Transition Elements IItjee

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Transition Element_1

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TRANSITION ELEMENTS

1. SYLLABUS

Definition, general characteristics, oxidation states and the color (excluding the details of electronic

transitions) and calculation of spin-only magnetic moment; Coordination compounds : Nomenclatureof mononuclear coordination compounds, cis-transitions isomerism, hybridization and geometries of mononuclear coordination compounds(linear, tetrahedral, square planar and octahedral).

2. INTRODUCTION

The elements lying between s and p-block elements of the periodic table are collectively known astransition or transitional elements. These elements either in their atomic state or in any of theircommon oxidation state have partly filled (n–1) d orbitals of (n–1)th main shell. In these elements thedifferentiating electron enters (n–1)d orbitals of (n–1)th main shell and as such these are called d-block elements.

The name transition is given to the elements on the basis of their position in the P.T and theirproperties i.e. they occupy a position between the highly electropositive elements on the left and theelectronegative elements on the right. Their properties are also intermediate of the s- and p-blockelementsIn the d-block, electrons are added to the penultimate shell, expanding it from 8 to 18 electrons. All thed-block elements are classified into four series i.e. 3d, 4d, 5d and 6d series corresponding to the fillingof 3d, 4d, 5d and 6d orbitals of (n–1)th main shell. Each of 3d, 4d and 5d series has ten elements while6d series has at present only one element viz. 90Ac whose valence shell configuration is 1 26d 7s .

3. ELECTRONIC CONFIGURATION AND IRREGULARITIES

The valence shell configurations of these elements can be represented by – , ,( – ) 1 10 01 2n 1d ns . Theobserved valence shell configuration of these elements differ from their predicted configuration. Theirregularities in the observed configurations of 5 1 10 1 10 0 10 0Cr(3d 4s ),Cu(3d 4s ),Mo(4d 5s ),pd(4d 5s )

10 1Ag(4d 5s ) and 10 1Au(5d 6s ) are explained on the basis of the concept that half-filled andcompletely filled d-orbitals are relatively more stable than other d-orbitals. Only this factor is notsufficient to explain the irregularities in observed electronic configuration. The net effect of all theforces, comprising (i) nuclear electronic attraction (ii) shielding of one electron by others from othernuclear charge (iii) inter electronic repulsion and (iv) exchange forces determines the stability of theelectronic configuration. It is not easy to explain why unlike 5 1Cr(3d 4s ) and 5 1Mo(4d 5S ) should

have the idealized electronic configuration 14 4 2(4f 5d 6s ). The configurations clearly show that strictly, According to the definition of d-block elements; Cu, Agand Au should be excluded from d-block elements, since these elements both in their atomic state (withconfiguration (n–1) 10 1d ns ) and in their +1 oxidation state (with configuration (n–1) 10d ), do not havepartly filled (n–1) d-orbitals. Similarly Zn, Cd and Hg which both in their atomic state [( – ) ]10 2n 1d ns

and in +2 oxidation state 10[(n–1)d ] do not contain partly filled [(n–1)d] orbitals, should also be

excluded from d-block elements. Similarly is the case with Pd atom with configuration 10 04d 5s . yet,in order to maintain a rational classification of elements, these elements (viz cu, Ag, Au, Zn, Cd, Hgand Pd) are also generally studied with d-block element.

Question 1: The mercurous ion is written as 22Hg while the cuprous ion is written as Cu .

Solution: 14 10 280 54Hg :[Xe] 4f 5d 6s

14 10 1Hg :[Xe]4f 5d 6s





In IHg , there is one unpaired electron and we expect (mercurous) salt to beparamagnetic. But magnetic moment of mercurous salt is zero indicating that it isdiamagnetic which is only possible when 6s electron has been used by two Hg atoms inbonding. Hence mercurous ion is 2

2Hg .

10 129Cu :[Ar]3d 4s

10

Cu :[Ar]3d

ICu (cuprous) salt is diamagnetic by experiment and thus it is not dimeric and is thusCu .

4. PHYSIOCHEMICAL PROPERTIES

The difference in the trends in the properties of d-block elements from those of s- and p-block elementsarises from a basic difference in their electronic configuration while in the building up of elementsfrom Li to F, the electrons are added to the outer most shell but in the case of transition metals, theelectrons are added to inner (n–1)d sub-shell. Although, the horizontal similarity amongst the d-blockelements is well marked, yet the chemistry of the elements of first transition series differ considerablyfrom that of the elements of the second and third transition series which are more similar to each other.

Some properties of d-block elemenetsProperty Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Coper ZincSymbol Sc Ti V Cr Mn Fe Co Ni Cu ZnAtomic No. 21 22 23 24 25 26 27 28 29 30Atomic weight 44.956 47.90 50.942 51.996 54.938 55.847 58.933 58.710 63.54 65.37Metallic radius

(pm)

164 147 135 130 135 126 125 125 128 137

Ionic radius

(pm)

81

(3+)

76, 68

(3+) (4+)

74, 60

(3+) (4+)

84, 69

(2+) (3+)

80, 66

(2+) (3+)

76, 64

(2+)

(3+)

74, 63

(2+)(3+)

72, 62

(2+)(3+)

96, 69

(1+)(2+)

74

(2+

Covalent radius

(pm)

144 132 122 118 117 117 116 115 117 125

Boiling point (K) 3000 3533 3673 2753 2370 3273 3173 3005 2868 1180Melting point (K) 1812 1948 2173 2163 1517 1808 1768 1726 1356 692Density

3 –310 kgm3.0 4.5 6.11 7.2 7.44 7.86 8.86 8.90 8.92 7.13

Electronegativity

(A/R)

1.2 1.3 1.45 1.55 1.6 1.65 1.7 1.75 1.75 1.65

Ionisation 1st 633 659 650 653 717 762 759 736 745 906Energy 2nd 1235 1309 1414 1591 1509 1561 1644 1751 1958 1732

(kJ–1mol ) 3rd 2388 2648 2866 2992 3259 2958 3230 3391 3556 3828

Electrode (III) (III) (IV) (II) (III) (II) (III) (II) (III) (II)(III) (II) (III) (II) (I) (II) (II)Potential (V)+ –2.1 –1.2

–1.63

–1.2

–0.86

–0.91

–0.74

–1.18

–0.28

–0.44

–0.04

–0.28

+0.4

–0.25 +0.52

+0.34–0.76

4.1 ATOMIC (COVALENT) AND IONIC RADII, ATOMIC VOLUME AND DENSITY

There is a gradual decrease in atomic radius and ionic radius on moving from left to right because,additional positive charges are placed on the nucleus and correspondingly electrons are added to the(n–1)d orbitals. As the electrons in the d-orbitals shield the ns electrons and also themselves from thenuclear charge incompletely, effective nuclear charge felt by them increases and hence a contraction insize occurs. The atomic radii for the elements from Cr to Cu are very close to one another because





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 atomicradii do not alter much on moving from Cr to Cu.

As we move from alkali metals to transition elements, radii decreases steeply but within transitionelements this rate of decrease is less due to an increase in interelectronic repulsion.

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 type oxidesof the transition element should be similar to CaO in many ways, although somewhat less basic andless soluble in water. Similarly the hydration energies of M2+ ions should be similar to but somewhatgreater than that of Ca2+ ion. This is borne out by facts, since the hydration energy of Ca2+ ion is 395kcal and the observed values of hydration energies for the elements Ti2+… Cu2+are between 446 kcaland 597 kcal.

The radii of M3+ ions are slightly greater than that of Ga3+ ion (0.62Å). Thus M2O3 oxides of transitionelements should be similar to but slightly less acidic (more basic) than Ga2O3 and the hydrationenergies of M3+ ions should be less than 1124 kcal which is the hydration energy of Ga3+ ion. Theobserved values of hydration energies for the series Sc3+… Fe3+are between 947 kcal and 1072 kcal.

Atomic volumes follows the same trend (decreasing) as the atomic size. While there is a general trendof increasing density across transition series because smaller the size, the higher is the density.

4.2 METALLIC CHARACTER

All the transition elements are metals, since the number of electrons in the outer-most shell is verysmall, 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. These metals are goodconductors of heat and electricity

4.3 MELTING AND BOILING POINTS

The m.p. and b.p. of transition elements are usually high in comparison with s- and p block elements. The m.p of elements (i.e.-transition elements) depend upon the strength of metallic bond whichincreases with the availability of the unpaired d-electrons to participate in the bonding bydelocalization. In first transition series, there is a sharp decrease of m.p. after the middle (Mn having 5unpaired electrons) in the series due to electron pairing. Thesoftness and low m.p. of Zn, Cd and Hg

(liquid) is due to pairing of all electrons. Periodic trends in the b.p. are similar to those in m.p.

4.4 IONISATION POTENTIAL

The first ionisation potentials of transitional elements lie between the values of those of s- and p-blockelements. 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 screeningeffect which shields the outer ns electrons from the inward pull of positive nucleus on the outer nselectrons. Thus the effects of the increasing nuclear charge and the shielding effect created due to theexpansion of (n-1)d orbital oppose each other. On account of these counter effects, the ionisationpotentials increase rather slowly on moving in a period of the first transition series.

a) 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 IE I for Zn is considerably higher. This is due tothe extra-stability of 3d10 level which is completely filled in Zn atom.





b) Second ionisationpotentials : The second ionisation potentials are seen to increase more or lessregularly with the increase of atomic number. The value of IE II for Cr and Cu are higher thanthose 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 I IIA group elementsthe 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. Thusmore energy will be required to remove an electron from a filled 4s-orbital in comparison to thatused to remove an electron from a 4p-orbital which is incompletely filled.

c) Electropositive character of transitional elements as compared to that of alkali metals andalkaline earth metals : The values of first ionisation potentials of transition elements in mostcases lie between those of s-and p-block elements. Thus the transition elements are lesselectropositive than the elements of I A and II A groups. Thus, although the transitional elementsdo form ionic compounds, yet they do not form ionic compounds so readily as the alkali andalkaline earth metals do. Also, unlike the alkali and alkaline earth metals, the transitional elementsalso have a tendency to form the covalent compounds under certain conditions. Generally thecompounds in which the transition elements show a smaller valency are ionic, while those in whicha higher valency is exhibited are covalent in character.

