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MODULE - 6 Chemistry
Notes
Chemistry of Elements
ou have already learnt in lesson 4 on periodic classification, that each period (except the first period) of the periodic table starts with the filling of ns subshell and ends with the filling of np subshell ( n is the principal quantum number and also the number of the period). The long form of the periodic table is based on the filling of electrons in various levels in order of increasing energy as given by Aufbau principle. In the fourth period, filling of the 4th shell commences with the filling of 4 s subshell followed by 3 d and 4 p subshells. For the first time, we come across a group of elements in which a subshell of the previous principal quantum number (3 d ) starts getting filled instead of the expected subshell 4 p. This group of elements that occurs in between the 4s and 4p elements is referred to as 3d elements or elements of first transition series (see periodic table). 4f Series consist of 14 members from Ce to Lu (At. No. 58-71), where the penultimate subshell, 4f subshell is filled up. They have general electronic configuration [Xe] 4 f 1-14^5 d 1,2^6 s^2. La is also included in this series: it is the prototype for the succeeding 14 elements. In this lesson you will study more about these elements and also about the preparation, properties and uses of potassium dichromate (K 2 Cr 2 O 7 ) and potassium permanganate (KMnO 4 ).
Objectives After reading this lesson, you will be able to: define transition metals and write their electronic configuration; list the general and characteristic properties of the transition elements; explain the properties of 3d transition series: metallic character, variable oxidation state, variation in atomic and ionic radii, catalytic properties, coloured ions, complex formation, magnetic properties, interstitial compounds and alloy formation; recall the preparation of potassium permanganate from pyrolusite ore; write the chemical equations illustrating the oxidizing properties of KMnO 4 in acidic, alkaline and neutral media (acidic: FeSO 4 , SO 2 , alkaline: KI and ethene, neutral: H 2 S and MnSO 4 );
d -BLOCK AND
f -BLOCK ELEMENTS
Y
d -Block and f -Block Elements MODULE - 6
Notes
recall the preparation of potassium dichromate from chromite ore;^ Chemistry of Elements
write the oxidation reactions of potassium dichromate with SO 2 and ferrous sulphate in acidic medium;
write electronic configuration of lanthanoides (4f-elements) and
explain lanthanoide contraction.
23.1 d -Block Elements
d -Block elements occupy the middle portion of the periodic table i.e. between s- and p- block elements. They include elements from groups 3 to 12. In these elements the outermost shell contains one or two electrons in their outer most i.e, ns orbital but the last electron enters into the inner d-subshell i.e. ( n -l) d orbital. The elements of the d-block are metallic in nature. Their general characteristic properties are intermediate between those of the s- block elements, on one hand and of the p-block elements on the other. We can say that d - block elements represent a change (or transition) from the most electropositive s -block elements to the least electropositive p -block elements and are, therefore, also named as transition elements.
Transition elements are elements in which the d subshell is partially filled either in atomic state or in ionic state.
There are four series of transition elements in the periodic table. The first transition series begins with scandium (At. No. 21) and ends at copper (At. No. 29) whereas the second, third and fourth series begin with yttrium (At. No. 39), lanthanum (At. No. 57) and actinium (At. No. 89) and end at silver (At. No. 47), gold (At. No. 79) and at the element having atomic number 112 (a synthetic element), respectively. These series are also referred to as 3d, 4d, 5d and 6d series, respectively. It may be noted that although elemental copper, silver and gold as well as Cul+, Ag1+^ and Au1+^ have a d^10 configuration but Cu2+^ has a 3 d^9 , Ag2+^ a 4 d^9 and Au3+^ a 5 d^8 configuration and that is why these elements are classified as transition elements. On the other hand, zinc, cadmium and mercury do not have partially filled d subshell either in the elemental state or in any of their common ions. These elements, therefore, are not transition elements. However, zinc, cadmium and mercury are often considered along with d- block elements.
Intext Questions 23.
- What are transition elements?
...................................................................................................................................
- How many elements comprise the first transition series? Give names of all these elements. ...................................................................................................................................
- Whereas copper is a transition element, zinc is not included amongst transition elements. Explain. ...................................................................................................................................
