1. IIT-JEE Syllabus
Transition elements (only the first row, 3d series): Definition, Werner’s approach to coordination compounds, general characteristic properties [viz., variable oxidation states, colour (excluding the details of electronic transition) calculation of spin-only magnetic moment, formation of complexes (stereochemistry excluded), nomenclature of simple coordination compounds, valence bond approach to define geometries of coordination compounds of linear, tetrahedral, octahedral and square planar geometries.
- Transition Elements – Introduction
2.1 Definition and Electronic Configurations of Atoms
The elements lying between s and p-block elements of the periodic table are collectively known as transition or transitional elements (T.E.’s): These elements either in their atomic state or in any of their common oxidation state have partly filled (n-1) d-orbitals of
(n-1)th main shell. In these elements the differentiating electron enters (n-1)d orbitals of (n-1)th main shell and as such these are called d-block elements.
The valence shell configurations of these elements can be represented by
(n-1)d1–10.ns0, 1, 2. The configurations clearly show that strictly, according to the definition of d-block elements, Cu, Ag and Au should be excluded from d-block elements, since these elements, both in their atomic state [with configuration (n-1)d10ns1] and in their +1 oxidation state [with configuration (n-1)d10], do not have partly filled (n – 1)d-orbitals. Similarly Zn, Cd and Hg which both in their atomic state [(n-1)d10ns2] and in +2 oxidation [(n-1)d10] do not contain partly filled (n-1)d orbitals, should also be excluded from d-block elements. Similar is the case with Pd atom with configuration 4d105s0. Yet, in order to maintain a rational classification of elements, these elements (viz Cu, Ag, Au, Zn, Cd, Hg and Pd) are also generally studied with d-block elements.
All the d-block elements are classified into four series viz 3d, 4d, 5d and 6d series corresponding to the filling of 3d, 4d, 5d and 6d orbitals of (n-1)th main shell. Each of 3d, 4d and 5d series has ten elements while 6d series has at present only one element viz Ac80 whose valence shell configuration is 6d1 7s2.
2.2 Irregularities in Configurations
The irregularities in the observed configurations of Cr (3d54s1), Cu (3d104s1), Mo (4d55s1), Pd (4d105s0), Ag (4d10 5s1) and Au (5d106s1) are explained on the basis of the concept that half-filled and completely filled d-orbitals are relatively more stable than other d-orbitals.
On the basis of the above concept it is, however, not easy to explain the irregularities found in the observed electronic configurations of the atoms of other elements, since one has to consider the net effect of so many other factors such as (i) nuclear electronic attraction (ii) shielding of one electron by several other electrons (iii) inter-electronic repulsion (iv) the exchange-energy forces etc. All these factors play an important part together in determining the final stability of an electronic configuration of an atom. It is not easy to explain why W unlike Cr(3d55s1) and Mo(4d5 5s1) should have the idealised electronic configuration (4f145d46s2).
The properties of transition elements of any given period are not so much different from one another as those of the same period of non-transition elements. The reason of this fact lies in the electronic configuration of transition elements. We know that electronic configurations of transition elements is invariably (n-1)d1–10 ns0 or 1 or 2 which indicates that (i) the electronic configurations of transition elements differ from one another only in the number of electrons in d orbitals in the (n-1)th shell and (ii) the number of electrons in the outermost shell, ns, is invariably 1 or 2.
- Physicochemical Properties
3.1 Metallic Character
All the transition elements are metals, since the number of electrons in the outermost shell is very small, being equal to 2. They are hard, malleable and ductile. They exhibit all the three types of structures: face centred cubic (fcc), hexagonal close packed (hcp) and body centred cubic (bcc). Metals of VIII and IB Groups are softer and more ductile than other metals. It appears that covalent and metallic bonding both exist in the atoms of transition metals. The presence of unfilled d-orbitals favours covalent bonding. These metals are good conductors of heat and electricity.
3.2 Melting and Boiling Points
The transition elements have very high melting and boiling points as compared to those of s and p block elements. Zn, Cd and Hg have relatively low values. The reason for these low values is that these metals have completely filled d-orbitals with no unpaired electron that may be available for covalent bonding amongst the atoms of these metals. The formation of covalent bonding occurs in the rest of the d-block elements on account of the presence of unfilled d-orbitals.
Although melting and boiling points show no definite trends in the three transition series, the metals having the highest melting and boiling points are towards the middle of each transition series.
3.3 Atomic (Covalent) and Ionic Radii
It will be seen that atomic and ionic radii values decrease generally, on moving from left to right in the period. This is due to the fact that an increase in the nuclear charge tends to attract the electron cloud inwards. The atomic radii for the elements from Cr to Cu are, however, very close to one another. This is due to the fact that simultaneous addition of electron to 3d-level exercise the reverse effect by screening the outer 4s-electrons from the inward pull of the nucleus. As a result of these two opposing effects, the atomic radii do not alter much on moving from Cr to Cu.
The ionic radii of M2+ and M3+ ions follow the same trends as their atomic radii. The radii of M2+ ions, although somewhat smaller than that of Ca2+ ion (0.99Å) are comparable with it. Thus MO oxides of the transition element should be similar to CaO in many ways, although somewhat less basic and less soluble in water. Similarly the hydration energies of M2+ ions should be similar to but somewhat greater than that of Ca2+ ion. This is borne out by facts, since the hydration energy of Ca2+ ion is 395 kcal and the observed values of hydration energies for the elements Ti2+ … Cu2+ are between 446 kcal and 597 kcal.
The radii of M3+ ions are slightly greater than that of Ga3+ ion (0.62Å). Thus M2O3 oxides of transition elements should be similar to but slightly less acidic (more basic) than Ga2O3 and the hydration energies of M3+ ions should be less than 1124 kcal which is the hydration energy of Ga3+ ion. The observed values of hydration energies for the series Sc3+ … Fe3+ are between 947 kcal and 1072 kcal.
3.4 Ionisation Potentials
The first ionisation potentials of transitional elements lie between the values of those of s- and p-block elements. The first ionisation potentials of all the transition elements lie between 5 to 10 electron volts. In case of transition elements the addition of the extra electron in the (n-1) d level provides a screening effect which shields the outer ns electrons from the inward pull of positive nucleus on the outer ns electrons. Thus the effects of the increasing nuclear charge and the shielding effect created due to the expansion of (n-1)d orbital oppose each other. On account of these counter effects, the ionisation potentials increase rather slowly on moving in a period of the first transition series.
3.4.1 First ionisation potentials
It is evident that the values for the first four 3d block elements (Sc, Ti, V and Cr) differ only slightly from one another. Similarly the values for Fe, Co, Ni and Cu also are fairly close to one another. The value of II for Zn is considerably higher. This is due to the extra-stability of 3d10 level which is completely filled in Zn atom.
3.4.2 Second ionisation potentials
The second ionisation potentials are seen to increase more or less regularly with the increase of atomic number. The value of III for Cr and Cu are higher than those of their neighbours. This is due to the fact that the electronic configurations of Cr+ and Cu+ ions have extra stable 3d5 and 3d10 levels.
There is a sudden fall in the values of ionisation potentials in going from II B (Zn-group elements) to IIIA sub-group. This sudden fall is explained on the basis that in case of IIIA group elements the electron to be removed is from a 4p-orbital which is incompletely filled, while in case of the II B group elements, the electron to be removed is from 4s-orbital which is completely filled. Thus more energy will be required to remove an electron from a filled 4s-orbital in comparison to that used to remove an electron from a 4p-orbital which is incompletely filled.
3.4.3 Electropositive character of transitional elements as compared to that of alkali metals and alkaline earth metals.
The values of first ionisation potentials of transition elements in most cases lie between those of s-and p-block elements. Thus the transition elements are less electropositive than the elements of I A and II A groups. Thus, although the transitional elements do form ionic compounds, yet they do not form ionic compounds so readily as the alkali and alkaline earth metals do. Also, unlike the alkali and alkaline earth metals, the transitional elements also have a tendency to form the covalent compounds under certain conditions. Generally the compounds in which the transitional elements show a smaller valency are ionic, while those in which a higher valency is exhibited are covalent in character.
3.5 Oxidation States
One of the most important property that distinguishes transition elements from s-and p-block elements is that they show variable oxidation states. s-and p-block elements have oxidation states either equal to their group number G or equal to (8-G). The transition elements on the other hand exhibit variable oxidation states.
This unique property is due to the fact that the energy levels of 3d, 4d and 5d orbitals are fairly close to those of 4s, 5s and 6s orbitals respectively and, therefore, in addition to ns electrons and variable number of (n-1) d electrons are also lost in getting various oxidation states.
- i) Minimum oxidation state: All the transition elements with the exception of Cr, Cu, Ag, Au and Hg which have a minimum oxidation state of +1 exhibit a minimum oxidation state of +2. In most cases this +2 oxidation state arises due to the loss of two s-electrons.
- ii) Maximum oxidation state: Each of the elements in groups III B to VII B can show the maximum oxidation state equal to its group number. Thus, Cr in group VIB shows a maximum oxidation state of +6 in Cr2O72– Most of the elements in VIII group show a maximum oxidation state equal to + 6. However, Ru and Os have a maximum oxidation state equal to +8 which is the highest oxidation state shown by any element.
iii) Relative stability of various oxidation states: The relative stabilities of various oxidation states of 3d-series elements can be correlated with the extra stability of 3d0, 3d5 and 3d10 configurations to some extent. Thus Ti4+ (3d0) is more stable than Ti3+ (3d1) and similarly Mn2+ (3d5) is more stable than Mn4+ (3d4). It may, however, be pointed out that such a generalisation for the relative stability of various oxidation states of 4d and 5d series elements is often rather difficult to realise.