4.5 ELECTRONEGATIVITY

Transition element have fairly low values of electronegativity. It increases from Sc to Cu with a fall atMn and Zn. However, this increase in electronegativity is much slower because the additional electronis being added to an inner shell which provides relatively good shielding to the outer electrons from thenucleus. The increasing electronegativity from Sc to Cu means that the elements become slightly lessmetallic and this is reflected in the increasing positive electrode potentials of their ions 2M and 3M .

4.6 ELECTRODE POTENTIAL

The potential difference set up in a 1M solution of metal ions at 298 K is called standard electrodepotential against hydrogen electrode as reference.

Electrode potential is a measure of the electro positive character and it decreases along first transition

series (except Cu which has negative value) and can react with acids producing hydrogen.

4.7 OXIDATION STATES

Transition elements exhibit a wide range of oxidation states differing usually by units of one. This isdue to the fact that(n–1) d electrons may get involved along with ns electrons in bonding, as electronsin (n–1) d orbitals are in an energy state comparable to ns electrons. There exists a general trend of lesser no of oxidation states at each end of the series and a higher no in the middle. The lesser no of oxidation states in the beginning of the series can be due to the presence of too few electrons to lose orshare towards the end of series it can be ascribed to the presence of too many electrons and thus fewerempty orbitals to share electrons with the ligands. Another feature is the reduced tendency of higheroxidation states towards the end of the series. This could be due to steady increase in the effectivenuclear charge along the series, thus pulling the d-orbitals into the electron core and not making themreadily available for bonding.

i) Minimum oxidation state. All the transition elements with the exception of Cr, cu, Ag, Au andHgwhich have a minimum oxidation state of +1 exhibit a minimum oxidation stateof +2. In mostcases 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 maximumoxidation state equal to its group number. Thus, Cr in group VIB shows a maximum oxidationstate of +6 in 2–

2 7CrO ion. Most of the elements in VIII group show a maximum oxidation sate

equal to +6. However, Ru and Os have a maximum oxidation state equal to +8 which is thehighest oxidation state shown by any element.





iii) Relative stability of various oxidation states. The relative stabilities of various oxidation statesof 3d- series elements can be correlated with the extra stability of 0 53d ,3d and 103dconfigurations to some extent. Thus 4 Ti 0(3d ) is more stable than 3 1 Ti (3d ) and similarly

2 5Mn (3d ) is more stable than 4 4Mn (3d ) . It may, however, be pointed out that such ageneralization for the relative stability of various oxidation states of 4d- and 5d- series elements isoften rather difficult to realise.

The higher oxidation states of 4d and 5d series elements are generally more stable than those of the elements of 3d series, e.g., Mo, Te (4d-series elements) and W, Re (5d-series elements) formthe oxyanions: – – – –, Re, ,VI 2 VII VI 2 VII

4 4 4 4Mo O Tc O W O O which are stable and in which thetransition elements concerned show their maximum oxidation states. The correspondingoxyanions of Cr and Mn namely VI 2–

4Cr O and VII –4Mn O are strong oxidizing agents.

Furthermore, the highet oxidation states of second and third row elements are encountered incompounds containing the more electronegative elements viz. F,O and Cl. Thus for thecompounds , ,VIII VIII

4 4Ru O Os O VI6W Cl and VI

6Pt F there are no analogs being formed by thefirst row elements. The lower oxidation states particularly +2 and +3 are important in thechemistry of aquated and complex ions of the 3d-series (i.e. first row) elements but these ions arenot 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 asub-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 higherattractive force exerted (on the electrons) by the nucleus. Higher oxidation states of these metalsare not formed by the actual loss of electrons but due to the formation of new hybrid orbitalsinvolving(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 ioniccharacter of the bond the basic character of the oxides decreases, e.g. MnO (oxidation state of Mn=+2) is basic, 2MnO (Mn 4) is amphoteric and 3MnO (Mn =+6) is acidic.

4.8 COLOUR

Compounds of transition elements are usually markedly coloured, in contrast to compounds of s- andp-block elements which are mostly white or colourless unless the anion is coloured. As you know,substances appear coloured when they absorb light of a particular wavelength in the visible region of the spectrum and transmit light of other wavelengths. The colour which we see is the colour of thetransmitted wavelengths. In other words, the colour of the compound observed by us is thecomplmentary colour of the colour absorbed by the compound.. You know that the transition metals assuch or in the form of ions have partly filled d-orbitals which are degenerate, i.e., they are of equalnergy. In transition metal complexes the d-orbitals do not remain degenerate, but these split into sets of orbitals of different energies. By absorbing energy, electrons can move from a d-orbital of lowerenergy to that of higher energy. This transition of electron from one d-orbital to another corresponds toa fairly small energy difference, therefore, light is absorbed in the visible region of spectrum. For

example, the aqua ion ( )32 6 Ti H O , which has one electron in the 3d orbital absorbs light of

wavelength in the yellow-green region of spectrum and therefore, appears reddish violet in colour.

Whenever the d-orbitals are completely filled or empty, there is no possibility of electronic transitionswithin the d-orbitals. In such cases, the ions will not show any colour. For example, the compounds of

, ,3 4Sc Ti Cu and 2Zn are white or colourless.

In the s- and p-block elements there cannot be any d-d transitions and the energy needed to promate sor p electron to a higher level is much greater and may correspond to ultraviolet region, in which casethe compound will not appear coloured to the eye.





Relationship between colour and wavelength

Question 2: ZnO is yellow when hot, but white when cold.

Solution: 2Zn has 103d E.C. hence all electrons paired thus form white salt. The Yellow colourof hot ZnO is due to defects in the solid structure which increase with temperature.

Exercise - 1 Cu is a 10d ion and colourless but 2Cu O is red and 2Cu S is black.

4.9 MAGNETIC PROPERTIES

When a substance is placed in a magnetic field of strength H, the intensity of the magnetic field in thesubstance may be greater than or less than H.

If intensity >H , then substance is paramagnetic (Electrons unpaired).

If intensity <H , then substance is diamagnetic (Electrons paired).

Substances which are weakly repelled by a magnetic field are called diamagnetic, white thesubstances which are weakly attracted by the magnetic field and lose their magnetism when removedfrom the field are called paramagnetic. If the force of attraction is very large and the permanentmagnetization is retained, the substance is said to be ferromagnetic, e.g., iron and some ironcompounds.

It should be noted that Fe,Co and Ni are ferromagnetic. Ferromagnetic materials may be regarded asspecial case of paramagnetism in which the moments of individual atoms become aligned and all pointin the same direction. It is also possible to get antiferromagnetism by pairing the moments of adjacentatoms 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 unpairedelectrons as follows

( )n n 2 B.M.

n =number of unpaired electrons

B.M. =Bohr Magneton, unit of magnetic moment

More the magnetic moment more is the paramagnetic behaviour.

Question 3: 33 6[Co(NH ) ] isdiamagnetic and 3–

6[CoF ] is strongly paramagnetic.

Wavelength absorbed in nm Colour absorbed Colour observed<400 UV region White/colourless

400—435 Violet Yellow-green435—480 Indigo Yellow

480—490 Green-blue Orange

490—500 Blue-green Red

500—560 Green Purple560—580 Yellow-green Violet

580—595 Yellow Indigo

595—605 Orange Green-blue

605—750 Red Blue-green

>750 Infra-red White/colourless





Solution: 3Co

33 6[Co(NH ) ]

3–6[CoF ]

2 3d sp

3 2sp d

diamagnetic due to pairedelectrons

Paramagnetic due to four unpaired electrons

3CO has 63d configuration with four unpaired electrons in ground state. In presence of 3NH

(strong ligand) all the unpaired electrons in 3Co get paired and thus 33 6[Co(NH ) ] has 2 3d sp

hybridization (octahedral structure), thus it is diamagnetic (no electron unpaired). –F is a weakligand hence six lone pairs of six –F are filled in outer d-orbitals of 3Co which has now four

electrons unpaired. Thus 3–6CoF has 3 2sp d hybridization in 3Co and is thus paramagnetic due

to unpaired electrons.

Exercise - 2 Classify each of the following complexes as either high or low spin. Explainyou answers.

–[ ( ) ] , . ;[ ( ) ] , .2 42 6 6Co H O 46 BM Co CN 19BM

–[ ( ) ] , . ;[ ( ) ] , .4 32 6 2 6Fe NO 00BM Fe H O 594 BM

4.10 COMPLEX FORAMTION

The chemistry of the transition metals is dominated by their tendency to form complex ions. This is

because the transition elements form small, highly charged ions which have vacant orbitals of suitableenergy to accept lone pairs of electrons donated by other groups or ligands. In case of transition metalsin high oxidation states, highly charged inos can strongly bind electrostatically a wide variety of negative or polar ligands. In the case of transition metals in low oxidation states, the electrons in the dorbitals become involved in bonding with ligands. The majority of transition metal ion complexescontain six ligands surrounding the central ion octahedrally. Some elements contain four ligands whichare either arranged tetrahedrally or less frequently at the corners of a square. Besides these geometries,other geometries like trigonal bipyramid, pentagonal bipyramid, etc., are also present occasionally. Thebonding between the ligand and the transition metal ion can either be predominantly electrostatic orcovalent or in many cases intermediate between the two extremes. Some of the typical complexes of thetransition meals are – –[ ( ) ] ,[ ( ) ] ,[ ( ) ]3 2 3

6 3 4 4Fe CN Ni NH Cu CN , [ ( ) ]23 4Cu NH etc.

4.11 CATALYTIC PROPERTIES

Many transition metals and their compounds have catalytic properties. These metals can function ascatalysts because they can utilize both d and s electrons for the formation of bonds between reactantmolecules and the surface catalyst atoms. This increases the concentration of the reactants at the catalystsurface and weakens the bonds in the reactant molecules with the result that the activation energy islowered. Compounds of transition metals are able to act as catalysts because of the case with which themetal can adopt different oxidation states and also because of their ability to form complexes. Some of the common catalysts used for important reactions.

