
should have been greater than that of Zr 4+ (which lies immediately
above it) by about 20 pm. But, the lanthanide contraction of about the same
magnitude almost cancels the expected increase. As a result, Hf 4+
and Zr 4+ have almost equal radii, being 80 and 81 pm respectively.
It is seen that the normal
increase in size from Sc → Y → La disappears after the lanthanides and the
pairs of elements such as, Zr – Hf, Nb – Ta, Mo – W, etc., have almost the same
size. The properties of these elements are also very similar. It is thus a direct consequence of
lanthanide contraction that the elements of the second and third transition
series resemble each other much more closely than do the elements of the first
and second transition series.
(ii)
Due
to lanthanide contraction, i.e.,
decrease of ionic size on moving from La3+ to Lu3+, the
covalent character in bonding increases in the direction La3+ → Lu3+.
As a result, the basic character of the lanthanide hydroxides (M(OH)3)
decreases with increase in atomic number.
Thus, La(OH)3 is the
most basic, while Lu(OH)3 is the least basic. This aspect has been
utilized in the separation of lanthanides from each other.
(6) Formation of complex salts and ions. Lanthanide ions (M3+)
have high charge, but due to their larger size, these cannot polarize the
neighbouring anion/molecule. As a result, these lanthanides do not show a
strong tendency towards complex formation.
(7) Colour of the salts and ions in solution.
Most of the lanthanide trivalent ions are coloured in solid as well as in
the solution phase. The ions containing x and (14 – x) electrons
show the same colour. The colour of the salts or ions is due to the f – f
transition of electrons.
Actinoides
The fourteen elements (atomic number
90–103) after actinium are called actinides. These are also called second
series of inner-transition elements. The general electronic configuration
of actinides is 5f 1–14 6d 0–1 7s2.
Names and the outer-electronic configurations of actinides are given below in
Table 9.11.
General Characteristics of Actinides
(2) Atomic and Ionic radii. The
radii for tripositive (M3+) and tetrapositive (M4+) ions
decrease in going from Th to Cm. This steady decrease is similar to that
observed in lanthanides and is called actinide contraction.
Fact: The actinide contraction is larger than
lanthanide contraction.
Reason: because in
lanthanoids electrons are filled in 4f orbital whose screening effect is more
stronger than 5f orbitals of actinoid elements.
(3) Colour of salts and ions in solution.
Most
of the salts of actinides having M3+ or M4+ ions are
coloured. Ions having 5f °, 5f 1 and 5f 7
configurations are colourless, while those containing 5f 2, 5f
3, 5f 4, 5f 5 and 5f 6
configurations are coloured.
Due to their
alloy-forming tendencies, actinides and lanthanides form many alloys
particularly with iron. These elements improve the workability of steel. A well
known alloy is ‘misch-metal’ which consists of a rare earth element (94
– 95%), iron (up to 5%) and traces of sulphur, carbon, calcium and aluminium.
The pyrophoric
alloys containing rare-earth metals are used in the preparation of ignition
devices, e.g., tracer bullets and shells and flints for lighters. This
alloy has normally the composition: cerium 40.5%, lanthanum + neodymium 44%,
iron 4.5%, aluminium 0.5% and the remainder calcium, silicon and carbon.
(i)
Cerium constitute about 30-50% of the
alloys of lanthanides. They are used fro scavenging oxygen and sulphur.
(ii)
Thorium is used in the manufacture of
fine rods of atomic reactor.
(iii)
Thorium salts are also used in
treatment of cancer.
(iv)
Uranium and Plutonium are used for
production of nuclear energy by the nuclear fission.
SUMMARY
- Electronic
configuration of transition metals is
(n-1)¹ ̄ ¹⁰ns¹
̄ ² .
•
The name of transition metal refers to d
block.
•
The metal of inner transition metal refers
to f block.
•
D block contains 3-12 groups elements
where d block are progressively filled,
whereas f block contain elements in
which the 4f and 5f orbitals are progressively filled.