The higher oxidation state of 4d and 5d series elements are generally more stable than those of the elements of 3d series, e.g., Mo, Tc (4d series elements) and W, Re (5d-series elements) form the oxyanions: MoVIO42–, TcVIIO42–, WVIO42–, ReVIIO4– which are stable and in which the transition elements concerned show their maximum oxidation states. The corresponding oxyanions of Cr and Mn namely CrVIO42– and MnVIIO4– are strong oxidising agents.
Furthermore, the highest oxidation states of second and third row elements are encountered in compounds containing the more electronegative elements viz. F, O and Cl. Thus for the compounds RuVIIIO4, OsVIIIO4, WVICl6 and PtVIF6 there are no analogs being formed by the first row elements. The lower oxidation states particularly +2 and +3 are important in the chemistry of aquated and complex ions of the 3d-series (i.e. first row) elements but these ions are not very important in the chemistry of second (i.e. 4d series) and third (5d-series) row elements. In short it may be said that in going down a sub-group the stability of the higher oxidation states increases while that of lower oxidation states decreases.
- iv) Formation of ionic and covalent compounds: Transition elements cannot form ionic compounds in higher oxidation states because the loss of more than three electrons is prevented by the higher attractive force exerted (on the electrons) by the nucleus. Higher oxidation states of these metals are not formed by the actual loss of electrons but due to the formation of new hybrid orbitals involving (n-1)d, ns and np orbitals.
The transition elements form ionic bonds in the lower oxidation states and the ionic character of the bond decreases with the increase in the oxidation state. With this decrease in the ionic character of the bond the basic character of the oxides decreases, e.g. MnO (oxidation states of Mn = +2) is basic, MnO2 (Mn = +4) is amphoteric and MnO3 (Mn = +6) is acidic.
Ionic and covalent compounds of transition elements are usually markedly coloured, in contrast to compounds of the s and p block elements which are often white and are generally not strongly coloured. Colour is associated with incomplete electron shells and the ability to promote an electron from one energy level to other. Exactly the right amount of energy to do this is obtained by absorbing the light of a particular wave length.
Illustration 1: Why Zn+2 salts are white while Ni2+ salts are blue
Solution: Zn+2 has completely filled d-orbitals (3d10) while Ni2+ has incompletely filled
In the transition elements, d-electrons are promoted to a higher energy level within the
d-subshell. This corresponds to a fairly small energy difference, and so light is absorbed in the visible region. If red light is absorbed then the transmitted light contains an excess of the other colours of the spectrum-particularly blue, so that the compound appears blue, for example Cu2+.
Exercise 1: [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless. Explain
3.7 Complex Formation
The transition elements have an unparallel tendency to form coordination compounds with Lewis Base, i.e., with groups which are able to donate an electron pair. These groups are called ligands. A ligand may be a neutral molecule such as NH3 or ion such as Cl– or CN– etc.
Co3+ + 6(NH3) ¾® [Co(NH3)6]3+
Fe2+ + 6CN– ¾® [Fe(CN)6]4–
The reason transition elements are good in forming complexes are:
- i) Small size and high effective nuclear charge
- ii) Availability of low lying vacant d–orbitals which can accept lone pair of electrons donated by a ligand.
3.8 Catalytic properties
Transition metals and their compounds are known to act as good catalyst due to the following reasons:
- i) Due to the their variable oxidation state, they form unstable intermediate compounds and provide a new path with lower activation energy for the reaction (Intermediate compound formation theory)
- ii) In some cases the finely divided metals or their compounds provide a large surface area for adsorption and the adsorbed reactants react faster due to the closer contact(Adsorption theory)
- TiCl3 Used as Ziegler – Natta catalyst
- V2O5 Converts SO2 to SO3 in the contact process for making H2SO4
- MnO2 Used as a catalyst to decompose KClO3 to give O2
- Fe Used in Haber – Bosch process for making NH3
- FeCl3 Production of CCl4 from CS2 and Cl2
- FeSO4 & H2O2 Fenton’s reagent
- PdCl2 Wacker process for the following conversion
C2H2 + H2O + PdCl2 ® CH3CHO + 2HCl + Pd
- Pd For hydrogenation (Phenol ® Cyclohexanone)
- Pt/PtO Adams catalyst used for reduction
- Pt SO2 ® SO3 contact process
- Pt Cleaning car exhaust fumes
- Cu In manufacture of (CH3)2SiCl2
- Cu/V Oxidation of cyclohexanol
- CuCl2 Deacon process for making Cl2 from HCl
- Ni Raney nickel
3.9 Magnetic Properties
When a substance is placed in a magnetic field of strength H, the intensity of the magnetic field in the substance may be greater than or less than H.
If the field in the substance is greater than H, the substance is paramagnetic. Thus paramagnetic materials attract lines of force and if it is free to move, a paramagnetic material will move from a weaker to a stronger part of the field. Paramagnetism arises
as a result of unpaired electron in the atom. If the field in the substance is less than H, the substance is diamagnetic. They tend to repel lines of force and move from a strong to weaker part of a magnetic field. In diamagnetic substance, electrons are paired up.
Exercise 2: Why does Mn (II) show maximum paramagnetic character amongst the bivalent ions of the first transition series?
It should be noted that Fe,Co and Ni are ferromagnetic. Ferromagnetic materials may be regarded as special case of paramagnetism in which the moments of individual atoms become aligned and all point in the same direction. It is also possible to get antiferromagnetism by pairing the moments of adjacent atoms which point in opposite directions. It occurs in salts of Fe3+, Mn2+ etc.
Paramagnetism is expressed in terms of magnetic moment, which is related to the number of unpaired electrons as follows
m = B.M.
n = number of unpaired electrons
B.M. = Bohr Magneton, unit of magnetic moment
More the magnetic moment more is the paramagnetic behaviour
Illustration 2: Calculate the magnetic moment of V3+.
Solution: The electronic configuration of V3+ is [Ar] 4s°3d2. In the d–orbitals there are 2 unpaired electrons
\ m = B.M.
= B.M. = 2.73 B.M
Illustration 3: Calculate the magnetic moments of Fe2+ and Fe3+
Solution: In Fe2+ there are 4 unpaired electrons.
In Fe3+ there are 5 unpaired electrons.
3.10 Formation of Alloys
As the transition elements have similar atomic sizes hence in the crystal lattice, one metal can be readily replaced by another metal giving solid solution and smooth alloys. The alloys so formed are hard and have often high melting point.
3.11 Interstitial Compounds
Transition metals form number of interstitial compounds, in which they take up atoms of small size e.g. H, C and N in the vacant spaces in their lattices. The presence of these atoms result in decrease in malleability and ductility of the metals but increases their tensile strength.
3.12 Compounds of Iron
3.12.1 Ferrous sulphate (Green vitriol), FeSO4.7H2O
It occurs in nature as copperas and commonly known as hara Kasis.
- a) Preparation
- i) By dissolving scrap Fe in dil. H2SO4
- ii) From Kipp’s waste which contains ferrous sulphate with some free H2SO4; the latter is neutralised with scrap iron forming FeSO4 and hydrogen.
iii) By the action of air and water on iron pyrites. The solution is treated with scrap iron to remove H2SO4 and to reduce Fe2(SO4)3 to FeSO4.
- b) Properties
- i) Hydrated and anhydrous FeSO4 are green and white in colour respectively. It is isomorphous with epsom salt, MgSO4.7H2O and ZnSO4.7H2 It effervesces on exposure to air.
- ii) Light green crystals of FeSO4 lose water and turn brown on exposure to air, due to oxidation.
iii) On heating at 300°C it gives anhydrous FeSO4 which on further heating gives Fe2O3 and SO2.
- iv) Like other ferrous salts, it takes up HNO3 forming brown coloured double compound, Fe(NO)SO4, nitroso ferrous sulphate (Ring test for nitrates).
- v) It decolourises acidified potassium permanganate and turns acidified dichromate green (reducing character).
- vi) It forms double salts with sulphates of alkali metals with general formula R2SO4.FeSO4.6H2 With ammonium sulphate, it forms a double salt known as ferrous ammonium sulphate or Mohr’s salt, FeSO4.(NH4)2SO4.6H2O. It does not effervesce. It ionises in solution to gives Fe2+, NH4+ and SO42– ions.
3.12.2 Ferric oxide, Fe2O3
- i) It occurs in nature as haematite.
- ii) Fe2O3 is a red powder, insoluble in H2O and not acted upon by air or H2O
iii) It is amphoteric in nature and reacts with acids and alkalies.
- iv) It is reduced to iron by H2,C and CO.
- v) It is used as a catalyst in the oxidation of CO to CO2 in the Bosch process.
3.12.3 Ferric chloride, FeCl3
- a) Preparation
- i) Hydrated ferric chloride (FeCl3.6H2O) can be prepared by dissolving iron, Fe(OH)3 or ferric oxide in dil. HCl.
- ii) Reaction of Fe with dry Cl2 gives anhydrous FeCl3,
- b) Properties
- i) Anhydrous salt is yellow, deliquescent compound and highly soluble in H2
- ii) Its aqueous solution is acidic due to hydrolysis.
iii) On heating it gives FeCl2 and Cl2.
- iv) It oxidizes H2S to S, SO2 to H2SO4, SnCl2 to SnCl4 and Na2S2O3 to Na2S4O6.
3.13 Copper, Silver and gold
- a) These metals are commonly called as coinage or currency metals. Their general electronic configuration is (n – 1) d10 ns1. These show variable valencies +1, +2 and +3.
- b) Gradation in properties
- i) The nobility increases from copper to gold.
- ii) The affinity for oxygen also decreases from Cu to Au.
iii) Copper forms a large number of salts followed by silver followed by gold.