3 TiCl Used as theZiegler-Natta catalyst in the production of polythene.





2 5V O Converts 2SO to 3SO in thecontact process for making 2 4H SO .

2MnO Used as a catalyst to decompose 3KClO to give 2O .

Fe Promoted iron is used in theHaber-Bosch processfor making 3NH .

3FeCl Used in the production of 4

CCl from2

CS and2

Cl .

4FeSO

and 2 2H O

Used asFenton’s reagent for oxidizing alcohols to aldehydes.

2PdCl Wacker process for converting 2 4 2 2C H H O PdCl to 3CH CHO 2HCl Pd

Pd Used for hydrogenation (e.g. phenol to cyclohexanone).

Pt/PtO Adams catalyst, used for reductions.

Pt Formerly used for 2 3SO SO in thecontact processfor making 2 4H SO .

Pt

Is increasingly being used in three stage-convertors for cleaning car exhaust fumes.Pt/Rh Formerly used in theOstwald processfor making 3HNO to oxidize 3NH to NO.

Cu Is used in the direct process for manufacture of 3 2 2(CH ) SiCl used to make silicones.

Cu/V Oxidation of cyclohexanol/cyclohexanone mixtures to adipic acid which is used to makenylon-66.

2CuCl Deacon processof making 2Cl from HCl.

Ni Raney nickel, numerous reduction processes (e.g. manufacture of hexamethylenediamine,production of 2H from 3NH , reducing anthraquinone to anthraquinol in the production of

2 2H O ).Nicomplexes

Reppe synthesis(polymerization of alkynes) e.g. to give benzene or cyclooctatetraene.

4.12 INTERSTITIAL COMPOUND Transition metals can trap some small atoms like hydrogen, boron, carbon, nitrogen, etc., in vacantspaces in their crystal lattice forming interstitial compounds. Carbon and nitrogen always occupyoctahedral holes; hydrogen is smaller and always occupies tetrahedral holes. As only transition metalsform such compounds, the d electrons are, therefore, presumably involved in the bonding. Thestructure of the meal often changes during the formation of such compounds. The composition of these compounds is generally non-stoichiometric, e.g., . .,173 056 TiH PdH , but may approach regular

stoichiometry and a regular structure, e.g., TiC and VN. The later transition elements of the first seriesform non-stoichiometric carbides with irregular structures, such as 7 3CrC , which are more reactive

than the interstitial carbides of the early transition elements. These interstitial compounds are of muchimportance, e.g., carbon steels are interstitial iron-carbon compounds in which the interstitial carbonprevents the iron atoms from sliding over one another, making iron harder, stronger but more brittle.

The presence of these atoms results in decrease in malleability and ductility of the metals but increasestheir tensile strength.





5. INNER TRANSITION ELEMENTS

The elements in which the additional electrons enters (n–2)f orbitals are called inner transitionelements. The valence shell electronic configuration of these elements can be represented as

(n–2) , ... , , ,( – )0 2 14 012 2f n 1d ns . These are also called f-block elementsbecause the extra electron goesto f-orbitals which belongs to (n–2)th main shell. 4f-block elements are also called lanthanides or rareearths. Similarly 5f-block elements are called actinides or actinones. The name Lanthanides and

actinides have been given due to close resemblance with Lanthanum and actinium respectively.Lanthanides constitutes the first inner transition series while actinides constitutes second innertransition series.

General Characteristics:

1. Electronic Configurationn 1 0 2[Xe]4f 5d 6s

or n 1 2[Xe]4f 5d 6s

2. Oxidationstate

They readily form 3M ions some of them also exhibit oxidation state of +2 and +4.3. Colouration

Ions of Lanthanides and actinides are coloured in the solid state as well as in aqueous solutionbecause of absorbation of light due to f–f transition since they have partly filled f-orbitals.

5.1 LANTHANIDE CONTRACTION

In lanthanides, the additional electron enters 4f-sub shell but not in the valence-shell namely sixthshell. The shielding effect of one electron in 4f-sub-shell by another in the same sub-shell (i.e. mutualshielding effect of 4f-electrons) is very little, being even smaller than that of d-electrons, because theshape of f-sub shell is very much diffused. The nuclear charge (i.e. atomic number), however,increases by unity at each step. Thus the nuclear charge increases at each step, while there is nocomparable increase in the mutual shielding effect of 4f-electrons. This results in that electrons in theoutermost shell experience increasing nuclear attraction from the growing nucleus. Consequently theatomic and ionic radii go on decreasing as we move from 57La to 71Lu .

5.2 CONSEQUENCE OF LANTHANIDE CONTRACTION

1. Atomic and ionic radii of post-Lanthanide elements : The atomic radii of second row transitionelements are almost similar to those of the third row transition element because the increase insize on moving down the group from second to third transition elements is cancelled by thedecrease in size due to lanthanide contraction.

2. High density of post lanthanide elements : It is because of very small size due to lanthanidecontraction.

3. Basic strength of oxides and hydroxides : Due to lanthanide contraction the decrease in size of

lanthanides ions, from 3La to 3Lu increases the covalent character (i.e. decreases the ionic

character) between 3Ln and –OH ions in Ln(III) hydroxides (Fajans rules). Thus 3La(OH) is

the most basic while 3Lu(OH) is the best basic.

Similarly, there is a decrease in the basic strength of the oxides.

4. Seperation of Lanthanides : Due to the similar size (Because of lanthanide contraction) of thelanthanides, it is difficult to separate them. But slight variation in their properties is utilized toseparate.



TransitionElement_10


6. COORDINATION COMPOUNDS

6.1 INTRODUCTION

The complexes show a wide variety of physical and chemical properties which are quite different fromnormal salts. These difference arise due to the difference in their structures.

Molecular or addition compounds : When solution containing two or more salts in stoichiometric

(i.e., simple molecular) proportions are allowed to evaporate, we get crystals of compounds known asmolecular or addition compounds.

2 2 2 2(carnallite)

KCl MgCl 6H O KCl.MgCl .6H O

2 4 2 4 3 2 2 4 2 4 3 2(potassiumalum)

K SO Al (SO ) 24H O K SO .Al (SO ) .24H O

4 3 2 4 3 2(tetramminecopper(II)sulphatemonohydrate)

CuSO 4NH H O CuSO .4NH .H O

2 2(potassiumferrocyanide)

Fe(CN) 4KCN Fe(CN) .4KCN

These are of two types depending on their behaviour in aqueous solution.

1. Double salts or Lattice compounds : The addition compounds having the followingcharacteristic are called double salts or lattice compounds.

a) They exist as such in crystalline state.

b) When dissolved in water, these dissociate into ions in the same way in which the individualcomponents of the double salts do.

2 2–4 4 2 4 2 4 4

Mohr'ssaltFeSO .(NH ) SO .6H O Fe (aq) 2NH (aq) 2SO (aq)

3 2–2 4 2 4 3 2 4 2

Potashalum

K SO .Al (SO ) .24H O 2K (aq) 2Al (aq) 4SO (aq) 24H O

2. Coordination (or complex) compounds. It has been observed that when solutions of Fe( 2CN)

and KCN are mixed together and evaporated, potassium ferrocyanide, 2Fe(CN) 4KCN is

obtained which in aqueous solution does not give test for the 2Fe and –CN ions, but gives the

test for K ion and ferrocyanide ion, 4–2Fe(CN) .

4–2 2 6Fe(CN) 4KCN Fe(CN) .4KCN 4K Fe(CN)

Thus we see that in the molecular compound like 2Fe(CN) 4KCN, the individual compounds lose

their identity. Such molecular compounds are calledcoordination (or complex) compounds. Thedifference between a double salt and a complex compound appears to be one of the degree ratherthan of a more fundamental unit.

A complex compound contains a simple cation and a complex anion or a complex cation and asimple anion or a complex cation and complex anion, e.g. IV IV

2 6 3 4 2 2K [Pt Cl ],[PT (NH ) Br ]Br

and III III3 6 2 4 3[CO (NH ) ][Cr (C O ) ] are all complex compounds. The term complex compound is

used synonymously with the term coordination compound.

In the above complex compounds the ions, –[ ] ,[ ( ) ] ,[ ( ) ]IV 2 IV 2 III 3

4 3 4 2 3 6Pt Cl Pt NH Br Co NH

and –[ ( ) ]II I 3

2 4 3Cr C O are called complex ions. Thus a complex ion is an electrically charged

radical which is formed by the union of a metal cation with one or more neutral molecules oranions.





7. IUPAC NOMENCLATURE OF COORDINATION COMPOUNDS

1. Naming of salt: If the complex is a salt, the cation is named first followed by the name of theanion. For example, in naming 2 6K [PtCl ] we shall name potassium first, followed by the name

of the anion. In another example of 3 4 2[Cu(NH ) ]Cl , the name of the cation, 23 4[Cu(NH ) ] ,

will be placed before chloride ion.

2. For the complex entity, whether it is in cationic, anionic or neutral form, the name of theligand(s) is put before the name of the metal atom. However, the reverse order is followed inwriting the formula of the compound. For example, in 3 4 2[Cu(NH ) ]Cl , we shall name the four

ammonia molecules first followed by copper and finally the presence of chloride is mentioned;but in writing the formula copper is written before ammonia molecules.

3. Naming of the negative ligands: The names of all anionic ligands end in o’ replacing the finale’ in endings. Sometimes endings are also changed. Thus the given ligands acquire the following

endings:

Ligands Endings

–F (fluoride) fluoro–

Cl (chloride) choro2–O (oxide) oxo–OCN (cyanate) cyanato

Cationic and neutral ligands have no special ending. There are a few exceptions. like aqua’ for

2H O , ammine for 3NH , carbonyl for CO, and nitrosyl for NO groups.