- Although Cu+, Ag+^ and Au+^ have d^10 configuration but Cu, Ag and Au are transition elements, why? ...................................................................................................................................
d -Block and f -Block Elements MODULE - 6
Notes
As can be seen, in case of zinc, the 30th electron goes to 4 s level and not 3 d level which^ Chemistry of Elements
is already full. Thus by definition, zinc cannot be called a member of d block elements. Besides, no compound of zinc is known to have a partially filled 3 d subshell. Thus it does not fit into the definition of a transition element either. Hence zinc cannot be rightly called either a d -block element or transition element. However, zinc and other members of group 12, viz., cadmium and mercury are discussed along with 3 d, 4 d and 5 d transition elements for the sake of convenience.
It is important to understand at this point, the process of ionization (i.e. oxidation) of transition elements. From what has been said above regarding filling of the orbitals, it is logical to conclude that during ionization electrons should be lost first from the (n-1) d subshells and then from the 4 s level. This, however, is not the case. The reason for the deviation from the expected behavior is that once the filling of the 3d subshell commences at scandium (At. No.21) energy of 3 d subshell decreases and becomes lower than that of 4 s subshell. Consequently, on ionization, the first row transition elements lose electrons from the 4 s subshell followed by the loss from 3 d level. For example vanadium (Z = 23) has electronic configuration V= [Ar]3 d^3 4 s^2 and the electronic configuration of V2+^ is [Ar]3 d^3 , Similarly electronic configuration of V3+^ and V4+^ are [Ar]3 d^2 and [Ar]3 d^1 , respectively. In some cases, however, for example scandium, all the electrons beyond the core of 18 electrons are lost in single step. It is important to note that though 3 d orbitals are of higher energy than 4 s orbitals (as is evident from the order of filling) the difference is so little that these are considered almost of same energy.
Intext Questions 23.
- Write the general electronic configuration of transition elements.
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- Write down the electronic configuration of the following elements in ground state: Sc, Cr, Cu and Zn.
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- Write down the electronic configuration of the following ions: Cr3+, Ti 4+, Ni3+^ and Cu2+.
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- Why the electronic configuration of Mn2+^ is 3 d^5 and not 3 d^2 4 s^2?
...................................................................................................................................
23.3 Physical Properties
Some important physical properties of d -block elements are listed in Table 23.2. Like s - block elements, d -block elements are also metals. But properties of these elements are markedly different from those of s -block elements. The interesting feature of the chemistry of transition elements is that similarities in the properties of transition elements are much more marked as compared to those in s -block. Almost all transition elements show typical metallic properties such as high tensile strength, ductility, malleability, high thermal and
MODULE - 6 Chemistry
Notes
Chemistry of Elements electrical conductivity and metallic lusture. All the transition elements have typical metallic
structure except mercury, which is liquid at room temperature.
Transition elements show high melting and boiling points. They typically melt above 1356 K. It is due to the small atomic size and strong interatomic bonding. All the transition elements are hard except zinc, cadmium and mercury. They show high enthalpy of atomization (Table 23.2). Densities of transition elements are very high as compared to those of s-block elements. The density of the elements in a given transition series increases across a period and reaches a maximum value at groups 8,9 and 10. This trend can be explained on the basis of small radii and close packed structure of the elements.