•
There are three series of transition
metals:-
•
3d series :- from Sc - Zn
•
4d series :- from Y - Cd
•
5d series :- from La - Hg
(not including Ce - Lu )
•
The name transition is given because of
there position between s and p
block.
•
The electronic configuration of :-
Cr :- 3d⁵4s¹
Cu :- 3d¹⁰4s¹
•
Zn , Cd and Hg are not regarded as
transition elements because orbitals are completely filled.
Similarity
between lanthanoids and actinoids
•
Both show +3 oxidation states.
•
F orbitals are progressively filled.
•
Same number of unpaired electrons.
•
Electropositive and highly active.
•
Contraction
Lanthanoids
contraction :-
•
The steady decrease in atomic size from
183 pm to 173 pm and in ionic size 103 pm to 85 pm of lantanoid elements with
increase in atomic number from ₅₈Ce
to ₇₁Lu is called lanthanoid
contriction.
•
Cause :-
with increasing atomic number and nuclear
charge ,the effective nuclear charge experienced by each 4f-electon increases.
F orbitals have poor shielding effect due to highly diffused shape .as a result
gradual decrease in size is observed.
some questions and
answers for practice
:-
- Silver
atom has completely filled d orbitals
(4d¹⁰)
in its ground state.how can you say that it is a transition element ?
Ans: because
silver can exhibit +2 oxidation state , which has completely filled d orbital.
2.why
do transition elements exhibit higher enthalpies of atomisation ?
Ans:
due to large number of unpaired electrons , strong bonding atoms results in
higher enthalpies of atomisation .
3.name a transition element which exhibits maximum
variable oxidation states ?
Ans:
mn because s and d orbitals take part in bond formation.
4.
What is meant by ‘disproportionation’ ?give an eg of disproportionation
reaction in aqueous solution.
Ans:
in disproportionation reaction ,an element undergoes oxidation and reduction
both.
Eg:- 2Cu⁺ à Cu + Cu²⁺
5.the
e⁰(m²⁺/m) for cu is
positive (+0.34 v). Explain ?
Ans:
the enthalpy of atomisation is very high for cu but its hydration energy is
very low. The high energy to transform Cu(s) to Cu²⁺(aq) is not balanced by
its hydration enthalpy. Thus ,for this conversion e⁰ value is positive.
.
The e⁰ value for Mn³⁺/Mn²⁺ couple is much more
positive than Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ couple.
Ans:
much larger third ionisation energy of mn due to extra stability of mn²⁺ (d⁵configuration) (where
required change is d⁵
to d⁴) is mainly responsible
for this.
7.calculate
the magnetic moment of divalent ion in aqueuos solution. If its atomic number
is 25.
Ans: m(27)
M²⁺= [Ar]3d⁷4s⁰
Magnetic
moment, µ = √n(n+2)
=√5*7=√35
=
5.92 bm
8.name
a member of lanthanoid series which is well known to exhibit +4 oxidation state
?
Ans
:- cerium (z=58).
A few more questions and answers for
practice :-
- Why
are Mn²⁺
compounds more stable than Fe²⁺ towards oxidation state to their +3
state ?
Ans:
Mn²⁺(3d⁵) half filled is stable
,while Fe²⁺
is 3d⁶ not stable.
2.to
what extent do the e.c. Decide the stability of oxidation states in the first
series of transition elements ?
Ans:
the presence of half filled or completely filled orbitals .
3.
Name the oxometal anions of first series of the transition metals in which the
metal exhibits the oxidation state equal to its group no. ?
Ans
: MnO₄̄¹ oxidation state is +7
CrO₄̄² oxidation state is +6
In
what way is the e.c. Of the transition elements different from that of the non
transition elements ?
Ans:
in transition elements d orbitals are
progressively filled whereas in non transition elements outer most s or p
orbitals are progressively filled.
5.what
are the different oxidation states exhibited by the lanthanoids ?
Ans
: common oxidation state of lanthanoid
is +3 ,+2 and +4.