4. Indication of the number of ligands: The number of ligands is indicated by adding prefixes di–,tri–, tetra–, penta–, hexa–, etc. for two, three, four, five, six, etc entities of the ligand. Forexample, 3 6 3[Co(NH ) ]Cl will be called hexaamminecobalt chloride.

If the ligands are big complicated groups, instead of di–, tri–, tetra–, penta– prefixes we usebis–,

tris–, tetrakis–, pentalkis– etc. For instance 3 3 2Cu(CH COCHCOCH ) is called

bis(acetylacetonato) copper.5. Order of Naming ligands: In any complex species, the ligands are quoted in alphabetical order,

without regard to charge, before the name of the central metal atom. Numerical prefixes indicatingthe number of ligands are not considered in determining that order. For example, a compound like

2 2[CoCl(NO )(en) ]Cl will be called chlorobis (ethylenediamine) nitrocobalt(III) chloride.

6. Oxidation state: The oxidation state of the metal ion in a complex is indicated by Roman (I), (II),(III) etc. or an Arabic (O) and placed in parenthesis immediately after thename of the metal. If,however, the complex species is an anion, the oxidation state of the metal is mentioned at the endof the name of the complex.

7. Naming of complex: The name of the complex anion always ends in ‘ate’ and the Latin name of the metal atom is used. No specific ending is used for neutral or cationic complex species. For

example, 2 6K [PtCl ] is called potassium hexachloroplatinate(IV) and 2K[Ag(CN) ] is named aspotassium dicyanoargentate(I).

8. A little space is given between the name of the cation and the anion. No space or hyphen is usedanywhere else.

9. Once the complex entity is completely identified according to the above rules, no mention of thenumber of cations or anions used for charge balancing is required. For example, 3 6 3[Co(NH ) ]Clis called hexaamminecobalt(III) chloride and not hexaamminecobalt(III) trichloride. Similarly,

2 6K [PtCl ] is named potassium hexachloroplatinate(IV) and not dipotassium

hexachloroplatinate(IV).





10. Ligands having more than one donor atom: If a ligand has more than one donor atom, theactual atom involved in the bond formation with the metal ion is indicated by putting italicizedsymbol of the atom after the name of the ligand. For example, –[ ( ) ]3

2 3 2A g S O is calleddithiosulphato s-argentate(I) ion. Some exceptions in this connection should be remembered. These are:–SCN thiocyanato

–NCS isothiocyanato

– 2NO nitro

–ONO nitrite

11. Naming of Bridging ligands: A bridging group is indicated by putting the Greek letter ` ’immediately before its name and separated by hyphens from other ligands. For example,

42 4 2 4[(H O) Fe Fe(H O) ]

OHOH

is called octaaqua- dihydroxodiiron(III)’ ion and the

formula could be written as 42 4 2 2 4[(H O) Fe( OH) Fe(H O) ] also.

12. Structural information may be given names and formulas by prefixes such as cis–, trans– etc.

3 2 2[Pt(NH ) Cl ] can be written as cis-dichlorodiammineplatinum(II) or trans-

dichlorodiammineplatinum(II), respectively.Note : When writing the formula of a complex, the central atom is listed first. The coordinated

groups (i.e., ligands) are listed in the order: formally anionic ligands, neutral ligandsfollowed by cationic ligands. Within each group, the ligands are listed alphabeticallyaccording to the first symbol.

Question 4: Write down IPUAC name of K 2 [Fe(CN)3 Cl2(NH3)2]

Solution: The positive part is named first followed by the negative part. In the negative part the namesare written in alphabetical order followed by metal. So the name isPotassiumdiammlnedichlorotricyano-N-ferrate (III).

Question 5: Write the IUPAC name of [Co(NH3)4(NO2)2]Cl.

Solution: Here in this case the positive part is the complex. So it is name first with ligands inalphabetical 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 in Tetraamminedinitrocobalt (III) chloride.

Exercise - 3 Write IPUAC name of [Pt(Py)4][PtCl4].

7.1 WERNER’S THEORY OF COORDINATION COMPOUNDS

It was only in 1893, that Werner presented a theory known as Werner’ coordination theory whichcould explain all the observed properties of complex compounds. More important postulates of thistheory are:

i) Most elements exhibt two types of valenceis : (a) primary valency and (b) secondary valency.

a) Primary valency. This corresponds to oxidation state of the metal ion. This is also called

principal, ionisable or ionic valency. It is satisfied by negative ions and its attachment with thecentral metal ion is shown by dotted lines.

b) Secondary or auxiliary valency. It is also termed as coordination number (usuallyabbreviated as CN) of the central metal ion. It is non-ionic or non-ionisable (i.e. coordinatecovalent bond type). This is satisfied by either negative ions or neutral molecules. Theligands which satisfy the coordination number are directly attached to the metal atom or ionand shown by thick lines. While writing down the formulae these are placed in thecoordination sphere along with the metal ion. These are directed towards fixed position inspace about the central metal ion, e.g. six ligands are arranged at the six corners of aregularoctahedron with the metal ion at its centre. This postulate predicted the existence of different types of isomerism in coordination complexes and after 19 years Werner actuallysucceeded in resolving various coordination examples into optically active isomers.





.3 3CoCl 6NH or

Co

H3N

H3N

Cl

NH3

NH3

Cl

ClNH3

NH3

–

[( ( ) ] ( )

II I 3

3 6 3Co NH Cl

Co

H3N

H2O

Cl

NH3

NH3

Cl

ClNH3

NH3

.3 3 2CoCl 5NH H Oor–

[( ( ) ( )] ( )

I II 3

3 5 2 3Co NH H O Cl

.3 3CoCl 5NH or

Co

H3N

H3N

Cl

NH3

ClNH3

NH3

Cl

–[( ( ) ] ( )I I I 23 5 2Co NH Cl Cl

.3 3 2CoCl 4NH H Oor–[( ( ) ]III

3 4 2Co NH Cl Cl

Co

H3N

H3N

Cl

Cl

NH3

Cl

NH3

. [ ( ) ]I II 03 3 3 3 3CoCl 3NH or Co NH Cl

Co

H3N

Cl

Cl

NH3Cl

NH3

ii) Every element tends to satisfy both its primary and secondary vlencies. In order to meet thisrequirement a negative ion may often show a dual behaviour, i.e. it may satisfy both primary andsecondary valencies since in every case the fulfillment of coordination number of the centralmetal ion appears essential.

Characteristic of Co(III) ammines

7.2 EFFECTIVE ATOMIC NUMBERS

The number of secondary valencies in the Werner’s theory is now called the coordination number of the central metal in the complex. This is the number of ligand atoms bonded to the central metal ion.Each ligand donates an electron pair to the metal ion, thus forming a coordinate bond. Transition

Ammines (i.e.complexes

No. of –Cl

ions

precipitatedas AgCl by

3AgNOsolution

Molar

conducti

vityrange–1(ohm )

Total No.

of ions

given bycomplex in

soln.

Charge

type on

ions

Ionic Formulation

3 3CoCl .6NH 3 430 4 (3+, 1–) III 3 –3 6 3[Co (NH ) ] (Cl )

3 3 2CoCl .5NH .H O 3 430 4 (3+, 1–) III 3 –3 5 2 2[Co (NH ) (H O)] (Cl )

3 3CoCl .5NH 2 250 3 (2+, 1–) III 2 –3 5 2[Co (NH ) Cl] (Cl )

3 3CoCl .4NH 1 100 2 (1+, 1–) III –3 4 2[Co (NH ) Cl ] Cl

3 3CoCl .3NH 0 0 – – III 103 3 3[Co (NH ) Cl ](Non.electrolyte)

(Non-electrolyte)





metals form coordination compounds very readily because they have vacant d orbitals which canaccommodate these electron pairs. The electronic arrangement of the noble gases is known to be verystable. Sidgwick, with his effective atomic number rule, suggested that electron pairs from ligandswere added until the central metal was surrounded by the same number of electrons as the next noble

gas. Consider potassium hexacyanoferrate(II) [ ( ) ]4 6K Fe CN (formerly called potassium

ferrocyanide). An iron atom has 26 electrons, and so the central metal ion 2Fe has 24 electrons. The

next noble gas Kr has 26 electrons. Thus the addition of six electron pairs from six –CN ligands adds

12 electtrons, thus raising the effective atomic number (EAN) of 2Fe in the complex 4–6[Fe(CN) ] to

36.

[24 (6 2) 36]

The EAN rule correctly predicts the number of ligands in many complexes. There are, however, asignificant number of exceptions where the EAN is not quite that of a noble gas. If the original metalion has an odd number of electrons, for example, the adding of electron pairs cannot result in a noblegas structure. The tendency to attain a noble gas configuration is a significant factor but not anecessary condition for complex formation. It is also necessary to produce a symmetrical structure(tetrahedral, square planar, octahedral) ireespective of the number of electrons involved.

Effective atomic numbers of some metals in complexs

AtomAtomicnumber

ComplexElectrons lostin ionformation

Electrons gainedby coordination

EAN

Cr 246[Cr(CO) ] 0 12 36

[Kr]

Fe 26 4–6[Fe(CN) ] 2 12 36

Fe 265[Fe(CO) ] 0 10 36

Co 27 33 6[Co(NH ) ] 3 12 36

Ni 28

4

[Ni(CO) ] 0 8 36

Cu 29 3–4[Cu(CN) ] 1 8 36

Pd 46 43 6[Pd(NH ) ] 4 12 54 [Xe]

Pt 78 2–6[Pt(Cl ] 4 12 86 [Rn]

Fe 26 3–6Fe(CN) ] 3 12 35

Ni 28 23 6[Ni(NH ) ] 2 12 38

Pd 46 2–4[PdCl ] 2 8 52

Pt 78 23 4[Pt(NH ) ] 2 8 84

7.3 SHAPES OF d ORBITALS

Since d orbitals are often used in coordination complexes it is important to study their shapes anddistribution in space. The five d orbitals are not identical and theorbitals may be divided into two sets. The three 2gt orbitals have identical shape and point between the axes, x, y and z. The two ge orbitals

have different shapes and point along the axes. Alternative names for 2gt and ge are d and drespectively.





x

2gt

orbitals

(d )

x y

y z z

xyd

xz

d

2zd

yz

d

2 2x –yd

x

y

z

ge

orbitals

(d )

8. ISOMERISM

Compounds that have the same chemical formula but different structural arrangement are called

isomers. Because of the complicated formula of many coordination compounds,the variety of bondtypes and the no of shapes possible, many different types of isomerism occur.