Table 23.2: Some important physical properties of 1st transition series Property Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic number 21 22 23 24 25 26 27 28 29 30 Outer electronic configuration 3 d^1 4 s^2 3 d^2 4 s^2 3 d^3 4 s^2 3 d^4 4 s^2 3 d^5 4 s^2 3 d^6 4 s^2 3 d^7 4 s^2 3 d^8 4 s^2 3 d^9 4 s^2 3 d^10 4 s^2 Atomic radius (pm) 160 146 131 125 129 126 125 124 128 133 Ionic radius M2+^ (pm) – 90 88 84 80 76 74 72 69 79 Ionic radius M3+^ (pm) 81 76 74 69 66 64 63 63 – – Crystal structure fcc hcp bcc bcc bcc bcc,fcc hcp,fcc fcc fcc hcp Density (g ml–1) 3.1 4.5 6.1 7.2 7.6 7.9 8.7 8.9 8.9 7. Melting point (K) 1817 1998 2173 2148 1518 1809 1768 1726 1356 693 Boiling point (K) 3003 3533 3723 2138 2423 3273 3173 3003 2868 1179 Stable oxidation states +3 +4 +3,+4,+5 +2,+3,+6 +2,+3,+4,+7 +2,+3 +2,+3 +2 +1,+2 + Ist ionization enthalpy (kJ mol–1) 632 659 650 652 717 762 758 736 745 906 Electronegatively 1.3 1.5 1.05 1.6 1.05 1.8 1.8 1.8 1.8 1. Heat of fusion (kJ mol–1) 15.9 15.5 17.6 13.8 14.6 15.3 15.2 17.6 13.0 7. Heat of vaporization (kJ mol–1) 338.9 445.6 443.6 305.4 224.7 353.9 389.1 380.7 338.9 114. Reduction potential (E^0 )M2+/M(V) – –1.63 –1.20 –0.91 –1.18 –0.44 –0.28 –0.25 +0.34 –0.
Atomic radii
The radii of the elements decrease from left to right across a row in the transition series until near the end, then the size increases slightly. On passing from left to right, extra protons are placed in the nucleus and extra electrons are added. The d- orbital electrons shield the nuclear charge poorly. Thus the effective nuclear charge increases and, therefore, electrons are attracted more strongly, hence contraction in size occurs. There is an increase in atomic radii with increase in atomic number in a given group, for example Ti (146 pm), Zr (157 pm) and Hf (157 pm). The very close similarity between the radii of elements of second and third transition series is a consequence of the filling of the 4 f - subshell (causing lanthanide contraction which you will study later in this lesson).
MODULE - 6 Chemistry
Notes
Chemistry of Elements have ionic bonds whereas bonds are essentially covalent in higher oxidation states. For
example, in case of permanganate ion, MnO 4 – , bonds formed between manganese and oxygen are covalent. Considering the acid base character of the oxides, it can be inferred that increase in oxidation state leads to decrease in basic character of the oxide and vice- versa. For example, MnO is a basic oxide whereas Mn 2 O 7 is an acidic oxide.
Since transition metals exhibit multiple oxidation states, their compounds in the higher oxidation states are strong oxidizing agents as they tend to accept electrons and come to stable lower oxidation states.
23.4.2 Magnetic Properties
Substances possess two types of magnetic behaviour, either diamagnetism or paramagnetism. Diamagnetic substances are either repelled or remain unaffected by an applied magnetic field whereas, paramagnetic substances are attracted towards the applied field.
There is a strong co-relation between the magnetic behaviour, electronic configuration and oxidation state. Paramagnetism arises due to the presence of unpaired electrons (Table 23.3). Since transition metal ions generally contain unpaired electrons a large number of transition metal ions exhibit paramagnetic behavior.
Magnetic moment () of paramagnetic material can be calculated (in B.M., Bohr Magneton) by using the expression: = (^) n n ( 2)where n is the number of unpaired electrons.
For example, Ni2+^ ion has two unpaired electrons (i.e. n = 2). The magnetic moment can be calculated as = (^) 2 (2 2)= 8 = 2.83 B.M The mangentic moments of some 3 d metals ions are listed in Table 23.3 which shows that greater the number of unpaired electrons, greater is the magnetic moment.
Table 23.3 : Magnetic moments of some ions of the transition elements: Ion Electronic configuration Number of unpaired Calculated magnetic electrons moments (B.M.)
Sc3+^3 d^0 0
Ti3+^3 d^1 1 1.
Ti2+^3 d^2 2 2.
V2+^3 d^3 3 3.
Cr2+^3 d^4 4 4.
Mn2+^3 d^5 5 5.
Fe2+^3 d^6 4 4.
Co2+^3 d^7 3 3.
Ni2+^3 d^8 2 2.