- iv) The ease with which the salts of these elements are reduced increases from Cu to Au.
3.13.1 Compounds of Copper
- i) Copper sulphate, cupric sulphate or blue vitriol, CuSO4.5H2
- i) By treating copper scrap or turnings, cuprous oxide, cupric oxide or malachite with H2SO4.
- ii) By roasting copper pyrites, CuFeS2 in air.
- i) It has 5 molecules of water of crystallisation; all of which can be removed on heating, to form colourless CuSO4 (again coloured with H2O).
- ii) At high temperature it forms cupric oxide.
iii) It forms double salts with alkali sulphates, e.g. K2SO4.CuSO4.6H2O
- iv) When treated with NH4OH, it first forms precipitate of cupric hydroxide
copper (II) sulphate (Schweitzer’s reagent), used for dissolving cellulose in the manufacture of artificial silk.
- v) It reacts with KCN forming a complex compound K3[Cu(CN)4].
- vi) It liberates iodine from soluble iodides.
Used in the preparation of Bordeaux mixture (CuSO4 solution + lime) which is used to kill moulds and fungi on wines.
3.13.2 Compounds Silver
(A) Silver Nitrate (Lunar caustic) AgNO3
- a) Preparation
- i) By dissolving Ag in warm dil. HNO3.
- b) Properties
- i) It is very soluble in H2O and when comes in contact with organic substances (e.g. skin, clothes, etc.) it produces burning sensation and is reduced to metallic silver which is white like the moon Luna hence its name Lunar caustic.
- ii) On heating above its melting point it decomposes to silver nitrite.
iii) When treated with soluble halides, it forms the corresponding silver halide.
- iv) When treated with alkali, it forms silver oxide which in case of NH4OH dissolves to form complex ion.
- v) It reacts with iodine and gives AgIO3 and AgI (when AgNO3 is in excess) or HIO3 and AgI (when I2 is in excess).
- vi) With a very dilute solution of Na2S2O3, it gives white precipitate which quickly changes to yellow, brown and finally black due to the formation of silver sulphide. With conc. solution of sodium thiosulphate, it does not give any precipitate.
- c) Uses – in volumetric analysis, photography and in silvering of mirrors.
- d) Silvering of mirrors – The process is based on the reduction of an ammonical solution of silver nitrate by some reducing agent like glucose, formaldehyde, tartarate, etc.
- B) Silver Bromide, AgBr
- a) Preparation – by adding AgNO3 solution to soluble bromide solution
- b) Properties
- i) It is insoluble in water and conc. acid but soluble in excess of strong solution of ammonia (cf. AgCl is soluble in dilute solution of NH4OH, AgI is insoluble in NH4OH solution).
- ii) Silver halides are also soluble in KCN and hypo solutions
iii) On heating, it melts to red liquid
- iv) It is used as the light sensitive material in photographic films. It is the most sensitive AgX to photo-reduction.
3.14 Zinc, Cadmium and Mercury
- a) These are the elements of group 12 having electronic configuration (n – 1) d10 ns2 and +2 oxidation state. In these elements the d-subshell is full, hence these are regarded as non-transition elements which is evident from the following characteristics.
- i) They do not show variable valency except mercury
- ii) Many of their compounds are white.
iii) Their melting and boiling points are very low.
- b) Unique structure of mercurous ion – Unlike Zn and Cd, Hg exhibits +1 as well as +2 oxidation state. Thus mercurous ion exists as Hg22+ and not as Hg+.
- c) Structure of mercurous ion – It consists of two atoms linked by a covalent bond
(–Hg – Hg –)2+ and explains the diamagnetic character of mercurous ion. If it were Hg+ (presence of an unpaired electron in 6s orbital) then mercurous salt should have been paramagnetic.
- d) Anomalous behaviour of mercury,
- i) It is liquid at ordinary temperature while Zn and Cd are solids.
- ii) It is less electropositive than hydrogen and therefore does not displace hydrogen from acids while Zn and Cd does.
iii) It does not form hydroxide or peroxide, while Zn and Cd do so
- iv) Mercuric oxide, on heating, gives metallic mercury and oxygen while oxides of Zn and Cd are stable towards heat.
- v) HgCl2 is covalent while zinc and cadmium chlorides are ionic. With NH3, HgCl2 gives a white ppt. of Hg(NH2)Cl, while Zn and Cd salts form complex ions, [M(NH3)4]2+.
4. Inner Transition Elements
The f-block elements are known as inner transition elements because they involve the filling for inner sub-shells (4f or 5f)
Lanthanides: It consists of elements that follows lanthanum and involve the filling of 4f subshell
4.1 Some Important Properties Of Lanthanides
- Electronic Configuration :
[Xe] 4fn+1 5d° 6s2 or [Xe] 4fn 5d1 6s2
- Oxidation State: Lathanides exhibit the oxidation state of +3. Some of them also exhibit the oxidation state of +2 and +4.
- Colouration: Many of the lanthanides ions are coloured in solid state as well as in solutions. The colour is due to the f-f transition since they have partly filled f-orbitals.
- Lanthanide Contraction: The steady decrease in the size of lanthanide ions (M3+) with the increase in atomic number is called lanthanide contraction.
Causes: As we move down the group from left to right in a lanthanide series, the atomic number increases and for every proton in the nucleus the extra electron goes to 4f orbital. The 4f orbital is too diffused to shield the nucleus effectively, thus there is a gradual increase in the effective nuclear charge experienced by the outer electrons. Consequently , the attraction of the nucleus for the electrons in the outermost shell increases with the increase of atomic number, thus the size decreases.
4.2 Consequence of Lanthanide Contraction
- Separation of Lanthanides: Due to the similar sizes of the lanthanides, it is difficult to separate them but due to lanthanide contraction their properties slightly vary (such as ability to form complexes). The variation in the properties is utilized to separate them.
- Basic Strength of Hydroxide: Due to the lanthanide contraction, size of M3+ ions decreases and there is increase in covalent character in M–OH and hence basic character decreases.
- Similarity of second and third transition series: The atomic radii of second row transition elements are almost similar to those of the third row transition elements because the increase in size on moving down the group from second to third transition elements is cancelled by the decrease in size due to the lanthanide contraction.
Illustration 4: Why Sm2+, Eu2+ and Yb2+ ions in solutions are good reducing agents but an aqueous solution of Ce4+ is a good oxidizing agent?
Solution: The most stable oxidation state of lanthanides is +3 hence ions in +2 state tend to change to +3 state by loss of electron and those in+4 state tend to change to +3 state by gain of electron.
- Co-ordination Chemistry
When solutions containing two (or) more simple stable compounds in molecular proportions are allowed to evaporate, crystals of new substances are obtained. These substances are termed molecular (or) addition compounds. Some common examples are
K2SO4 + Al2(SO4)3 + 24H2O ¾® K2SO4×Al2(SO4)3×24H2O
CuSO4 + 4NH3 ¾® CuSO4×4NH3
The molecular (or) addition compounds are of two types
- i) Double salts: The addition compounds which are stable in solid state only but are broken down into individual constituents when dissolved in water are called double salts. Their solutions have the same properties as the mixture of individual compounds.
When Mohr’s salt, [FeSO4×(NH4)2SO4×6H2O] is dissolved in water it exhibits the properties of FeSO4 and (NH4)2SO4 i.e. they produce Fe2+, NH4+ and SO42– ions in solution. Thus each ion has its identity in double salt.
- ii) Co-ordination compounds: The addition compounds in which some of the constituent ions (or) molecules lose their identity and when dissolved in water they do not break up completely into individual ions are called co-ordination compounds. The properties of their solutions are different than those of their constituents. When CuSO4.4NH3 are dissolved in water there is hardly any evidence for the presence of Cu2+ ions (or) NH3
A new ion [Cu(NH3)4]2+ is furnished in which NH3 molecules are directly linked with the metal ion
CuSO4×4NH3 [Cu(NH3)4]2+ + SO42–
Thus, a co-ordination compound may be defined as a molecular compound that results from the combination of two (or) more simple stable molecular compounds and retains its identity in the solid as well as in dissolved state. The properties of such compounds are totally different than individual constituents. A co-ordination compound contains very often but not always a complex ion.
A complex ion may be defined as an electrically charged radical which consists a central metal atom (or) ion surrounded by a group of ions (or) neutral molecules (or) both.
5.2 Iupac Nomenclature for Co-ordination Compounds
- Positive part of a co-ordination compound is named first and is followed by the negative part.
- In naming the complex ion, the name of the ligands are given in alphabetical order regardless of their charge followed by metal.
- When there are several ligands of the same kind, we normally use the prefixes di, tri, tetra, penta and hexa to show the number of ligands of that type. An exception occurs when the name of the ligand includes a number (di, tri etc). To avoid confusion in such cases bis, tris and tetrakis are used instead of di, tri and tetra and the name of the ligand is placed in brackets.
- i) Negative ligands end in ‘O’, for example:
F– – Fluoro H– – hydrido OH– – hydroxo
Cl– – Chloro I– – iodo NO2– – nitro.
- ii) Neutral groups have no special endings, Examples include NH3 ammine, H2O aqua, CO carbonyl and NO nitrosyl. The ligands N2 and O2 are called dinitrogen and dioxygen. Organic ligands are usually given their common names for example phenyl, methyl, ethylenediamine, pyridine, triphenylphosphine.
iii) Positive ligands end in – ium e.g. NH2-NH3+ hydrazinium.
The spellings of ammine with two m’s distinguishes it from organic amines.
- The oxidation state of the central metal is shown by Roman numeral in brackets immediately following its name (i.e. no space, e.g. titanium(IV) ).