[A] STRUCTURAL ISOMERSIM

(I) Polymerization Isomerism: This is not true isomerism because it occurs between compoundshaving the same empirical formula, but different molecular weights. For example,[Pt(NH3)2Cl2], [Pt(NH3)4][PtCl4], [Pt(NH3)4][Pt(NH3)Cl3]2.

(II ) Ionization Isomerism : This type of isomerism is due to the exchange of groups between thecomplex ion and the ions outside it. [Co(NH3)5Br]SO4 is red – violet. An aqueous solution givesa white precipitate of BaSO4 with BaCl2 solution, thus confirming the presence of free

24SO 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 withAgNO3, thus confirming the presence of free Br– ions.

(II I) Hydrate Isomerism : Three isomers of CrCl3.6H2O are known. From conductivitymeasurements and quantitative precipitation of the ionized chlorine, they have been given thefollowing 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)

(IV) Linkage Isomerism : Certain ligands contain more than one atom which could donate anelectron pair. In the

2NO ion, either N or O atoms could act as the electron pair donor. Thusthere is the possibility of isomerism. Two different complexes [Co(NH3)5NO2]Cl2 have beenprepared, each containing the

2NO group in the complex ion.

(V) Coordination Isomerism : If the complex is a salt having both cation and anion as complexions then the ligands can exchange position between the cation and the anion. This will result inthe formation of coordination isomers. For example

3 2 4 3 2 2 4 2 4 2[Co(en) ][Cr(C O ) ]and[Co(en) (C O )][Cr(en)(C O ) ]

2 2 4 2 4 2 3 2 4 3[Cr(en) (C O )][Co(en)(C O ) and[Cr(en) ][Co(C O ) ]

(VI ) Coordination Position Isomerism : If in a multinuclear complex the distribution of ligandsaround the metal centres changes it will result in a different isomer. Such an isomerism is calledcoordination position isomerism. Some typical examples are :





3 4 3 2 2 2 3 3 3 3 2[(NH ) Co Co(NH ) Cl ]Cl and [Cl(NH ) Co Co(NH ) Cl]Cl 2

2

NH

O2

2

NH

O

3 2 2 3 3[(R P) Pt PtCl ] and [Cl(R P) Pt Pt(R P)Cl] Cl

Cl

Cl

Cl

(VI I) Electronic Isomerism: The complex

3 5 2[Co(NH ) NO]Clexists in two forms. One is black

paramagnetic while the other is pink and diamagnetic. Thorough structural studies show that theblack isomer is a Co(II) complex containing neutral NO group whereas the pink one is a Co(III)

complex where –NO ion is present. This kind of isomerism is known as electronic isomerism.

[B] STEREO ISOMERISMS

Stereoisomers have the same bonds but the arrangement of atoms in space is different.Stereoisomerism can be divided into two kinds: geometrical and optical.

(I) Geometrical isomerism or cis-trans isomerisms: It occurs when ligands can assume differentpositions around rigid bonds with the metal ion. For example, the compound 3 2 2Pt(NH ) Cl has

a square planar structure. The two possible arrangements are.

Pt

3H N

3H N Cl

Cl

Pt

3H N

3NHCl

Cl

In the cis-isomer the two ammonia molecules are next (cis) or adjacent to each other whereas inthe trans-isomer the two ammonia molecules are across (trans) to each other. This type of isomerism is not possible for complexes with coordination number 2 (linear molecule), 3(trigonal planar) and 4 (tetrahedral geometry).

For square planar complexes 4 3Ma ,Ma b or 3Mab where a and b etc., are monodentate

ligands, again the geometrical isomerism is not possible. However, square planar complexes of the type 2 2Ma b , 2Ma bc,Mabcd and 2M(AA) , 2M(AB) where AA and AB represent

symmetrical and unsymmetrical chelating agents – do give geometrical isomers. A fewexamples are given below:

Type Compound Isomers

2 2Ma b 3 2 2Pt(NH ) Cl

Pt

3H N

3

H N Cl

Cl

Pt

3H N

3NHCl

Cl

2Ma bc 3 2Pt(NH ) ClBr

Pt

3H N

3H N Br

Cl

Pt

3H N

3NHBr

Cl





Mabcd3 5 5Pt(NH )(C H N)(Cl)(Br)

Pt

3H N

Br Cl

Pt

3H N5 5NC H

Cl

5 5NC H

Br

Pt

3H N

5 5NC H

Cl

Br

Bridged binuclear planar complexes like [( ) ]3 2 2Pt PEt Cl may exist in three isomeric forms:

ClPt

ClPt

PEt3

Et3P Cl Cl

ClPt

ClPt

Cl

Et3P Cl PEt3

Et3PPt

ClPt

Cl

Et3P Cl Cl

trans– cis– unsymmetrical

Six coordinated octahedral complexs of the type , , , ,4 2 3 3 3 2 3 2 2Ma b Ma b Ma b c Ma bcd Ma b cd ,

,2Ma bcde Mabcdef would all give geometrical isomers. Systems with one or two bidentate ligands

and rest monodentate would also give geometrical isomers. Thus we see that with this geometry a lagenumber of isomers are possible whether they can be isolated or separated is a different question whichdepends on so many factors. As we increase the number of different ligands, the possible number of isomers increases. For example, 4 2Ma b type of complex would give only two isomers cis–and trans–.

Similarly for 3 3Ma b type of complex we again get two isomers facial (fac–) and meridional (mer–)

isomers. In the former (fac–) three ligands of one type form one triangular face of the octahedron andthe other three on the opposite face. In the latter (mer–) one set of these ligands are arranged around anedge of the octahedron whereas the other set occupies the opposite edge as shown in figure.

M

a

a b

b

b

a

Mab

b

a

b

a

Facial and meridional isomers of 3 3Ma b complex

(II ) Optical isomerism: Two isomers whichhave almost identical physical and chemical propertieslike mp, bp, density, colour etc., but differ in the way they rotate the plane-polarised light arecalled optical isomers. Such optically active compounds exist in pairs and are known asstereoisomers or enantiomers. These isomers aer non-superimposable mirror images of eachother. Hence, any molecule which contains either a centre of symmetry or a plane of symmetrywill not show optical isomerism.

Optical isomerism is rarely observed in square planar complexes. Tetrahedral complexes of thetype [ ( ) ]2M AB do give optical isomers as shown in figure especially where M =Be, B etc.

M

A

B

A

B

M

A

B

A

B





However, optical isomerism is very common with octahedral complexes of the type( ) , ( ) , ( ) ,3 2 2 2M AA M AA ab M AA a b ( )( ) 2M AA BB a etc. A few typical examples are shown in

figure.

Co

2H N

2H N

2H N

2H N

2H C

2H C

2CH

2CH

2CH

2CH

Co

N

2NO

2NON

N

N

en

en

Co

N

2O N

2O N N

N

N

en

en

Co

2H N

2NH

2H C

2H C

2CH

2CH

2CH

2CH3 3

2NH

2NH

2H N

2H N2NH

2NH

Question 6: How do you distinguish between the following pairs of isomers?i) [ ( ) ] [ ( ) ]B3 5 3 5Cr NH Br Cl and Cr NH Cl r

ii) [ ( ) ][ ( ) ] [ ( ) ][ ( ) ]3 6 2 6 3 6 2 6Co NH Cr NO and Cr NH Co NO

Solution: i) The isomers can be distinguished by using 3AgNO reagent. One gives curdy

precipitate of AgCl soluble in ammonia while the other will form light yellowprecipitate of AgBr partly soluble in ammonia.

ii) The isomers can be distinguished by passing their aq. solutions through a cationexchanger. In the isomer 3 6 2 6[Co(NH ) (Cr[NO ) ] the cation 3

3 6[Co(NH ) ]

will be replaced by H ions from the resin. The resulting solution will thuscontain 3 2 6H [Cr(NO ) ]. In the other case, the resulting solution after passage

through the exchanger will contain 3 2 6H [Co(NO ) ]. On adding KCl solution,3–

2 6[Co(NO ) ] ion will give yellow precipitate of potassium cobaltinitrate,

3 2 6K [Co(NO ) ]. Hence the resulting solutions obtained from the two isomers

can be distinguished.

9. HYBRIDIZATION AND GEOMETRY

9.1 FORMATION OF AN OCTAHEDRAL COMPLEX

Let us consider the case of six ligands forming an octahedral complex. For convenience, we mayregard the ligands as being symmetrically positioned along the axes of a Cartesian co-ordinate systemwith 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 ionbecause 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 2y2x2zdandd

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

–2 2 2z x yd and d at a higher energy than the other three.

2y2x2zd,d

dxy, dyz, dxz

0

[A] Factors affecting the magnitude of 0

i) Oxidation state of the metal ion: The magnitude of 0 increases with increasing ionic charge onthe central metal ion. As the ionic charge on the metal ion increases greateris the attraction for theligands, greater the repulsion and hence greater the magnitude of 0.

ii) Nature of the ligands: Based on experimental observation for a wide variety of complexes, it ispossible to list ligands in order of increasing field strength in a spectrochemical series. Although itis not possible to form a complete series of all ligands with a single metal ion, it is possible toconstruct 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 spectraand 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 greatesteffect. To the contrary most anionic ligands lie at the low end of the spectrochemical series. Furthermore, 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.