Cu2+^3 d^9 1 1.
d -Block and f -Block Elements MODULE - 6
Notes
Compounds containing Sc3+, Ti4+, V5+, Cr6+, Mn7+^ and Cu+^ ions are diamagnetic since^ Chemistry of Elements
these ions do not contain any unpaired electron.
23.4.3. Colour of Ions and Compounds
Most of the compounds of d -block elements are coloured or they give coloured solution when dissolved in water (Table 23.4). This property of transition elements is in marked contrast to that of the s- and p -block elements, which often yield white compounds. In transition metal compounds colour is generally associated with incomplete (n-1) d subshell of the transition metal. When white light, which has colored constituents, interacts with a substance, a part of it is absorbed by the substance. For example, if red portion of white light is absorbed by a substance, it would appear blue (the complementary colour of red). This is observed in case of copper sulphate solution. Since most compounds of transition elements are coloured, there must be energy transition, which can absorb some of the energy of the visible light. The colour of transition metal ions containing unpaired electrons is attributed to electronic transitions from one energy level to another in the d -subshell. In these metals the energy difference between the various d -orbitals is in the same order of magnitude as the energies of the radiation of white light (A. = 4000 to 8000 A).
Table 23.4 : Colours of hydrated ions of some transition elements
Hexahydrated ion of Number of d electrons Color of solid/solution Ti3+^1 Violet V3+^2 Blue V2+^3 Violet Cr3+^3 Green
Mn3+^4 Violet Fe3+^5 Yellow/colorless Mn2+^5 Yellow/colorless Fe2+^6 Pale green Co2+^7 Pink
Ni2+^8 Green Cu2+^9 Blue
23.4.4 Alloy and Interstitial Compound Formation
In the Table 23.2 it may be observed that the atomic size of the elements of first transition series is quite close to each other. Thus, in the crystal lattice, anyone of these elements can easily replace another element of similar size forming solid solutions and smooth alloys. Transition elements, therefore, form a number of alloys. Cr, V and Mn are used to produce alloy steel and stainless steel, copper forms brass, bronze etc. Besides, transition metals also form a number of interstitial compounds in which they take up atoms of small size, like hydrogen, carbon and nitrogen etc. These are located in the vacant spaces of metal lattices and are bound firmly there in. The products thus obtained are hard and rigid. For example, steel and cast iron become hard due to formation of an interstitial compound with carbon. In such compounds, malleability and ductility may marginally decrease but
d -Block and f -Block Elements MODULE - 6
Notes
Chemistry of Elements
Intext Questions 23.
- Why do transition elements act as good catalysts?
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- Name some of the common catalysts you have studied.
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- Which of the following compounds are expected to be diamagnetic: CrCl 3 , ScCl 3 , CuSO 4 , CoCl 2 , TiCl 4 and ZnCl 2?
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- Which of the following do you expect to be colored and why, Cr+^ and Cu+?
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- Name any two alloys of transition elements.
...................................................................................................................................
- Calculate in B.M., magnetic moments expected for the following ions:
V4+, Ni3+, V4+, Ni3+, Cr3+^ and Ti4+.
...................................................................................................................................
23.5 Important Compounds of Transition Elements
The preparation, properties and applications of two important compounds of transition elements viz. K 2 Cr 2 O 7 and KMnO 4 which are widely used in industry and laboratory are discussed below:
23.5.1 Potassium Dichromate (K 2 Cr 2 O 7 )
Mineral chromite (FeO.Cr 2 O 3 ) is the starting material for the manufacture of all chromates and dichromates. Soluble chromates are prepared using alkali metal oxides, hydroxides or carbonates whereas insoluble chromates are made by double decomposition of soluble chromates.
Large Scale Production of Potassium Dichromate from Chromite ore
A mixture of finely powdered chromite, sodium carbonate and quick lime is heated in a reverberatory furnace in free supply of air. Carbon dioxide is evolved and sodium chromate is formed. The function of quick lime is to keep the mass porous and prevent fusion.