- Sometimes a ligand may be attached through different atoms. Thus M-NO2 is called nitro and M-ONO is called nitrito. Similarly SCN group may bond M-SCN thiocyanato or M-NCS isothiocyanato. These may be named systematically thiocyanato-S or thiocyanato-N to indicate which atom is bonded to the metal. This convention may be extended to other cases where the mode of linkage is ambiguous.
- When writing the formula of complexes, the complex ion should be enclosed by square brackets. The metal is named first, then the co-ordinated groups are listed in the order; negative ligands, neutral ligands; positive ligands (and alphabetically according to the first symbol with in each group).
- If a complex contains two or more metal atoms, it is termed polynuclear. The bridging ligands which link the two metal atoms together are indicated by the prefix m. If there are two or more bridging groups of the same kind, this is indicated by di-m, tri-m etc. Bridging groups are listed alphabetically with the other groups unless the symmetry of the molecule allows a simpler name. If a bridging group bridges more than two metal atoms it is shown as m3, m4, m5 or m6 to indicate how many atoms it is bonded to.
- The complete metal name consists of the name of the metal, followed by-ate if the complex is an anion, which in turn is followed by the oxidation number of the metal, indicated by roman numerals in parenthesis. (An oxidation state of zero is indicated by 0 in parenthesis). When there is a latin name for the metal, it is used to name the anion (except for mercury). These names are given in the following table.
|Latin name||Anion name|
Illustration 5: Write down the IUPAC name of the complex K4[Fe(CN)6].
Solution: Firstly the +ve part should be named followed by the negative part in which the name of the ligand should be given in alphabetical order with the metal in the negative part ending in – ate (oxidation state in parenthesis) Thus the name is Potassiumhexacyano – C-ferrate (II). Hyphen C is shown to indicate CN– bonded via carbon
Illustration 6: Write down the IUPAC name of K2[Fe(NC)3Cl2(NH3)2].
Solution: The positive part is named first followed by the negative part. In
the negative part the names are written in alphabetical order followed
by metal. So the name is Potassiumdiamminedichlorotricyano-N-
Illustration 7: Write the IUPAC name of [Co(NH3)4(NO2)2]Cl
Solution: In the earlier two examples the negative part was the complex part while in this case the positive part is the complex. So it is named first with ligands in alphabetical order followed by metal (but not ending in –ate as the metal belong to the positive part of the complex). This is followed by the negative part.
So the name is Tetraamminedinitrocobalt (III) chloride.
Illustration 8: Write down the IUPAC name of [Co(NH3)2Cl(en)2]Cl2
Solution: The approach is same as the earlier one with the exception that in case of -en which is actually ethylene diammine the term bis – comes to indicate two – en groups instead of – bi
The name is Diamminechlorobis (ethylene diamine) cobalt(III) chloride
Illustration 9: Write IUPAC name of [Pt(Py)4][PtCl4].
Solution: Here both the positive and negative part has the same metal. Procedure is same as earlier for the IUPAC name. Tetrapyridineplatinum(II) tetrachloroplatinate(II).
Illustration 10: Write the IUPAC name of [Fe(NH3)4O2C2O4]Cl
Solution: In this charge on the complex part is +1. The ligand oxalato has a charge of –2, so iron should be in +3 state meaning O2 to be neutral. Now had O2 been superoxo (O2–) or peroxo (O2– – ) the negative charge of the ligands should have been –3 and –4 respectively. In that case Iron has to be +4 and +5 which is not possible. So O2 will behave as a neutral ligand and IUPAC name is Tetraammineoxalatodioxygeniron (III) chloride.
5.3 Werner’s Co-ordination Theory
Werner’s Co-ordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. Werner was able to explain the nature of bonding in complexes, and he concluded that in complexes the metal shows two different sorts of valency;
- Primary Valencies: The complex commonly exists as a positive ion. The primary valency is the number of charges on the complex ion. The complex [Co(NH3)6]Cl3 actually exists as [Co(NH3)6]3+ and 3Cl–. Thus the primary valency is 3, as there are three ionic bonds.
- Secondary Valencies: These are directional. In modern terms the number of secondary valencies equals the number of ligand atoms coordinated to the metal. This is now called the coordination number. Secondary valencies are directional, and so a complex ion has a particular shape, e.g. the complex ion [Co(NH3)6]3+ is octahedral. Whenever the complex is formed the secondary valency should always be satisfied. This he proved with the following example.
CoCl3.6NH3 + AgNO3 ¾® 3AgCl
CoCl3.5NH3 + AgNO3 ¾® 2AgCl
CoCl3.4NH3 + AgNO3 ¾® 1AgCl
Werner deduced that in CoCl3. 6NH3 the three chlorines acted as primary valencies, and the six ammonias as secondary valencies. In modern terms the complex is written [Co(NH3)6]Cl3. The three Cl– are ionic and hence are precipitated as AgCl by AgNO3. In the second complex two moles of AgCl is precipitated indicating the presence of two chlorine atoms in the outer sphere i.e., in order to fulfill the secondary valency one chlorine from the outer sphere drifts into the inner sphere. Similar is the case with the third complex. So actual structure becomes
[Co(NH3)6]Cl3, [Co(NH3)5 Cl]Cl2, [Co(NH3)4 Cl2]Cl,
5.4 Bonding in Co-ordination Compounds
5.4.1 Factors governing complex formation
- i) Transition metals have greater positive charge which attract the negatively charged ligands to form stable complexes.
- ii) Transition metals also have vacant-d-orbitals which can accommodate the lone pair of electrons donated by the ligands in co-ordinate bond formation.
5.4.2 Effective Atomic Number
Co-ordinate bonds are formed between ligands and the central metal ion in a complex; that is, a ligand donates an electron pair to the metal ion. Co-ordination compounds are formed very readily by the transition metals since they have vacant-d-orbitals which can accommodate donated electron pairs.
The number of co-ordinate bonds which can be formed largely depends on the number of vacant orbitals of suitable energy. In many cases ligands are added until the central metal in the complex possesses (or) shares the same number of electrons as the next inert gas. The total number of the electrons on the central metal in the complex including those gained by bonding, is called effective atomic number (EAN).
Thus by forming complexes many metals obtain an EAN of the next inert gas. However a significant number of exceptions are known where EAN is one (or) two units more than the corresponding inert gas.
Fe Þ atomic number is 26
and forming a complex, [Fe(CN)6]4–
the number of electrons lost = 2
the number of electrons gained = 12 Þ EAN = 36
The tendency to attain an inert gas configuration is a significant factor but not a necessary condition for complex formation, because it is also necessary to produce a symmetrical structure irrespective of the number of electrons involved.
5.4.3 The d-orbitals
Since d-orbitals are frequently used in co-ordination complexes it is important to study their shapes and distribution in space. There is no unique way of representing the five d-orbitals, but the most eminent representations are shown below.
In fact there are six wave functions that can be written for orbitals having the typical four-lobed form. In as much as there can be only five d-orbitals having any physical reality, one of them is conventionally regarded as a linear combination of two others, the and . Thus these latter two orbitals have no independent existence, but the can be thought of as having the average properties of the two.
Therefore since both have high electron density along the z axis, the orbital has a large fraction of its electron density concentrated along the same axis. Also, since one of the component wave functions has lobes along the y-axis and the other along the x-axis, the resultant orbital has a torus of electron density in the xy plane. The xy component which is often referred to as a ‘doughnut’ (or) a ‘collar’, is frequently neglected in pictorial representations, especially when an attempt is being made to portray all five d-orbitals simultaneously.
The five d-orbitals in an isolated gaseous metal ion are degenerate. If a spherically symmetric field of negative charges is placed around the metal, the orbitals will remain degenerate, but all of them will be raised in energy as a result of repulsion between the negative field and the negative electrons in the orbitals. If the field results from the influence of real ligands the symmetry of the field will be less than spherical and the degeneracy of the d-orbitals will be removed.
5.5 Formation of an octahedral complexes
Let us consider the case of six ligands forming an octahedral complex. For convenience, we may regard the ligands as being symmetrically positioned along the axes of a Cartesian co-ordinate system with the metal ion at the origin. To simplify the situation, we can consider an octahedral complex as a cube, having the metal ion at the centre of the body and the 6 ligands at the face centres.
and if we take the metal ion as the origin of a Cartesian co-ordinate, the ligands will be along the axes. As in the case of a spherical field, all of the d-orbitals will be raised in energy relative to the free ion because of negative charge repulsions. However it should be pictorially obvious that not all of the orbitals will be affected to the same extent. The orbitals lying along the axes will be more strongly repelled than the orbitals with lobes directed between the axes (dxy, dxz, dyz). The d-orbitals are thus split into two sets with the at a higher energy than the other three.
5.5.1 Factors affecting the magnitude of D0
- i) Oxidation state of the metal ion: The magnitude of D0 increases with increasing ionic charge on the central metal ion. As the ionic charge on the metal ion increases greater is the attraction for the ligands, greater the repulsion and hence greater the magnitude of D0.
- ii) Nature of the ligands: Based on experimental observation for a wide variety of complexes, it is possible to list ligands in order of increasing field strength in a spectrochemical series. Although it is not possible to form a complete series of all ligands with a single metal ion, it is possible to construct one from overlapping sequences, each constituting a portion of the series:
I– < Br– < S2– < SCN– < Cl– < N3–, F– < urea, OH– < ox, O2–, H2O < NCS– < py, NH3 < en < bpy, phen < NO2– < C6H5– < CN– < CO.
The spectrochemical series and other trends described allow one to rationalise differences in spectra and permit some predictabiltiy. If the splitting of the d-orbitals resulted simply from the effect of point charges (ions (or) dipoles), one should expect that anionic ligands would exert the greatest effect. To the contrary most anionic ligands lie at the low end of the spectrochemical series. Further more, OH– lies below the neutral H2O molecule and NH3 produces a greater splitting than H2O. Despite its imperfections, the basic theory can be used to interpret a number of effects in co-ordination chemistry.