[B] Outer orbital and Inner orbital complexes

Consider the complexes [CoF6]3– and[Co(NH3)6]

3+

The electronic configuration of Co3+ ion is

3d 4s 4p 4d

↿⇂ ↿ ↿ ↿ ↿

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 theelectrons in 2y2x2z

dandd

orbitals are very less (or) in other words

we can say that the electrons in 2zd and 2y2x

d

cannot move away

from the approaching ligands since they have insufficient energy topair 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 (one4s, three 4p and two 4d). Since outer d orbitals are used, this is an outer orbit complex. The orbitals arehybridised 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 3delectrons.

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 differencebetween the two sets of orbitals is much greater and this energydifference is sufficient to allow the electrons in 2y2x2z

dandd

orbitals to move into the half filled dxz, dxy and dyz orbitals, eventhough this pairing requires energy. We can also view this like, theligand repel the electrons in higher energy level to an extent suchthat they get paired up against Hund's rule

So,

The 2zd and 2y2x

d

orbitals become vacant.

The six ligands each donate a lone pair to thefirst six vacant orbitals, which are: two 3d, one4s and three 4p. Inner d-orbitals are used andso this is an inner orbital complex. The orbitalare hybridised and written d2sp3 to denote theuse of inner orbitals.

Since, the orginal unpaired electrons have been forced to pair off, there is a low spin complex and is in factdiamagnetic.

The inner and outer orbital complexes may be distinguished by magnetic measurements. Since the outerorbital complexes use high energy levels, they tend to be more reactive. The inner orbitals are sometimescalled inert orbitals.





Distribution of d-electrons in 2gt and ge sets in strong(er)

and weak (er) octahedral ligand fields.

IonsStrong(er) field (low-spin or spin-paired

complexes) 0( P) Weak(er) field (high-spin or spin free

complexes) 0( P)

p q

2g gt eConfiguration n s

p q

2g gt econfiguration n S

1d2d3d

1 02g gt e

2 02g gt e

3 02g gt e

1

2

3

½

1

3/2

1 02g gt e

2 02g gt e

3 02g gt e

1

2

3

½

1

3/2

4d5d6

d7d

4 02g gt e

5 02g gt e

6 0

2g gt e6 12g gt e

2

1

0

1

1

½

0

1/2

3 12g gt e

3 22g gt e

4 2

2g gt e5 22g gt e

4

5

4

3

2

5/2

2

3/2

8d9d10d

6 22g gt e

6 32g gt e

6 42g gt e

2

1

0

1

½

0

6 22g gt e

6 32g gt e

6 42g gt e

2

1

0

1

½

0

9.2 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 2zd and 22 yx

d

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

3d 4s 4p

↿⇂ ↿⇂ ↿⇂ ↿ ↿

Consider, a Ni2+ion with one electron in the 22 yxd

orbital and one

in the 2zd orbital. The approach of ligands along x, y and z axeswill result in increase in the energy of these orbitals. Because the

2y2xd

orbital is attacked by four ligands and the 2z

d by only two,

the energy of 2y2xd

orbital will increase most. If the ligands have

enough strong field, the electrons will be forced out of the

2y2xd

orbital into the 2z

d .





Thus four ligands can approach along x, –x, y and –y axes. A ligand approaching in the z (or) – zdirection attempting to donate a lone pair meets the very strong repulsive forces from a completelyfilled 2Z

d orbital. Thus only four ligands succeed in bonding to the metal.

9.3 FORMATION OF TETRAHEDRAL COMPLEXES

A regular tetrahedron is related to a cube with an atom at the centre and four of the eight cornersoccupied by ligands.

yx

z

The directions x,y and z point to the centre of the faces. The 2Zd and 2y2x

d

orbitals point along x,y

and z axis and dxy, dyz and dxz orbitals point in between x,y andz.

The directions of approach of the ligands does not coincide exactly with either the 2zd and 2y2x

d

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 2

Z

d and 2y

2x

d

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

3d 4s 4p

↿ ↿ ↿ ↿ ↿

Since Cl– ion is a weak field ligand it is unable to pair the unpaired electrons and hence, the Cl – ionuses 4s and 4p orbitals to form atetrahedral complex of sp3 hybridisation.

10. CHELATION

Polydentate ligands whose structures permit the attachment of their two or more donor atoms (or sites)to the same metal ion simultaneously and thus produce one or more rings are called chelate or

chelating ligands (from the Greek for claw) or chelating groups. However, it should be noted that everymultidentate ligand is not necessarily a chelating ligand — the coordinating atoms of the ligand may beso arranged that they cannot be coordinated to thesame metal atom to produce a ring structure. Thus

— — —2 2 2 2NH CH CH NH is a chelating ligand, while

2 2 2N — CH — CH — NH

2 2—(CH ) —

2 2—(CH ) —

is not, although both are diamines. The formation of such rings is termed chelation and the resultingring structures have been called chelate rings or simply chelates by Morgan and Drew. These are alsocalled chelated complexes or cyclic complexes and the term chelation is also called cyclisation.

10.1 CLASSIFICATION OF CHELATESSince the polydentate ligands may be attached to the central metal ion through two kinds of functionalgroups namely acidic and coordinating groups to form covalent and coordinate linkages respectively,the classification of chelates follows the number and kind of linkages by which the metal ion isattached with the ligands. The covalent bonds are formed by the replacement of one or more H-atomsfrom the acidic groups present in the ligand by the metal atom. The examples of most common gropsof this type are —COOH (carboxyl), — 3SO H (sulphonic), —OH (enolic hydroxyl) and =NOH

(oxime). Coorinating linkages, without the replacement of H, are formed by the donation of an electronpair from the ligands. The examples of most common groups of this type are — 2NH (primary,





secondary and tertiary amines), =NOH (oxime), —OH (alcoholic hydroxyl), =CO (carbonyl) and—S— (thio ether).

Ordianrily such chelate rings as mentioned above are most stable, because of reducd strain, when theyhave 5 or 6 members includingthe metal ion. The enhanced stability of complexes containing chelatedligads (i.e., multidentate ligands) is known as the chelate effect.

11. SOLVED OBJECTIVE QUESTIONS

Question1: Which electronic configuration represents a transition element?

(a) 2 2 6 2 6 10 2 61s 2s 2p 3s 3p 3d 4s 4p

(b) 2 2 6 2 6 10 2 61s 2s 2p 3s 3p 3d 4s 4p

(c) 2 2 6 2 6 2 21s 2s 2p 3s 3p 3d 4s

(d) 2 2 6 2 6 21s 2s 2p 3s 3p 4s .

Solution: (c)

Question2: Transition metals are less reactive because of their(a) High I.P. and low melting point(b) High I.P. and high melting point(c) Low I.P. and low melting point

(d) Low I.P. and high melting point.Solution: (b)

Question 3: The stable oxidation states of Mn are(a) +2, +3 (b) +2, +7(c) +3, +7 (d) +3, +5.

Solution: (b).

Question 4: Which is not true statement about FeO?(a) it is non-stoichiometric and is metal deficient(b) it is basic oxide(c) its aqueous solution changes to ( )3FeOH and then to .( )2 3 2 nFe O H O by

atmospheric oxygen(d) it gives red colour with KCNS.

Solution: (d)

Question 5: Paramagnetism is given by the relation ( )2s s 1 magnetons where s’ is the

total spin. On this basis, the paramagnetism of Cu ion is:(a) 3.88 magnetons (b) 2.83 magnetons(c) 1.41 magnetons (d) zero.

Solution: (d)

Question 6: Fe is made passive by:(a) dil 2 4H SO (b) dil HCl

(c) aqua regia (d) conc. 2 4H SO

Solution: (c)

Question7: Maximum magnetic moment is shown by:(a) 5d (b) 6d

(c) 7d (d) 8d .Solution: (a)

Question8: –4MnO is of intense pink colour, though Mn is in (+7) oxidation state. I t is due to

(a) oxygen gives colour to it(b) charge transfer when Mn gives its electron to oxygen(c) charge transfer when oxygen gives its electron to Mn making it Mn (+VI ) hence

coloured(d) None is correct.





Solution: (c)

Question9: Out of – – –, , , , ,3 2 24 4 4 4 4 4SiCl TiCl PO SO CrO CCl isostructrual are:

(a) ,4 4SiCl TiCl (b) – –,2 24 4SO CrO

(c) both (d) none of these.Solution: (c)

Question 10: Increasing basic properties of ,2 2 TiO ZrO and 2HfO are in order:(a) 2 2 2 TiO ZrO HfO (b) 2 2 2HfO ZrO TiO

(c) 2 2 2HfO TiO ZrO (d) 2 2 2ZrO TiO HfO .

Solution: (a)

Question 11: 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

Solution: (d)

Question12: CrO3 dissolves in aqueous NaOH to give(a) Cr

2O

7

2– (b) CrO4

2–

(c) Cr(OH)3 (d) Cr(OH)2Solution: (b)

Question13: Transition elements are coloured due to(a) small size (b) metallic nature(c) unpaired d-electrons (d) none of these

Solution: (c)

Question14: 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.

Solution: (c)Question15: Pt, Pd and Ir are called noble metals because

(a) Alfred 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

Solution: (c)

12. SOLVED SUBJECTIVE QUESTIONS

Question 1: Calculate the magnetic moments of Fe2+and Fe3+

Solution: In 2Fe there are 4 unpaired electrons.

4(4 2) 4 6 24 4.89B.M.

In 3Fe there are 5 unpaired electrons.

5(5 2) 5 7 35 5.91B.M.

Question 2: When Zn is added to an acidified solution of 2 2 7K Cr O , the colour of the solution

changes from orange to green, then to blue and over a period of time, back to green.Write equations for this series of reactions.

Solution: Zn reduces 2–2 7CrO (orange)/H to 3Cr (green) and then to 2Cr (blue), but

atmospheric 2O oxides 2Cr (blue) back to 3Cr (green):





2 –Zn Zn 2e 2– – 3

2 7 2Cr O 14H 6e 2Cr 7H O 3 – 2Cr e Cr

2 32 2

air

4Cr O 4H 4Cr 2H O

Question 3: a) CO and –CN are toxic. Explain by a suitable example.

b) –22 8S O (peroxy disulphate ion) oxidizes Ag in the presence of excess

pyridine. Write chemical reaction.