4FeO.Cr 2 O 3 + 8Na 2 CO 3 + 7O 2 2Fe 2 O 3 + 8Na 2 CrO 4 + 8CO 2 Chromite
The mass after roasting is extracted with water, which dissolves soluble sodium chromate leaving behind insoluble ferric oxide. After concentrating the solution containing sodium chromate, concentrated sulphuric acid is added.
2Na 2 CrO 4 + H 2 SO 4 Na 2 Cr 2 O 7 + Na 2 SO 4 + H 2 O
MODULE - 6 Chemistry
Notes
Chemistry of Elements Sodium sulphate produced, crystallizes out and is removed. On further concentrating the
solution, deliquescent red crystals of sodium dichromate separate out slowly on cooling. When a hot saturated solution of sodium dichromate is mixed with a saturated solution of potassium chloride, sodium chloride separates out, followed by separation of garnet red triclinic crystals of potassium dichromate.
Na 2 Cr 2 O 7 + 2KCl K 2 Cr 2 O 7 + 2NaCl
Since potassium dichromate is moderately soluble in cold water (100 gL–1^ at 298 K) but easily soluble in hot water (1000 g L–1) at 373 K, it is readily purified by recrystallization from water.
Physical Properties
K 2 Cr 2 O 7 forms orange red prismatic crystals. Its specific gravity is 2.676 and its melting point is 696 K. It is moderately soluble in cold water but highly soluble in hot water and insoluble in alcohol.
Chemical Properties
- Since chromium forms stable compounds in low oxidation states as well, potassium dichromate in which oxidation number of chromium is +6, acts as a powerful oxidizing agent. For this reason, it is used as a primary standard in volumetric analyses. In acidic solutions, one molecule of potassium dichromate furnishes three atoms (i.e. six equivalents) of available oxygen as follows:
K 2 Cr 2 O 7 + 4H 2 SO 4 Cr 2 (SO 4 ) 3 + K 2 SO 4 + 4H 2 O + 3O
The available oxygen then oxidizes ferrous, iodide ions and sulphur dioxide as follows:.
2FeSO 4 + H 2 SO 4 + [O] Fe 2 (SO 4 ) 3 + H 2 O
2HI + [O] H 2 O + I 2
SO 2 + [O] + H 2 O H 2 SO 4
These reactions can also be shown as ionic equations.
In acidic solution, the oxidizing action of K 2 Cr 2 O 7 can be represented as follows:
Cr 2 O 7 2–^ + 14H+^ + 6e–^ 2Cr3+^ + 7H 2 O
The ionic equation for the reducing action of Fe(II) can be represented as:
Fe2+^ Fe3+^ + e–
The complete ionic equation may be obtained by adding the half reaction of dichromate ion to the half reaction of Fe(II):
Cr 2 O 7 2–^ + 14H+^ + 6e–^ 2Cr3–^ + 7H 2 O
(Fe2+^ Fe3+^ + e–^ ) 6
Cr 2 O 7 2–.+ 14H+^ + 6Fe2+^ 2Cr3+^ + 6Fe3+^ + 7H 2 O
Similarly the reactions of dichromate with iodide ion and sulphur dioxide can be written as given below:
MODULE - 6 Chemistry
Notes
Chemistry of Elements 2. Write down the formula of chromite ore.
- How is sodium dichromate converted into potassium dichromate?
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- What happens when potassium dichromate is heated with an alkali metal chloride and concentrated sulphuric acid?
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- Give the molecular formula of chrome alum. What are its uses?
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- Why does dichromate act as an oxidizing agent?
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- What happens when a base is added to dichromate?
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- What is the oxidation state of chromium in (i) K 2 CrO 4 and (ii) K 2 Cr 2 O 7?
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23.5.2 Potassium Permanganate (KMnO 4 )
Pyrolusite ore (MnO 2 ) is the starting material for the manufacture of potassium permanganate. Pyrolusite is first converted into potassium manganate which is then oxidized to potassium permanganate.
Conversion of pyrolusite into potassium manganate
When pyrolusite is fused with hydroxide of sodium or potassium in the presence of air manganite first formed is converted into a dark green mass of corresponding manganate as follows :
MnO 2 + 2KOH K 2 MnO 3 + H 2 O Potassium manganite
2K 2 MnO 3 + O 2 2K 2 MnO 4 Potassium manganate
The dark green mass of potassium manganate is dissolved in a small quantity of cold water to form a dark green solution from which dark green crystals of potassium manganate may be obtained on concentraion.