5.5.2 Outer orbital and Inner orbital complexes
Consider the complexes [CoF6]3– and [Co(NH3)6]3+
The electronic configuration of Co3+ ion is
In a weak ligand field such as [CoF6]3–, the approach of the ligand causes only a small split in the energy level.
|Since the ligand is a weak field ligand, its repulsions with the electrons in orbitals are very less (or) in other words we can say that the electrons in and cannot move away from the approaching ligands since they have insufficient energy to pair up with the electrons in dxy, dyz and dxz orbitals.|
Thus there are no vacant orbitals in the 3d shell and the ligands occupy the first six vacant orbitals (one 4s, three 4p and two 4d). Since outer d orbitals are used, this is an outer orbit complex. The orbitals are hybridised and are written sp3d2 to denote this. Since none of the electrons has been forced to pair off, this is a high spin complex and will be strongly paramagnetic because it contains 4 unpaired 3d electrons.
Under the influence of a strong ligand field as in the complex [Co(NH3)6]3+, the approach of the ligand causes a greater split in the energy level.
|Since, the split is very high, we can say that the energy difference between the two sets of orbitals is much greater and this energy difference is sufficient to allow the electrons in orbitals to move into the half filled dxz, dxy and dyz orbitals, even though this pairing requires energy. We can also view this like, the ligand repel the electrons in higher energy level to an extent such that they get paired up against Hund’s rule|
|The and orbitals become vacant. The six ligands each donate a lone pair to the first six vacant orbitals, which are: two 3d, one 4s and three 4p. Inner d-orbitals are used and so this is an inner orbital complex. The orbital are hybridised and written d2sp3 to denote the use of inner orbitals.|
Since, the orginal unpaired electrons have been forced to pair off, theis is a low spin complex and is in fact diamagnetic.
The inner and outer orbital complexes may be distinguished by magnetic measurements. Since the outer orbital complexes use high energy levels, they tend to be more reactive. The inner orbital are sometimes called inert orbitals.
Exercise 3: All the octahedral complexes of Ni2+ must be outer orbital complexes. Explain
5.6 Formation of a square planar complex
If the central metal ion has eight d-electrons, these will be arranged as
In a weak octahedral ligand field, a regular octahedral complex is thus formed by using outer d-orbitals.
However, under the influence of a strong ligand field, the electrons in the and orbitals may pair up, leaving one vacant d-orbital, which can accept a lone pair from a ligand.
For example consider [Ni(CN)4]2–
The electronic configuration of Ni2+ ion is
|Consider, a Ni2+ ion with one electron in the orbital and one in the orbital. The approach of ligands along x, y and z axes will result in the energy of these orbitals increasing. Because the orbital is attacked by four ligands and the by only two, the energy of orbital will increase most. If the ligands have enough strong field, the electrons will be forced out of the orbital into the . Thus four ligands|
can approach along x, –x, y and –y axes. A ligand approaching in the z (or) – z direction attempting to donate a lone pair meets the very strong repulsive forces from a completely filled orbital. Thus only four ligands succeed in bonding to the metal.
Exercise 4: Three geometrical isomers of the square planar complex [Pt(NH3)(H2O(py)(NO2]+ are possible. What are they?
5.7 Formation of Tetrahedral complexes
A regular tetrahedron is related to a cube with an atom at the centre and four of the eight corners occupied by ligands.
The directions x,y and z point to the centre of the faces. The and orbitals point along x,y and z axis and dxy, dyz and dxz orbitals point in between x,y and z.
The directions of approach of the ligands does not coincide exactly with either the and orbitals (or) the dxy, dyz and dxz orbitals. The approach of ligands raises the energy of both sets of orbitals, but since the dxy,dyz and dxz orbitals correspond more closely to the position of the ligands, their energy increases most and the and orbitals are filled first. This is opposite to what happens in octahedral complexes.
Consider, the complex ion, [FeCl4]–. The electronic configuration of Fe3+ ion is
Since Cl– ion is a weak field ligand it is unable to pair the unpaired electrons and hence, the Cl– ion uses 4s and 4p orbitals to form a tetrahedral complex of sp3 hybridisation.
Exercise 5: [Ni(CN)4]2– is square planar and diamagnetic whereas (NiCl4]2– is tetrahedral and paramagnetic. Explain
Some of the factors which favour complex formation have already been mentioned. Small highly charged ions with suitable vacant orbitals of right energy, the satisfaction of the effective atomic number rule and the attainment of a symmetrical shape, all favour complex formation.
In some complexes a group occupies more than one co-ordination position, that is more than one atom in the group is bonded to the central metal.
For example ethylene diamine (en) forms a complex with copper ions
In this complex the copper is surrounded by four –NH2 groups, each nitrogen atom donating a lone pair of electrons and forming a co-ordinate bond. Thus each ethylene diamine molecule is bonded to the copper in two places, hence it is called a bidentate group (or) ligand. A ring structure is thus formed and such ring structures are called chelates. Normally chelate complexes are more stable than similar non-chelated complexes.
The more rings that are formed, the more stable the complex is chelating agent with three –, four – and six – donor atoms are known and are termed tri –, tetra – and hexa– dentate ligands.
Chelate compounds are even more stable when they contain a system of alternate double and single bonds. The p electron density is delocalised and spread over the ring, which is said to be stabilized by resonance.
Compounds that have the same chemical formula but different structural arrangements are called isomers. Because of the complicated formulae of many coordination compounds, the variety of bond types and the number of shapes possible, many different types of isomerism occur.
- Polymerization Isomerism
This is not true isomerism because it occurs between compounds having the same empirical formula, but different molecular weights. Thus [Pt(NH3)2Cl2], [Pt(NH3)4][PtCl4], [Pt(NH3)4] [Pt(NH3)Cl3]2.
5.9.2 Ionization Isomerism
This type of isomerism is due to the exchange of groups between the complex ion and the ions outside it. [Co(NH3)5Br]SO4 is red – violet. An aqueous solution gives a white precipitate of BaSO4 with BaCl2 solution, thus confirming the presence of free ions. In contrast [Co(NH3)5SO4]Br is red. A solution of this complex does not give a positive sulphate test with BaCl2. It does give a cream – coloured precipitate of AgBr with AgNO3, thus confirming the presence of free Br– ions.
5.9.3 Hydrate Isomerism
Three isomers of CrCl3.6H2O are known. From conductivity measurements and quantitative precipitation of the ionized chlorine, they have been given the following formulae:
[Cr(H2O)6]Cl3 violet (three ionic chlorines)
[Cr(H2O)5Cl]Cl2.H2O green (two ionic chlorines)
[Cr(H2O)4Cl2].Cl.2H2O dark green (one ionic chlorine)
5.9.4 Linkage Isomerism
Certain ligands contain more than one atom which could donate an electron pair. In the ion, either N or O atoms could act as the electron pair donor. Thus there is the possibility of isomerism. Two different complexes [Co(NH3)5NO2]Cl2 have been prepared, each containing the group in the complex ion.
Exercise 6: How many isomers are there for the complex [Co(NH3)Cl2]Cl
5.9.5 Geometric (or) Stereomerism
In disubstituted complexes, the substituted groups may be adjacent (or) opposite to each other. This gives rise to geometric isomerism. The square planar complexes such as [Pt(NH3)2Cl2] can be prepared in two forms.
This sort of isomerism can also occur in square planar chelate complexes if the chelating group is not symmetrical, for example the complexes between glycine and platinum.
Similarly disubstituted octahedral complexes, such as [Co(NH3)4Cl2]+ exist in cis and trans forms.
Illustration 11: How will you distinguish between the following isomer pairs
- a) i) [CoBr(NH3)5]SO4 and
- ii) [Co(SO4) (NH3)5]Br
- b) i) [Cr(H2O)6]Cl3 and
- ii) [CrCl(H2O)5]Cl2H2O
Solution: a) Isomer (i) gives white ppt of BaSO4 with BaCl2 whereas isomer (ii) does not form a ppt.
- b) The water molecule in isomer (ii) is lost easily on heating whereas the water molecule in isomer (i) are not lost easily, being coordinated to the central atom.
5.9.6 Optical Isomerism
If a molecule is asymmetric, it cannot be superimposed on its mirror image. The two forms have the type of symmetry shown by the left and right hands and are called an enantiomeric pair. The two forms are optical isomers and are called either dextro (or) laevo (d or l ) depending on the direction they roate the plane of polarized light.
Exercise 7: How do you distinguish between the following pairs of isomers?
- a) [Cr(NH3)5Br]Cl and [Cr(NH3)5Cl]Br
- b) [Co(NH3)6] [Cr(NO2)6] and [Cr(NH3)6 [Co(NO2)6]
- Solution to Exercises
Exercise 1: Ti has one electron in the d-orbital (3d1) which can absorb energy corresponding to yellow wavelength and jump from lower energy level to higher energy level. But Sc3+ has no electron in d-orbital.
Exercise 2: Mn+2 has maximum number of unpaired electrons.
Exercise 3: The Ni2+ electron configuration is
During rearrangement only one 3d orbital may be made available by pairing the electrons. Thus inner d2sp3 hybridisation is not possible.
Exercise 5: CN– being a strong ligand pairs up the electron of the 3rd level. So one inner d-orbital is available along with 4s and 4p and the mode of hybridisation is dsp2 and hence square planar. Cl– being a weak mode of hybridisation is sp3 and hence geometry is tetrahedral.