Solution: a) CO is toxic because it forms a complex with iron of haemoglobin in the blood,and this complex is more stable than oxy-haemoglobin. This prevents thehaemoglobin in the red blood corpuscles from carrying oxygen round the body. This causes an oxygen deficiency, leading to unconsciousness and then death.

The extreme toxicity of cyanides is due to –CN complexing with metals incytochrome-enzymes and haemoglobin in the body, thus preventing normalmetabolism.

b) 2– 2 2–2 82Ag S O 2Ag 2SO

2 22

red

Ag 4Py [Ag(Py) ]

Question 4: 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 treatedwith ammonia solution.

iii) The compounds of Zn, Cd and Hg are usually white.

iv) A drak blue precipitate is formed when NaOH solution is added to CuSO4

solution. The precipitate darkens on heating.

Solution: i) On heating hydrated 3FeCl , anhydrous 3FeCl is not formed as water of

crystallization reacts to form 2 3FeO and HCl.ii) 2 2Hg Cl reacts with 3NH to form a mixture of mercury and mercuric

aminochloride which is a black substance.

2 2 3 2

black

Hg Cl NH Hg Hg(NH )Cl HCl

iii) In the compounds of zinc metals, 2M ions possess the penultimate d-orbitalsdoubly occupied i.e., 10(n–1)d configuration. There is no d-d transition. Hencethe compounds of zinc metals are colourless.

iv) NaOH reacts with 4CuSO when dark blue precipitate of 2Cu(OH) is formed. This precipitate of heating forms CuO which is black in colour. Hence, the colourdarkens on heating.

4 2 2 4CuSO 2NaOH Cu(OH) Na SO

2 2BlackCu(OH) CuO H O

Question 5: Explain the followingi) Copper (l) 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.iv) The species [CuCl4]

2– exist but [Cul4]2– does not.

Solution: i) Cu(l) salts under go disproportionation in aqueous solution22Cu Cu Cu





ii) –I ion is a stronger reducing agent in comparision to Cl– ion. 3Fe is easilyreduced by –I ion

3 – 222Fe 2l 2Fe l

iii) The hydration energy of AgF is higher than its lattice energy. Hence it is solublein water. The hydration energy values of other halides are smaller than theirlattice energy values. Hence these halides are insoluble in water.

iv)–

I is a stronger reducing agent than–

Cl ion. It reduces2

Cu

ion into Cu

ion.Hence cupric iodide is converted into cuprous iodide. Thus the species 2–

4[Cul ]does not exist.

Question 6: Potassium pentacyanocobaltate (II) is diamagnetic in solid state but paramagnetic inaqueous solution. Why?

Solution: 2 5K [CO(CN) ] Potassium pentacyanocobaltate(II)3 6Co :d configuration

As –CN is a strong ligand hence it forces the electrons to pair up. There are no

unpaired electrons left so diamagnetic. But in aq. solution the ligand is 2H O which

is a weak one in comparision to –CN . There are 4 unpaired electrons and henceparamagnetic.

Question 7: FeSO4 solution mixed with (NH4)2SO4 solution (in the molar ration 1:1) gives the testof Fe2+ ion but CuSO4 solution mixed with liquid NH3 (the molar ration 1:4) does notgive the test of Cu2+. Explain why?

Solution: 4FeSO does not from any complex with 4 2 4(NH ) SO instead, they form a double salt

4FeSO 4 2 4 2(NH ) SO .6H O which dissociates completely into ions but 4CuSOcombines with 3NH to from the complex 3 4 4[Cu(NH ) ]SO in which complex ion,

23 4[Cu(NH ) ] does not dissociate to give 2Cu ions.

Question8: Explain why a green solution of potassium manganate (VI) 2 4K MnO turns purple

and a brown solid is precipitated when 2CO is bubbled into the solution.

Solution: 2CO in aqueous solution is acidic

–2 2 2 3 3CO H O H CO H HCO

In presence of H , 2–4MnO disproportionates to –

4MnO (purple, by oxidation) and

2MnO (brown solid, by reduction)

2– –4 4 2 2

VI VII IV

3MnO 4H 2MnO MnO 2H O

Question 9: When mercury is oxidized with a limited amount of oxidizing agent (i.e., an excess of

Hg) then IHg comounds are formed. If there is an excess of oxidizing agent, thenIIHg compounds are formed.

2 –Hg 2e Hg E 0.85V 2 –2Hg 2e 2Hg E 0.79V

Solution: The standard reduction potentials are so close that oxidizing agents like 3HNO will

convert Hg to 2Hg rather than Hg(I) if the oxidizing agent is present in excess.

The reduction potential diagram shows that 22Hg is stable to disproportionation by a

small margin under standard conditions.2 2

2Hg Hg Hg E 0.06V





Question 10: In an atmosphere with industrial smog, Cu corrodes to a basic sulphate( )2 2 4Cu OH SO cu and basic carbonate ( )2 2 3Cu OH CO . Propose a seriese of

chemical reaction to describe this corrosion

Solution: 2 3 2 2 2 42Cu H O SO O Cu (OH) SO

2 3 2 2 2 32Cu H O CO O Cu (OH) CO

Fortunately this corrosion product forms a tough adherent coating that protects theunderlyingmetal. Overall reaction is a combination of oxidation-reduction, acid-baseand precipitation reaction

Question 11: Why is hydrochloric acid not used to acidify a permangnate solution is volumetricestimation of Fe2+and C2O4

2–.

Solution: This is because a part of the oxygen produced from KMnO4 +HCl is used is oxidizingHCl to Cl2

4HCl +2(O) 2H2O +Cl2

Question 12: 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 withlower 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. Thecomplimentary colour, viz blue is reflected.

Question 13: 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.

Question 14: K 2 [Pt Cl6] is well known compound whereas corresponding Ni compound is notknown. State a reason for it.

Solution: This is because Pt4+ is more stable than Ni4+as the sum of four ionization energies of Ptis less than that of Ni.

Question 15: 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 wemove along any transition series the nuclear charge increases and size decreases.

13. OBJECTIVE ASSIGNMENTS

LEVEL – I

1 Which of the following electronic configuration belong to transition elements?(a) KL, 2 6 6 13s ,p d ,4s (b) KL, 2 6 10 2 33s p d ,4s p

(c) KL, 2 6 10 2 13s p d ,4s p (d) KLM, 2 6 10 2 14s p d ,5s p .

2. The no. of d-electrons present in 2Fe ions is

(a) 6 (b) 4(c) 8 (d) 3.

3. Adam’s catalyst is(a) Ft and PtO (b) Pt(c) Pt and 2PtO (d) 2PtO and PtO

4. Which is the common oxidation state of the first transition series of elements?(a) +2 (b) +6(c) +8 (d) +4.





5. If orange-red colour is absorbed from white-light, the observed colour is(a) Yellow (b) Orange(c) Blue (d) Green.

6. An example of double salt is(a) Bleaching powder (b) 4 6K [Fe(CN) ](c) Hypo (d) Potash alum.

7. 2 Ti is purple while 4 Ti is colourless, because:(a) 2 Ti has 23d configuration(b) 4 Ti has 23d configuration(c) There is no crystal field effect in 4 Ti

(d) 4 Ti is very small cation when compared to 2 Ti and hence does not absorb any radiation.

8. 4FeSO solution gives brown colour ring in testing nitrates or nitrites. this is

(a) 22 5[Fe(H O) NO] (b) 2

2 5 2[Fe(H O) NO ]

(c) 22 4 2[Fe(H O) (NO) ] (d) 2

2 4[Fe(H O) NO] .

9. Which shows maximum magnetic moment among the bivalent ions of the first transition series?(a) 2Fe (b) 2Co

(c) 2Ni (d) 2Mn

10. Which is not true statement:(a) ions of d-block elements are coloured due to d—d transition(b) ions of f-block elements are coloured due to f—f transition(c) 3 4

2 6 2 6[Sc(H O) ] ,[Ti(H O) ] are coloured complexes

(d) Cu is colourless ion

11. Magnetic moment [Ag(CN)2]– is zero, the no. of unpaired electrons is

(a) Zero (b) One(c) Two (d) Three

12. Which of the following is not a complex(a) NiSO4.(NH4)2SO4.6H2O (b) [Co(NH3)4Cl2]Cl(c) K 2[Ni(CN)4] (d) [Pt(NH3)Cl2Br]Br

13. Which one is a monodentate ligand(a) NC (b) NH3

(c) H2O (d) All

14. The complex [Cr(H2O)4Br2]Cl gives the test for(a) Br– (b) Cl–

(c) Cr3+ (d) Br– and Cl– both

15. In the reaction

4KCN +Fe(CN)2 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-

16. Oxidation state of Cobalt in the complex [Co(NO2)6]3– is

(a) +2 (b) +3(c) +1 (d) +2 and +3

17. Coordination number of platinum in [Pt(NH3)4Cl2]++ion is

(a) 4 (b) 2(c) 8 (d) 6





18. 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

19. Low spin complex is formed by(a) sp3d2 hybridisation (b) sp3d hybridisation(c) d2sp3 hybridisation (d) sp3 hybridisation

20. 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–

LEVEL – II

1. In 2 5Na [Fe(CN) NO], sodium nitroprusside,

(a) oxidation state of Fe is +2 (b) this has NO as ligand(c) both correct (d) none is correct.

2. Stability of Cu and Ag halide complexes are in order:(a) I >Br >Cl >F (b) F >Cl >Br >I

(c) Cl >F >I >Br (d) Br >I >Cl >F.

3. An acidic solution contains 2 2Cu ,Pb and 2Zn . If hydrogen sulphide gas is passed through thissolution, the precipitate will contain:(a) CuS and ZnS (b) PbS and ZnS(c) CuS and PbS (d) CuS, PbS and ZnS.

4. Magnetic moment of Cr(Z =24), 2Mn (Z 25)andFe (Z =26) are x, y, z. They are in order:(a) x <y <z (b) x =y <z(c) z <x =y (d) x =y =z.