Conversion of potassium manganate to potassium permanganate :
Any of the following methods can be used for preparing potassium permanganate.
- When green concentrated solution of potassium manganate is gently warmed, or largely diluted with water, the green color changes to pink owing to the formation of potassium permanganate. Potassium manganate is stable in alkaline solutions or in
d -Block and f -Block Elements MODULE - 6
Notes
pure water. But even a trace of acid, like carbonic acid, is enough to bring about its^ Chemistry of Elements
disproportionation :
3MnO 4 2–^ + 4H+^ 2MnO 4 –^ + MnO 2 + 2H 2 O
- Potassium manganate may also be converted into potassium permanganate by oxidation either chemically with chlorine or ozone or electrolytically at the anode.
Chemical oxidation:
2K 2 MnO 4 + Cl 2 2KMnO 4 + 2KCl
2K 2 MnO 4 + O 3 H 2 O 2KMnO 4 + 2KOH + O 2.
Anodic oxidation:
MnO 4 2–^ MnO 4 –^ + e–^ (at anode)
green purple
Physical properties:
Potassium permanganate forms dark purple red rhombic prisms. It is sparingly soluble in water (5.31 g in 100 mL at 298K) giving a deep purple colored solution which is opaque until very dilute. The crystals on heating evolve oxygen and form a black powder of potassium manganate and manganese dioxide.
2KMnO 4 K 2 MnO 4 + MnO 2 + O 2
Chemical properties:
Potassium permanganate is a powerful oxidizing agent. The action is different in acidic, neutral and alkaline solutions.
(i) In acidic solution, two molecules of permanganate furnish five atoms of oxygen as follows :
2KMnO 4 + 3H 2 SO 4 K 2 SO 4 + 2MnSO 4 + 3H 2 O + 5O
In ionic form the equation is:
MnO 4 –^ + 8H+^ + 5e–^ Mn2+^ + 4H 2 O
Ferrous sulphate is oxidized to ferric sulphate by acidified potassium permanganate.
2KMnO 4 + 8H 2 SO 4 + 10FeSO 4 K 2 SO 4 + 2MnSO 4 + 5Fe 2 (SO 4 ) 3 + 8H 2 O
or
MnO 4 –^ + 8H+^ + 5Fe2+^ Mn2+^ + 5Fe3+^ + 4H 2 O
Sulphur dioxide is oxidized to sulphuric acid:
2KMnO 4 + 5SO 2 + 2H 2 O K 2 SO 4 + 2MnSO 4 + 2H 2 SO 4
or
2MnO 4 –^ + 5SO 2 + 2H 2 O 2Mn2+^ + 5SO 4 2–^ + 4H+
(ii) In neutral solution the main reaction is:
d -Block and f -Block Elements MODULE - 6
Notes
7. What are the oxidation states of Mn in MnO^ Chemistry of Elements
2 , K 2 MnO 4 and KMnO 4? ...................................................................................................................................
23.6 f -Block Elements (Lanthanoides)
In addition to d -block elements, there are two rows of elements shown separately at the bottom of the periodic table. The elements from La to Lu (14 elements) are called lanthanoides. They are characterised by the filling up of the anti penultimate 4 f orbitals. They are extremely similar to each other in properties. Earlier these were called the rare earths. This name is not appropriate because many of these elements are not particularly rare. Now these elements are known as inner transition elements (because they form transition series within the d -block transition elements) or lanthanoids.
23.6.1 Electronic Configuration
Lanthanum is the first member of the third transition series, and it has one 5 d and two 6 s electrons. The next element is cerium, which while still retaining two 6 s electrons, has two electrons in the 4 f orbitals and none in the 5 d orbitals. There are 7 separate 4 f orbitals, each of which can accommodate two electrons with opposite spins. The atoms
of the elements from cerium to lutetium have two to fourteen electrons in 4 f - orbitals,
respectively. These elements constitute the first inner transition series known as lanthanides
and, although lanthanum itself does not possess any 4 f electrons, it is customary to
include this element in this series.