Exercise 6: No structural isomers because co ordination number of orbital is 6 but two geometrical isomers. viz Cl2 Cis and trans.
Exercise 7: a) The isomers can be distinguished by using AgNO3 reagent. One gives curdy precipitate of AgCl soluble in ammonia while the other will form light yellow precipitate of AgBr partly soluble in ammonia.
- b) The isomers can be distinguished by passing their aq. solutions through a cation exchanger. In the isomer [Co(NH3)6] [Cr(NO2)6] the cation [Co(NH3)6]3+ will be replaced by H+ ions from the resin. The resulting solution will thus contain H3[Cr(NO2)6]. In the other case, the resulting solution after passage through the exchanger will contain H3[Co(NO2)6]. On adding KCl solution, [Co(NO2)6]3– ion will give an yellow precipitate of potassium cobaltinitrate, K3[Co(NO2)6]. Hence the resulting solutions obtained from the two isomers can be distinguished.
- Solved Problems
Problem 1: Why is hydrochloric acid not used to acidify a permangnate solution is volumetric estimation of Fe2+ and C2O42–.
Solution: This is because a part of the oxygen produced from KMnO4 + HCl is used is oxidizing HCl to Cl2
4HCl + 2(O) ® 2H2O + Cl2
Problem 2: Why is copper sulphate pentahydrate coloured?
Solution: In the presence of H2O as ligand d-orbitals of Cu(II) ions split into two sets, one with lower energy and the other with higher energy. From the white light falling on it, red wavelength is absorbed or excitation of electron from lower to higher energy level. The complimentary colour, viz blue is reflected.
Problem 3: Most of the transition metals do not displace hydrogen from dilute acids. Why?
Solution: This is because most of the transition metals have negative oxidation potentials.
Problem 4: K2 [Pt Cl6] is well known compound whereas corresponding Ni compound is not known. State a reason for it.
Solution: This is because Pt4+ is more stable than Ni4+ as the sum of four ionization energies of Pt is less than that of Ni.
Problem 5: Why have the transition elements high enthalpy of hydration?
Solution: This is due to their small size and large nuclear charge. This is so because when we move along any transition series the nuclear charge increases and size decreases.
Problem 6: Though copper, silver and gold have completely filled sets of a d-orbitals yet they are considered as transition metals. Why?
Solution: These metals in their common oxidation states have incompletely filled d-orbitals e.g. Cu2+ has 3d9 and Au3+ has 5d8 configuraton.
Problem 7: Draw structure of
- I) Zeise’s salt anion [PtCl3(h2 – C2H4)]–
- II) [Al(CH3)3]2
Problem 8: Gives the IUPAC names of the following Co-ordination compounds :
- i) [Co(NH3)6]Cl3 ii) K3[Fe(C2O4)3]
- ii) [CoCl(NH3)5]Cl2 iv) [Co(NH3)5Br]SO4
- v) Na2[Fe(CN)5NO] vi) [Ir(Ph3P)2(CO)Cl]
vii) (NH4)3[Co(C2O4)3] viii) [Al(OH)(H2O)5]SO4
- ix) [V(H2O)6]Cl3 x) [Co(NH3)4Cl2][Cr(CN)6]
Solution: i) Hexammine Cobalt (III) chloride
- ii) Potassium trioxalate ferrate (III)
iii) Pentammine chlorocoablt (III) chloride
- iv) Pentammine bromocobalt (III) sulphate
- v) Sodiumpentacyanonitrosylferrate(III)
- vi) Carbonylchlorobis(triphenylphosphine)iridium(I)
- ix) Hexaaquavanadium(III) chloride
- x) Tetraamminedichlorocobalt(III) hexacyanochromate(V)
Problem 9: Arrange the following compounds in order of increasing molar conductivity:
(a) K[Co(NH3)2(NO2)4] (b) [Cr(NH3)3(NO2)3]
(c) [Cr(NH3)5 (NO2]3 [Co(NO2)6]2 (d) Mg[Cr(NH3) (NO2)5]
Solution: The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest conductivity to highest conductivity are:
b < a < d < c
Problem 10: Write the formula of
- a) Tetrachlorocuprate(II)ion
- b) Dichlorotetraaquochromium(III)chloride
- c) Bromo Chlorotetra ammine Cobalt(III) sulphate
- d) Diammine Silver (I) hexacyano ferrate (II)
- e) Dichlorobis (ethylenediammine) chromium (III) tetrachloro palladate (II)
- f) Aluminium tetrachloro aurate (III)
Solution: a) [CuCl4]2–
- b) [Cr(H2O)4Cl2]Cl
- c) [Co(NH3)4BrCl2]SO4
- d) [Ag(NH3)2]4[Fe(CN)6]
- e) [Cr(en)2Cl2]2[PdCl4]
- f) Al[AuCl4]3
Problem 1: The transition elements are so named because
(A) they have partly filled d-orbitals
(B) their properties are similar to other elements
(C) their properties are different from other elements
(D) they lie between the s- and p-blocks
Problem 2: CrO3 dissolves in aqueous NaOH to give
(A) Cr2O72– (B) CrO42–
(C) Cr(OH)3 (D) Cr(OH)2
Problem 3: Transition elements are coloured due to
(A) small size (B) metallic nature
(C) unpaired d-electrons (D) none of these
Problem 4: Choose the correct answer for transition elements
(A) they have low melting points
(B) they do not exhibit catalytic activity
(C) they exhibit variable oxidation states
(D) they exhibit inert pair effect.
Problem 5: Pt, Pd and Ir are called noble metals because
(A) Altered Nobel discovered them
(B) they are found in active states
(C) they are inert towards many common reagents
(D) they are shining, lustrous and pleasant to look at
Problem 6: The complexes [Cr(NH3)6] [Cr(SCN)6] and [Cr(NH3)4(SCN)2] [Cr(NH3)2(SCN)4] are examples of
(A) linkage isomers (B) ligand isomers
(C) co-ordination isomers (D) ionisation isomers
Problem 7: The complex Cr2[CoCl4] has
(A) square planar structure with all electrons paired
(B) tetrahedral structure with three unpaired electrons
(C) square planar structure with one unpaired electron
(D) tetrahedral structure with oine unnpaired electron
Problem 8: The splitting of d-orbitals in tetrahedral complexes is
(A) not comparable to octahedral complexes since the d-orbitals in tetrahedral complexes does not split.
(B) reverse of the octahedral complexes
(C) same as in octahedral complexes
(D) none of these
Problem 9: Which of the following cannot act as a bridging ligand?
(A) CO (B) Cl–
(C) H2O (D) NH3
Problem 10: The type of magnetism exhibited by [Mn(H2O)6]2+ ion is
(A) diamagnetism (B) paramagnetism
(C) ferromagnetism (D) antiferromagnetism
Problem 11: Carbonate ion(CO32–) acts as a
(A) tridentate ligand (B) bidentate ligand
(C) monodentate ligand (D) tetradentate ligand
Problem 12: Which of the following co-ordination number is unknown for Co3+?
(A) 5 (B) 4
(C) 6 (D) zero
Problem 13: AgCl dissolves in NH4OH due to the formation of
(A) [Ag(NH4)2]Cl (B) [Ag(NH4)3]Cl
(C) [Ag(NH3)2]Cl (D) [Ag(NH3)2]OH
Problem 14: A group of atoms can function as a ligand only when
(A) it is a small molecule (B) it has an unshared electron pair
(C) it is a negatively charged ion (D) it is a positively charged ion
Problem 15: Which of the following is an organometallic compound?
(A) Lithium methoxide (B) Lithium acetate
(C) Lithium dimethylamide (D) Methyllithium
- Assignments (Subjective Problems)
Level – I
- Give the co-ordination number of central atom in the following
- i) [Cu(NH3)4]2+
- ii) [CuBr4]2–
- iv) [Cr(NH3)4CO3]+
- v) Na[Hg(CN)2]
- vi) Fe4[Fe(CN)6]3
- ix) [ZnCl4]2–
- x) [PtCl4]2–
- xi) [Co(en)(H2O)(Br)]Cl
- xv) K3[Fe(C2O4)3]
- Write the formula for each of the following complexes.
- i) Tetraaquadichloroiron(III) ion.
- ii) Dicyanobis(ethylenediamine) cobalt (III) chlorate
iii) Dichloro tetraammineplatinum (IV) ion
- iv) Potassium hexafluoronickelate (IV)
- v) Bromotriammineplatinum (II) nitrite
- vi) Diamminechloroethylenediaminenitroplatinum(IV) chloride
vii) Tetracyanoaurate(III) ion
viii) Hexammine cobalt (III) chloride
- ix) Diamminebis(ethylenediamine)cobalt (III) chloride
- x) Potassium tetraoxomanganate (VII)
- xi) Hexaaquairon (II) sulphate
xii) Potassium tetracyanonickelate (III)
xiii) Potassium trioxoalatoaluminate (III)
xiv) Chloronitrodiammineplatinum (II)
- xv) Potassium hexacyanoferrate (II)
xvi) Chlorodiammineplatinum (II) ion
xvii) Dichlorobis (ethylenediamine) cobalt (II) ion
xviii) Dichlorotetramminecobalt (III) ion
xix) Potassium pentacyanonitrosylcobaltate (III)
- xx) Sodium ethylenediaminetetraacetato chromate (II)
- Name the following complexes according to the IUPAC system of nomenclature
- i) Na[Au(CN)2]
- ii) [Co(en)3]Cl3
- iv) TiCl4(Et2O)2
- v) Na[Mn(CO)5]
- vi) Cr(acac)3
- ix) Ca2[Fe(CN)6]
- x) [CrCl2(H2O)4]NO3
- xi) [CoCl(NO2)(NH3)4]NO3
- xv) [Co(NH3)6]Cl3
- xx) K4[Fe(NC)6]
- Co(II) is stable in aqueous solution but in the presence of strong ligand and air, it can get oxidized to Co(II). Explain
- [NiCl4]2– is paramagnetic while (Ni(CO)4] is diamagnetic. Why?