6.predict
which of the following will be coloured in aqueous solution Ti³⁺ , V³⁺ , Cu⁺, Sc³⁺ , Mn²⁺ , Fe³⁺ and Co²⁺ .give reasons for each.
Ans:
Ti³⁺, V³⁺,Mn²⁺ , Fe³⁺, Co²⁺ as they contain unpaired
electrons.
.
Compare the chemistry of actinoids with that of lanthanoids with special
reference to :-
i)
Electronic configuration
ii)
Atomic and ionic sizes
iii)
Oxidation state
iv)
Chemical reactivity
Ans.
I) e.c :- in lanthanoids 4f orbitals are progressively filled whereas in
actinoids 5f are progressively filled
Ii)oxidation
state :- lanthanoids show +3 state some shows +2 and +4 .
but actinoids show +3,+4,+5,+6,+7 oxidation
state. +3 and +4 are common.
Iii)atomic
and ionic sizes :- decreases from left to right in both.but decreases more in
case of actinoids.
Iv)
chemical reactivity :- actinoids are more reactive than lanthanoids.
9.
How would you account for the following :-
I)
of the d4 species,Cr²⁺
is strongly reducing while Mn(iii) is strongly oxidising.
Ii)
co(ii) is stable in aqueous solution but in presence of complexing reagents it
is easily oxidised.
Iii)
the d¹ configuration is very unstable in ions
Ans
: i)Cr²⁺ is reducing agent ,it
change to Cr³⁺
by loosing electron, Cr³⁺
is more stable due to half filled state. Mn³⁺
is oxidising agent, reduced to mn²⁺
which is half filled and very stable.
Ii)Co(2)
is oxidised to Co(3) because Co(3) is more stable .
Iii)
because after loosing electron it become more stable.
10.
What is meant by ‘disproportionation’ ?give an eg of disproportionation
reaction in aqueous solution.
Ans:
in disproportionation reaction ,an element undergoes oxidation and reduction
both.
Eg:- 2Cu⁺ à Cu + Cu²⁺
11.
Which metal in first series of transition metal exhibits +1 oxidation state
most frequently and why ?
Ans
: Cu exhibits +1 oxidation state ,by loosing one electron ,the cation ion
aquires a stable configuration of d orbital.
12.give
examples and suggest reasons for the following features of the transition metal
chemistry :-
I)
The lowest oxide of transition metal is
basic ; the highest amphoteric/acidic.
II)
A transition metal exhibits highest
oxidation state in oxides and flourides.
III)
Iii) the highest oxidation state is
exhibited in oxo-anions of a metal.
IV)
Ans : i) in lowest oxidation state ,ionic
bond is formed ,less number of electrons are involved. Oxide donate electrons
and behave like a base. In the highest oxidation state , covalent bond is
formed .in high oxidation state more electrons are involved in bonding.oxide
can gain electron, behave like lewis acid.
Ii)because
oxygen and fluorine are strong oxidising agent ,higly electronegative element.
Iii)
due to high electronegativity of oxygen.
13.what
are alloys ? Name an important alloy which contains some of lanthanoid metals.
Mention its uses.
Ans:
alloy are homogenous mixture of two or more metals . Misch metal is an alloy
,which contains 45% lanthanoid metals (iron 5% ,traces of S,C,Ca,Al).
Use-
Used
to produce bullets ,shells and lighter flint. 3% misch metal to mg used in
making jet engine parts.
14.
What are inner transition elements ? Decide which of the following atomic
numbers are atomic numbers of inner transition elements : 29,59,74,95,102,104.
Ans-
lanthanoids and actinoids are called inner transition elements because inner f
orbitals are progressively filled. z=
58 to 71 are lanthanoids , z= 90 to 103 are actinoids , so atomic no. 59,95,102
belong to inner transition elements.
15.
The chemistry of actinoid is not smooth as that of the lanthanoids .justify
this statement by giving some examples from the oxidation state of these
elements.