The filling up of the 4 f orbitals is regular with some exceptions (Table 23.6); the element europium has the outer electronic configuration 4 f^75 s^25 p^65 d^06 s^2 and the next element gadolinium has the extra electron in the 5 d orbital. The element ytterbium has a full compliment of 4 f electrons (4 f^145 s^25 p^65 d^06 s^2 ) and the extra electron in the lutetium atom enters the 5d orbitals (4 f^145 s^25 p^65 d^16 s^2 ). Except for lanthanum, gadolinium and lutetium, which have a single 5d electron, the lanthanoides do not have electrons in the 5 d orbitals.
Table 23.6: Electronic configuration of lanthanides Element Symbol Z Electronic configuration Lanthanum La 57 [Xe]4 f^05 d^16 s^2 Cerium Ce 58 [Xe]4 f^26 s^2 Praseodymium Pr 59 [Xe]4 f^36 s^2 Neodymium Nd 60 [Xe]4 f^46 s^2 Promethium Pm 61 [Xe]4 f^56 s^2 Samarium Sm 62 [Xe]4 f^66 s^2 Europium Eu 63 [Xe]4 f^76 s^2 Gadolinium Gd 64 [Xe]4 f^75 d^16 s^2 Terbium Tb 65 [Xe]4 f^96 s^2 Dysprosium Dy 66 [Xe]4 f^106 s^2 Holmium Ho 67 [Xe]4 f^116 s^2 Erbium Er 68 [Xe]4 f^126 s^2 Thulium Tm 69 [Xe]4 f^136 s^2 Ytterbium Yb 70 [Xe]4 f^146 s^2 Lutetium Lu 71 [Xe]4 f^145 d^16 s^2
MODULE - 6 Chemistry
Notes
Chemistry of Elements
23.6.2 The lanthanoide contraction
Each succeeding lanthanoide differs from its immediate predecessor in having one more electron in the 4 f orbitals (except for some exceptions as discussed above) and one extra proton in the nucleus of the atom. The 4f electrons constitute inner shells and are rather ineffective in screening the nucleus; thus there is a gradual increase in the attraction of the nucleus for the peripheral electrons as the nuclear charge increases, and a consequent contraction in atomic radius is observed. For example, the ionic radii of the +3 cations decrease steadily from a value of 115 pm for La3+^ to a value of 93 pm for Lu3+. The regular decrease in atomic radii with increase in atomic number is known as lanthanoide contraction.
The lanthanoide contraction considerably influences the chemistry of the elements, which succeed the lanthanides in the periodic table; for instance the atomic radii of zirconium (At. No. 40) and hafnium (At. No. 72) are almost identical and the chemistry of these two elements is strikingly similar. Incidentally, the density of hafnium (which immediately follows the lanthanides) is almost twice the density of zirconium (which is in the same group).
Intext Questions 23.
- How many elements constitute lanthanoide series?
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- Why Zr and Hf show almost same properties?
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- Write down the electronic configuration of the following in the ground state: Gd, Lu, Ho, Er.
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- Write down the electronic configuration of the following ions: Eu3+, Y63+, Ce4+.
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What You Have Learnt
Transition elements have partially filled d -orbitals either in atomic or ionic state.
They show general electronic configuration (n-1) d 1–10n s 1,2.
They show high M.P. and B.P. due to strong inter-atomic bonding.
They show variable oxidation states.
They form colored ions and compounds.
They show paramagnetic behaviour.
They form complexes.
MODULE - 6 Chemistry
Notes
Chemistry of Elements 15. Discuss the main characteristic features of the transition elements with special
reference to their atomic size, variable oxidation states, magnetic and catalytic properties.
- Explain the trends of variations of:
(a) melting and boiling points.
(b) atomic radius in the first transition series.
- A solution of KMnO 4 on reduction yields either a colorless solution or a brown precipitate or a green solution depending on the pH of the solution. What different stages of the reduction do these represent and how are they carried out?