Level – II
- In the complex ion [Co(NH3)3(H2O)2Cl]2+
- a) Identify the ligands formula and charge on each one of them
- b) Write the geometry of the complex ion
- Which tripositive lanthanide ion(s) is (are) most likely to react with chromium (II) chloride?
- The pale red colour of [Co(H2O)6]2+ ion changes into intense blue colour by the addition of chloride ion. Why?
- Ethylenediamine tetracetic acid, EDTA in the from of its calcium dihydrogen salt, is administered as an antidote for lead poisioning. Explain why this reagent might be an effective medicine. Why is the calcium salt administered rather than the free acid?
- The blue coloured solution of [CoCl4]2– ion changes to pink on reaction with HgCl2, why?
- Explain the following:
- a) Gold is not attacked by common acids but dissolves in aqua regia.
- b) Copper dissolves in aqueous KCN solution with the evolution of hydrogen.
- Square planar complexes with co-ordination number of four exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?
- FeSO4 solution mixed with (NH4)2SO4 solution (in the molar ratio 1:1) gives the test of Fe2+ ion but CuSO4 solution mixed with liquid NH3. (the molar ration 1 :4) does not give the test of Cu2+. Explain why?
- A coordination compound has the formula CoCl3.4NH3. It does not liberate ammonia but precipitates chlorides ions as silver chloride. Give the IUPAC name of the complex and write is structural formula.
- Each of the compounds [Pt(NH3)6]Cl4, [Cr(NH3)6]Cl3, Co(NH3)4×Cl3 and K2[PtCl6] has been dissolved in water to make its 0.001 M solution. Rank then in order of their increasing conductivity in solution.
Level – IIi
- State the hybridisation of Pt in the following
- a) [PtCl4]2–
- b) [Pt(NH3)2Cl2]
- c) [Pt(NH3)6]Cl4
- NH4+ion does not form complexes. Why?
- i) A ferrous salt turns brown in air
- ii) Copper hydroxide is soluble in ammonium hydroxide but not in sodium hydroxide.
- A solution containing 2.665g of CrCl3×6H2O is passed through a cation exchanger. The chloride ions obtained in solution react with AgNO3 and give 2.87g of AgCl. Determine the structure of the compound.
- Two compounds have the empirical formula Cr(NH3)3×(NO2)3. In aqueous solution one of these conducts electricity while the other does not. Deduce their probable structures.
- Answer the following
- i) Which one of Fe2+and Fe3+ ions is more paramagnetic and why?
- ii) Which of the following ions are expected to coloured and why?
Fe2+, Mn2+, Cr3+, Cu+, Se3+, Ti+
iii) Name the two elements of first transition series which have abnormal electronic configuration and why?
- Explain the following
- i) Anhydrous FeCl3 cannot be obtained by heating hydrated FeCl3.
- ii) The colour of mercurous chloride changes from white to black when treated with ammonia solution.
iii) The compounds of Zn, Cd and Hg are usually white.
- iv) A dark blue precipitate is formed when NaOH solution is added to CuSO4 The precipitate darkens on heating.
- v) Cuprous chloride is insoluble in water and dil. HCl but dissolves in conc. HCl.
- Explain the following
- i) Zinc becomes dull in moist air
- ii) A little acid is always added in the preparation of aqueous ferrous sulfate solution.
iii) Mercuric chloride and stannous chloride cannot exist as such if present together in an aqueous solution.
- Explain the following
- i) The species [CuCl4]2– exist but [CuI4]2– does not.
- ii) A solution of K2CrO4 changes colour on being acidified.
- i) Copper (I) salts are not known in aqueous solutions.
- ii) Ferric iodide is very unstable but ferric chloride is not.
iii) Silver fluoride is fairly soluble in water while other silver halides are insoluble.
- Assignments (Objective Problems)
Level – I
- Magnetic moment [Ag(CN)2]– is zero, the no. of unpaired electrons is
(A) Zero (B) One
(C) Two (D) Three
- Which of the following is not a complex
(A) NiSO4.(NH4)2SO4.6H2O (B) [Co(NH3)4Cl2]Cl
(C) K2[Ni(CN)4] (D) [Pt(NH3)Cl2Br]Br
- Which one is a monodentate ligand
(A) (B) NH3
(C) H2O (D) All
- The complex [Cr(H2O)4Br2]Cl gives the test for
(A) Br– (B) Cl–
(C) Cr3+ (D) Br– and Cl– both
- In the reaction
4KCN + Fe(CN)2 ¾¾® Product
The product formed can give the test of
(A) Fe2+ (B) CN–
(C) K+ and [Fe(CN)6]4– (D) CN– and [Fe(CN)6]3-
- Oxidation state of Cobalt in the complex [Co(NO2)6]3– is
(A) +2 (B) +3
(C) +1 (D) +2 and +3
- Coordination number of platinum in [Pt(NH3)4Cl2]++ ion is
(A) 4 (B) 2
(C) 8 (D) 6
- Which of the following is copper (I) compound
(A) [Cu(H2O)4]2+ (B) [Cu(CN)4]3–
(C) [Cu(NH3)4]2+ (D) All of these
- Low spin complex is formed by
(A) sp3d2 hybridisation (B) sp3d hybridisation
(C) d2sp3 hybridisation (D) sp3 hybridisation
- Which of the following is a high spin complex
(A) [Co(NH3)6]3+ (B) [Fe(CN)6]4–
(C) [Ni(CN)4]2– (D) [FeF6]3–
- K2[NiF6] exhibits d2sp3 hybridisation. The number of unpaired electrons in this compound is
(A) 3 (B) 2
(C) 1 (D) 0
- Which of the following ions is colourless?
(A) V3+ (B) Cu2+
(C) Ti4+ (D) Fe2+
- For Hf4+ the ionic radius is 0.86 Å which is almost the same as that of Zr4+. This is due to
(A) Hf4+ forming compounds having lower degree of ionic character.
(B) difference in co-ordination number of Zr4+ and Hf4+
(C) lanthanide contraction
(D) diagonal relationship
- Which of the following is a hexadentate ligand?
(A) DMG (B) en
(C) ox (D) EDTA
- An octahedral MA4B2 type complex gives
(A) 3 geometrical isomers (B) 2 geometrical isomers
(C) 1 geometrical isomer (D) 4 geometrical isomers
Level – II
- Which of the following complexes produces three moles of silver chloride when its one mole is treated with excess of silver nitrate
(A) [Cr(H2O)3Cl3] (B) [Cr(H2O)4Cl2]Cl
(C) [Cr(H2O)5Cl]Cl2 (D) [Cr(H2O)6]Cl3
- Which is an example of coordination isomer
(A) [Co(NH3)5NO2]Cl2 and [Cr(NH3)5ONO]Cl2
(B) [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(C) [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4
(D) [Co(NH3)4(H2O)Cl]Cl2 and [Co(NH3)4Cl2]Cl.H2O
- The number of chloride ions which would be precipitated when the complex PtCl4.4NH3 is treated with silver nitrate is
(A) Four (B) One
(C) Three (D) Two
- The number of ions per mole of the complex CoCl3.5NH3 in aqueous solution will be
(A) Four (B) Nine
(C) Three (D) Two
- In the complex [Co(en)2Cl2]Br, the coordination number and oxidation state of Cobalt are
(A) 6 and +3 (B) 3 and +3
(C) 4 and +2 (D) 6 and + 1
- Which has maximum paramagnetic nature
(A) [Cu(H2O)4]2+ (B) [Cu(NH3)4]2+
(C) [Mn(H2O)6]2+ (D) [Fe(CN)6]4–
- Which is a poor electrolytic conductor in solution
(A) K2[PtCl6] (B) [Co(NH3)3(NO2)3]
(C) K4[Fe(CN)6] (D) [Cu(NH3)4]SO4
- What is the charge on the complex [Cr(C2O4)2(H2O)2] formed by Cr(III)
(A) +3 (B) +1
(C) +2 (D) –1
- Coordination number of Cr in CrCl3.5H2O is six. The volume of 0.1 N AgNO3 needed to ppt. the chlorine in outer sphere in 200 ml of 0.01 M solution of the complex is
(A) 140 ml (B) 40 ml
(C) 80 ml (D) 20 ml
- Which of the following is a complex of a metal other than transition metal
(A) Vitamin B12 (B) Haemocyanin
(C) Haemoglobin (D) Chlorophyll
- The complex CoCl3.3NH3 on ionisation gives
(A) No Cl– ions (B) 3 Cl– ions
(C) 2 Cl– ions (D) 1 Cl– ion
- Which of the following complexes has a shape different from others
(A) [NiCl4]2– (B) Ni(CO)4
(C) [Ni(CN)4]2–` (D) [Zn(NH3)4]+
- Which of the following is not coloured
(A) Na2CuCl4 (B) Na2CdCl4
(C) K4Fe(CN)6 (D) K3Fe(CN)6
- The total no. of possible isomers for the complex [Cu(NH3)4] [PtCl4] is
(A) 3 (B) 6
(C) 5 (D) 4
- In which of the following, the magnetic moment would be maximum
(A) [Co(NH3)6]Cl3 (B) K4[Fe(CN)6]
(C) K3[Fe(CN6] (D) [Fe(H2O)6]SO4
- Answers to Objective Assignments
Level – I
- A 2. A
- D 4. B
- C 6. B
- D 8. B
- C 10. D
- D 12. C
- C 14. D
Level – II
- D 2. B
- D 4. C
- A 6. C
- B 8. D
- B 10. D
- A 12. C
- B 14. B
Transition Elements &
LEVEL – I
- i) +2 ii) +2
iii) +4 iv) +3
- v) +1 vi) +2
vii) +3 viii) 0
- ix) +2 x) +2
- xi) 4 xii) 8
xiii) 4 xiv) 6
- xv) 6 xvi) 6
xvii) 4 xviii) 6
- i) [FeCl2(H2O)4]+
- ii) [Co(en)2(CN)2]ClO3
- iv) K2[NiF6]
- v) [Pt (NH3)3Br]NO2
- vi) [Pt(NH3)2Cl(en)NO2]Cl2
- ix) [Co(NH3)2(en)2]Cl3
- x) K[MnO4]
- xi) [Fe(H2O)6]SO4
- xv) K3[Fe(CN)6]
- xx) Na2[Cr(CH3COO)4(en)]
- i) Sodiumdicyanoaurate(I)
- ii) Tris(ethylenediamine)cobalt(III)chloride
iii) chlorotriphenylphosphinepalladium(II)-m-dichlorochlorotriphenylphosphinepalladium (II)
- iv) Tetrachlorobis(diethylether)titanium(IV)
- v) Sodiumpentacarbonylmanganate(I)
- vi) Tris(acetylacetanato)chromium(III)
vii) Hexaaquamanganese (II) ion
viii) Hexaamminenickel (II) chloride
- ix) Calciumhexacyanoferrate (II)
- x) Tetraaquadichlorochromium (II) nitrate
- xi) Tetraaminechlorornitrocobalt (III) nitrate
xii) Potassium amminetrichloro platinate (II)
xiii) Tetrapyridine platinium (II) tetrachloroplatinate (II)
xiv) Sodium bisthiosulphatoargentate(I)
- xv) Hexaamminecobalt(III) chloride
xix) Tris(ethylenediamine)chromium(III) chloride.