Ans:
lanthanoids exhibits +2,+3,+4 oxidation state out of these +3 is most common
lanthanoids shows +3 oxidation state.
16.
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.
Ans:
Ir¹⁰³ is the best actinoid.
Its electronic configuration is [Rn]⁸⁶5f¹⁴6d¹7s².the possible
oxidation state is +3.
d and f block elements HOTS
Q1. Among the first transition metals, which divalent metal ion
has maximum paramagnetic character and why?
Ans: Mn2+,
because of maximum number of unpaired electrons.
Q2. Give the chemical equation for the reaction between MnO4−
and C2O4−.
(Ans) 2MnO4− +
16H+ + 5C2O4−2 → 2Mn2+ +
8H2O + 10CO2.
Q3. Which is the stronger reducing agent Cr2+ or Fe2+
and why?
(Ans) Cr2+ is stronger
reducing agent than Fe2+ because E0 Cr3+/
Cr2+ is -0.41V and E0 Fe3+/Fe2+
is +0.77 V so Cr2+ is easily oxidized to Cr3+ but Fe2+
is not easily oxidized to Fe3+ .
Q4. In the series Sc to Zn the enthalpy of atomization of zinc is
lowest why?
(Ans) In this series all the
elements have one or more unpaired electrons except zinc. Its outer electronic
configuration is 3d104s2.This shows that the atomic inter
metallic bonding in zinc is weakest. so enthalpy of atomization is lowest.
Q5. Which of the 3d series of the transition metals exhibits the
largest number of oxidation states and why?
(Ans) Mn show largest number
oxidation state. Its electronic configuration is 3d54s2.
The energy of 3d and 4s are close. So, it has maximum number of electrons to
loose or share and it shows maximum number of oxidation states from +2 to +7.
Q6. Copper I compounds are white and diamagnetic but copper II
compounds are coloured and paramagnetic. Why?
(Ans) In copper I ion all orbitals
are completely filled so its compounds are white and diamagnetic. The
electronic configuration of copper II ion is 1s22s22p63s23p63d9.
it has one unpaired electron so it is paramagnetic and forms blue coloured
compounds.
Q7. How does the ionic and covalent character of the
compounds of a transitional metal vary with its oxidation states?
(Ans) As the oxidation states
increases more and more valence shell electrons are involved in bonding. The
atomic core becomes less shielded and the force of attraction on the electrons
increased because of this ionic character of bonds decreases with increase in
oxidation state.
Q8. The E0 (M2+/M) value for copper is
positive (+ 0.34). What is the possible reason for this?
(Ans) E0 (M2+
/ M) for any metal is the sum of the enthalpy changes taking
place in the following steps:
M(s) +
Δa H -→M(g), ΔaH = enthalpy of
atomization
M(g) + Δi H →M2+(g), ΔiH = ionization enthalpy
M2+(g) + aq → M2+(aq) +Δhyd H,ΔhydH = hydration enthalpy
Copper has high enthalpy of atomization and low enthalpy of hydration so E0 (Cu2+ / Cu) is positive.
M(g) + Δi H →M2+(g), ΔiH = ionization enthalpy
M2+(g) + aq → M2+(aq) +Δhyd H,ΔhydH = hydration enthalpy
Copper has high enthalpy of atomization and low enthalpy of hydration so E0 (Cu2+ / Cu) is positive.
Q9. Calculate the “spin only” magnetic moment of M2+(aq)
ion (Z= 27).
(Ans) Electronic configuration of M
atom is 1s22s22p63s23p63d74s2.
It has three unpaired electrons in d orbitals. Magnetic moment = √ n(n+2) BM
= √3 (3+2)
= √15
3.87 BM
Q10. How does +2 oxidation state becomes more and more stable in
the first half of the first row transition elements with increasing atomic
number?
(Ans) The
sum of first and second ionization enthalpies increases with increasing atomic
number so the standard reduction potentials become less and less negative.
Hence the +2 oxidation state becomes more and more stable.
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