- A black color compound [X] of manganese when fused with KOH under atmospheric oxygen gave a green colored compound [Y]. When the compound [Y] was treated with an oxidizing agent (chlorine or ozone), it gave a purple colored solution [Z]. Identify X, Y, Z and write the chemical equation.
- Compound [A] of chromium when treated with sodium carbonate in the presence of atmospheric oxygen gave a yellow colored compound [B]. Compound [B] on treatment with acid gave an orange colored compound [C]. [B] can also be obtained by treatment of [C] with alkali. Identify the compound A, B, C and write the chemical equations.
- Why do transition elements form a large number of alloys and interstitial compounds?
- What are lanthanides? Why are they called inner transition elements?
- What is lanthanide contraction and what are its consequences?
- Write the electronic configurations of the following in ground state:
Eu, Ho and Gd.
- Describe two oxidizing properties of potassium dichromate.
- Describe two oxidizing properties of potassium permanganate.
Answers to Intext Questions
23.
- Transition elements are defined as “Elements whose atoms have partially filled d -orbitals either in the atomic or in ionic state (common oxidation state).
- 10 elements constitute the first transition series. These are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn.
- Since zinc does not have partially filled d -orbitals either in atomic or in ionic state.
- Because there are partially filled d -orbitals in Cu2+, Ag2+^ and Au3+.
23.
- General electronic configuration of transition elements is: (n–1) d 1–10^ n s 1 or 2.
- Sc = [Ar]3 d^14 s^2 , Cr = [Ar]3 d^54 s^1 , Zn = [Ar]3 d^104 s^2 , Cu = [Ar] 3 d^10 4 s^1
d -Block and f -Block Elements MODULE - 6
Notes
3. Cr3+^ = [Ar]3 d^3 , Ti4+^ = [Ar]3 d^0 , Ni3+^ = [Ar]3 d^7 and Cu2+^ = [Ar]3 d^9^ Chemistry of Elements
- Because less amount of energy is required to remove an electron from 4 s instead of 3 d orbital. It is due to the fact that after Sc, 3 d , becomes lower in energy than 4 s.
23.
- due to strong interatomic bonding.
- due to increase in effective nuclear charge.
- due to small size. Size does not increase in the same proportion as the atomic mass.
23.
- due to variable oxidation states.
- V 2 O 5 (contact process for H 2 SO 4 ) and iron (Haber’s process)
- ScCl 3 (3 d^0 ), TiCl 4 (3 d^0 ) and ZnCl 2 (3 d^10 ).
- Cr+^ because it has partially filled d- orbital i.e. 3 d^5.
- Nichrome and brass.
- V4+ 3 d^1 n n (^2) = n (1 2) 3 1.73 B.M., Ni3+ 3 d^7 n = 3, 3 3 (^2) = 15 =
3.87 B.M
23.
- Chromite ore, Na 2 CO 3 and O 2
- FeO.Cr 2 O 3.
- Na 2 Cr 2 O 7 + 2KCl K 2 Cr 2 O 7 + 2 NaCl
- K 2 Cr 2 O 7 + 4NaCl = 6H 2 SO 4 2KHSO 4 + 4NaHSO 4 + 2CrO 2 Cl 2 + 3H 2 O
- KCr(SO 4 ) 2 .12H 2 O or K 2 SO 4 .Cr 2 (SO 4 ) 3 .24H 2 O, tanning of leather and dyeing fabrics.
- For Cr, +3 is the stable oxidation state but in K 2 Cr 2 O 7 the oxidation state of Cr is +6.
- Cr 2 O 7 2–^ + 2OH–^ 2CrO 4 2–^ + H 2 O. Dichromate changes to chromate.
- (i) +6 (ii) +6.
23.
- By oxidation with ozone or chlorine
2K 2 MnO 4 + O 3 + H 2 O 2KMnO 4 + 2KOH + O 2
2K 2 MnO 4 + Cl 2 2KMnO 4 + 2KCl
- Alkaline:
2MnO 4 –^ + H 2 O + I–^ 2MnO 2 + 2OH–^ + IO 3 –