- xx) Potassium hexacyano-N-ferrate(II)
- CO(II) has the configuration 3d7 i.e. it has three unpaired electrons. Water being a weak ligand, the unpaired electrons do not pair up. In the presence of strong ligands and air, two unpaired electrons in 3d pair up and the third unpaired electron shifts to higher energy sub shell from where it can be easily lost and hence shows an oxidation state of III.
- In [Ni(CO)4], Ni is in zero oxidation state whereas is (NiCl4]2–, Ni is in +2 oxidation state. In the presence of ligand CO, the unpaired electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons.
Level – II
- a) Ligand formula: NH3, H2O, Cl–
Charge: 0 0 –1
- b) As there are six Monodentate ligands attached to the central cobalt ion, the complex will have octahedral geometry.
- The lanthanoids which will most likely react with this strong reducing agent are those with stable +2 oxidation states to which they can be reduced –Sm, Eu and Yb.
- On addition, the pale red coloured octahedral complex [Co(H2O)6]2+ changes into tetrahedral [CoCl4]2– ion. In Tetrahedral complexes, the d–d absorption bands are considerably more intense than in octahedral complexes.
- EDTA coordinates lead ion in the body, in which form it is passed out of the body without harmful effects. The calcium salt is used so that any excess EDTA will not remove calcium ions from the body.
- [CoCl4]2– reacts with HgCl2 to give pink coloured complex Co[HgCl4]
CoCl42–+ HgCl2 ¾® Co[HgCl4] + 2Cl–
- a) The oxidation is aided by the complexation of the product gold (III) as
- b) The stability of ion is so great that the E value for oxidation is reduced to a point where H2O can oxidise the copper.
- Tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands attached to the central metal atom are same with respect to each other.
- FeSO4 does not form any complex with (NH4)2SO4 instead, they form a double salt FeSO4 (NH4)2SO4.6H2O which dissociates completely into ions but CuSO4 combines with NH3 to from the complex [Cu(NH3)4]SO4 in which complex ion, [Cu(NH3)4]2+ does not dissociate to give Cu2+ ions.
- Remembering that co-ordination number of CO is 6 the formula of the complex will be [CoCl2(NH3)4]Cl. The name will be tetramminedichlorocobalt (III) chloride.
- In aq. solution (0.001 M) the complexes will dissociate to give the ions.
[Pt(NH3)6]Cl4 [Pt(NH3)6]4+ + 4Cl– (5 ions)
[Cr(NH3)6]Cl3 [Cr(NH3)6]3+ + 3Cl– (4 ions)
Co(NH3)4×Cl3 [Co(NH3)4Cl2]+ + Cl– (2 ions)
K2[PtCl6] [PtCl6]2– + 2K+ (3 ions)
Thus, their conductivities (in solution) will increase with increasing number of ions liberated
Co(NH3)4×Cl3 < K2PtCl6 < Cr(NH3)6Cl3 < Pt(NH3)6Cl4
(least conducting) (most conducting)
Level – IIi
- a) dsp2
- b) dsp2
- c) d2sp3
- NH4+ ion does not act as a ligand because nitrogen atom has no lone pair of electrons which it can donate to metal atom.
- i) A ferrous salt turns brown in air due to oxidation to ferric salt
- ii) Cu(OH)2 dissolves in NH4OH by forming a complex.
Cu(OH)2 + 4NH2OH ¾® [Cu(NH3)4](OH)2 + 4H2O
Cu(OH)2 is insoluble in NaOH as no such complex is formed
- 143.45g of AgCl contains 35.45 g Cl– ions
\ 2.87g of AgCl will contain = 0.709 g Cl– ions
266.35 g CrCl3×6H2O contains n ´ 35.45 g of ionisable Cl– ions (where n = no. of Cl– ions outside the coordination sphere).
Thus, 2.665g CrCl3×6H2O will contain g
Also = 0.709 Þ n » 2
Keeping in view the octahedral geometry of the complex, its structure may be written as
- The compound which conducts electricity means it gives ioins in aq. solution. Keeping in view the six co-ordination number of chromium and assuming that the H2O molecules enter into the co-ordination sphere, its possible structures can be one of the following.
- i) [Cr(NH3)3(NO2)2H2O]NO2
- ii) [Cr(NH3)3NO2(H2O)2](NO2)2
The compound which does not conduct electricity means it does not give ions in solution. Hence it is neutral (or) uncharged complex. Its structure may be written as
- i) Fe3+ is more paramagnetic than Fe2+ as Fe3+ consists five unpaired electrons while Fe2+ possesses four unpaired electrons.
- ii) Any ion of transition elements which possesses unpiared d-electrons, i.e. d – d transition is possible shows a characteristic colour (n –1)d0 (or) (n–1)d10 configuration does not involve d–d transition and hence, is colourless
Fe2+, Mn2+ and Cr3+ are coloured, while Cu+, Se3+ and Ti4+ are colourless.
iii) Chromium and copper. Chromium attains 3d54s1 configuration in which all the d-orbitals are unpaired in order to get extra stability. Copper attains 3d104s1 configuration in which all the d-orbitals are paired in order to get extra-stability.
- i) On heating hydrated FeCl3, anhydrous FeCl3 is not formed as water of crystallisation reacts to form Fe2O3 and HCl.
- ii) Hg2Cl2 reacts with NH3 to form a mixture of mercury and mercuric aminochloride which is a black substance
iii) In the compounds of zinc metals, M2+ ions possess the penultimate d-orbitals doubly occupied i.e., (n-1)d10 configuration. There is no d-d transition. Hence the compounds of zinc metals are colourless.
- iv) NaOH reacts with CuSO4 when dark blue precipitate of Cu(OH)2 is formed. This precipitate of heating forms CuO which is black in colour. Hence, the colour darkens on heating.
CuSO4 + 2NaOH ¾® Cu(OH)2 + Na2SO4
Cu(OH)2 ¾® CuO + H2O
- v) Cu2Cl2 dissolves in conc. HCl due to the formation of a copper complex
Cu2Cl2 + 2HCl ¾® 2H[CuCl2]
- i) When zinc is exposed to moist air, the surface is affected with the formation of a film of basic zinc carbonate on it. Due to this zinc becomes dull.
4Zn + 3H2O + CO2 + 2O2 ¾® ZnCO3×3Zn(OH)2
- ii) Ferrous sulfate is a salt of a weak base and a strong acid. Thus its hydrolysis occurs when it is dissolved in water and solutioin becomes turbid due to formation of ferrous hydroxide
FeSO4 + 2H2O Fe(OH)2 + H2SO4
Addition of a small amount of acid shifts the equilibrium towards left and prevents hydrolysis.
iii) HgCl2 is an oxidizing agent while SnCl2 is a reducing agent. When both are present, a redox reaction occurs forming Hg and stannic chloride as final products.
SnCl2 + 2HgCl2 ¾® Hg2Cl2 + SnCl4
Hg2Cl2 + SnCl2 ¾® 2Hg + SnCl4
- i) I– is a stronger reducing agent than Cl– ion. It reduces Cu2+ ion into Cu+ ion. Hence cupric iodide is converted into cuprous iodide. Thus the species [CuI4]2– does not exist.
- ii) On being acidifed, the chromate ion dimerises to form dichromate ions which are orange coloured.
2CrO42– + 2H+ ¾® Cr2O72– + H2O
- i) Cu(I) salts under go disproportionation in aqueous solution
2Cu+ ¾® Cu2+ + Cu
- ii) I– ion is a stronger reducing agent in comparison to Cl– Fe3+ is easily reduced by I– ion
2Fe3+ + 2I– ¾® 2Fe2+ + I2
iii) The hydration energy of AgF is higher than its lattice energy. Hence it is soluble in water. The hydration energy values of other halides are smaller than their lattice energy values. Hence these halides are insoluble in water.