5. Select incorrect statement(s)

(a) ionization energies of 5d-elements are greater than those of 3d and 4d elements(b) Cu(I) is diamagnetic while Cu(II) is paramagnetic(c) 3

2 6[Ti(H O) ] is coloured while 32 6[Sc(H O) ] is colorless

(d) transition elements cannot form complexes.

6. AgCl and NaCl are colourless. NaBr and NaI are also colourless but AgBr and AgI are coloured. This is due to :(a) Ag polarizes –Br and –I (b) Ag has unpaired d-orbital

(c) Ag depolarizes –Br and –I (d) None is correct.

7. The lowest degree of paramagnetism is shown by(a) 4 2MnSO .4H O (b) 4 2FeSO .6H O

(c) 4 2CuSO .5H O (d) 4 2NiSO .6H O.

8. Which is correct statement?(a) in less acidic solution 2 2 7K Cr O and 2 2H O give violet coloured diamagnetic

–2[CrO(O )(OH)] ion

(b) in alkaline 2 2 3 8H O ,K CrO (with tetraperoxo species) 3–2 4[Cr(O ) ] is formed

(c) in ammoniacal solution, 3 3 4(NH ) CrO is formed(d) all correct.





9. The structures of 4Ni(CO) and 3 2 2Ni(PPh ) Cl are(a) square planar(b) tetrahedral and square planar respectively(c) tetrahedral(d) square planar and tetrahedral respectively.

10. The hybridization states of the central atoms in the complexes 3–

6Fe(CN) , 4–

6Fe(CN) , and

3–2 6Co(NO ) are

(a) 2 3 3 4 2d sp ,sp andd sp respectively (b) 2 3 3 3 2d sp ,sp dandsp d respectively

(c) 2 3 3 2 2d sp ,sp d anddsp respectively (d) all 2 3d sp .

11. Among the following aquated metal ions, which has the highet degree of paramagnetism?(a) 3

2 6[Cr(H O) ] (b) 22 6[Fe(H O) ]

(c) 22 6[Cu(H O) ] (d) 2

2 2[Zn(H O) ] .

12. Which of the following is a high-spin (spin-free) complex?(a) 3

3 6[Co(NH ) ] (b) 4–6Fe(CN) ]

(c) 3–6[CoF ] (d) 23 6[Zn(NH ) ] .

13. The formation of the complex ion 33 6[Co(NH ) ] involves the 3 2sp d hybridization of 3Co .

Therefore the complex ion should have(a) an octahedral geometry (b) a tetrahedral geometry(c) a square-planar geometry (d) a square-antiprismic geometry.

14. Which of the following will produce a white precipitate upon reacting with 3AgNO ?

(a) 3 6 3[Co(NH ) ]Cl (b) 3 3 3[Co(NH ) Cl ](c) 2 2 2K [Pt(en) Cl ] (d) 2 4[Fe(en) ]Cl .

15. Which of the following is -bonded organometallic compound?

(a) Ferrocene (b) Diethyl zinc(c) Ethylmagnesium iodide (d) None of these.

16. Which of the following complexes produces three moles of silver chloride when its one mole istreatedwith excess of silver nitrate(a) [Cr(H2O)3Cl3] (b) [Cr(H2O)4Cl2]Cl(c) [Cr(H2O)5Cl]Cl2 (d) [Cr(H2O)6]Cl3

17. 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

18. The number of chloride ions which would be precipitated when the complex PtCl4.4NH3 is treatedwith silver nitrate is(a) Four (b) One(c) Three (d) Two

19. The number of ions per mole of the complex CoCl3.5NH3 in aqueous solution will be(a) Four (b) Nine(c) Three (d) Two

20. 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









14. SUBJECTIVE ASSIGNMENTS

LEVEL - I

1. Why Zn+2 salts are white while Ni2+salts are blue?

2. Why the transition elements have high enthalpy of hydration?

3. Explain briefly why zinc and cadmium are soft metals?

4. Explain, the existence of 4OsO in terms of trends in oxidastion states.

5. Explain the following

a) Mercury is a liquid

b) 2 TiO is white but 3 TiCl is violet

6. Determine the coordination number and the oxidation state of the transition metal ion in each of thefollowing complexes:

a) 4–6[Fe(CN) ]

b) 3–6[Fe(CN) ]c) 3 2 2[Pt(NH ) Cl ]

d) 2–4[CuF ]

7. Yttrium with chlorine does not form YCl or 2 YCl but only 3 YCl . How does this agree with the trendsin stability of oxidation states?

8. Predict the spin only magnetic moment for:

a) 2Fe

b) 7Mn

c) Cu

d) 3 Ti

9. [NICl4]2– is paramagnetic while [Ni(CO)4] is diamagnetic. Why?

10: [Ti(H2O6]3+ is coloured whreas [Sc(H2O)6]

3+ is colourless, though it is a transition metal complex,Explain the observation.

LEVEL - II

1. Write the formula ior each of the following complexes.

i) Ammonium heptafluorozirconate (IV)

ii) Tetraaquadichloroiron (III) ion.iii) Dichlorobis(ethylenediamine) chromium (III) tetrachloropalladate(II)

iv) Dicyanobis(ethylenediamine) cobait (III) chlorate

v) Dichloro tetraammineplatinum (IV) ion

vi) Bromotriammineplatinum (II) nitrite

vii) Diaquatetrachlorochromate(III) ion

viii) Bis(cyclopentadienyl)iron (II)

ix) Hexaaquairon (II) sulphate





x) Potassium hexacyanoferrate (II)

xi) dichlorotetramminecobalt (III) ion

xii) Carbonylchlorobis (triphenylphoshine) iridium(l)

xiii) Tetraamminedichlorocobalt(III) hexacyanochromate(V)

xiv) -hydroxo-bis (pentammine chromium) (III)

xv) tri- -carbonyl-bis (tricarbonyl iron)2. Name the following complexes according to the IUPAC system of nomenclature

i) 3 4 2 3 2[Co(NH ) (H O)Br](NO )

ii) 3 3[Co(en) ]Cl

iii) 3 5 3 5 5[(NH ) Cr— OH — Cr(NH ) ]Cl

iv) 2 3 2Ni(CO) (PPh )

v) 4 2 2 TiCl (Et O)

vi) 33 6[Cr(NH ) ]

vii) 4 4[Pt(Py) ][PtCl ]

viii) 2–4[Zn(NCS) ]

ix) 3–4[Cd(SCN) ]

x) 3 2 3[Co(en)(NH ) ClBr]NO

3. Why 4CuSO is blue while 4ZnSO is white?

4. In 22 6[Co(H O) ] the observed magnetic moment is higher than the spin-only value. Explain the

reason for this in the space provided below.

5. Write down all possible isomers of 2 2 2

[Co(en) Cl ]NO and give their names.

6. The complex 2 3 2[PdCl (PPh ) ] gives geometrical isomers and is found to be diamagnetic whereas an

analogous compound 2 3 2[NiCl (PPh ) ] does not give any geometrical isomers and is found to beparamagnetic. Explain these observations?

7. Each of the compounds 3 6 4 3 6 3 3 4 3[Pt(NH ) ]Cl ,[Cr(NH ) ]Cl ,Co(NH ) .Cl and 2 6K [PtCl ] has beendissolved in water to make its 0.001 M solution. Rank then in order of their increasing conductivity insolution.

8. Explain the following:

i) The complex 2–4[CuCl ] exists, but 2–

4[Cul ] does not.

ii) Gold is not attacked by common acids but dissolves in aqua regia.

iii) Copper dissolves in aqueous KCN solution with the evolution of hydrogen.iv) Zinc becomes dull in moist air.

9. The magnetic moment of 3–6[Mn(CN) ] is 2.8 B.M and that of 2–

4[MnBr ] is 5.9 B.M. What are thegeometries of these complex ions?

10. How will you distinguish between the following isomer pairs

a) i) [CoBr(NH3)5]SO4 and

ii) [Co(SO4) (NH3)6]Br

b) i) [Cr(H2O)6]Cl3 and

ii) [CrCl(H2O)6]Cl2H2O





Level-I II

1. A coordination compound has the formula CoCl3.4NH3. It does not liberate ammonia but precipitateschloride ions as silver chloride. Give the IUPAC name of the complex and write is structural formula.

2. In acidic solution silver (II) oxide first dissolves to produce 2Ag . This is followed by oxidation of

2H O to 2O and the reduction of 2Ag to Ag . Write chemical equation for the dissolution and

redox reactions.

3. An aqueous solution of 3 3CoCl .6NH is found to be a better conductor of electricity than an aqueous

solution of 3 3CoCl .4NH . No ammonium ions are detected by the addition of 2 4H SO to the solutionof either of these compounds. Can you explain the difference in the conductivities of two solutions?

4. Answer the following

i) Which one of 2Fe and F 3e ions is more paramagnetic and why?

ii) Which of the following ions are expected to coloured and why?2 2 3 3 4Fe ,Mn ,Cr ,Cu ,Se ,Ti

iii) Name the two elements of first transition series which have abnormal electronic configuration andwhy?

5. Why is hydrochloric acid not used to acidify a permanganate solution in volumetric estimation of 2Fe

and 2–2 4C O .

6. Both 2–2 7CrO and –

4MnO solutions can be used to titrate 2Fe in acidic mediuim. Suppose you hae

available 0.1 M solutions of each. For a given sample of 2Fe solution, if a given titration requires24.5 mL of 0.1 M 2–

2 7Cr O solution, how many mL of 0.1 M –4MnO solution would have been

required if it had been used instead?

7. a) When Mn 2(OH) is made by adding an alkali to a solution containing 2Mn ions, the precipitatequickly darkens, and eventually goes black.What might be the chemical giving the black colour,

and how it is made ?b) Dimercury (I) odide, 2 2Hg I is a greenish colour and is precipitated if iodide ions are added to a

solution of dimercury (I) sulphate, 2 4HgSO . Likewise the red mercury (II) iodide, 2HgI , is

precipitated from a solution of mercury (II) sulphate, 4HgSO . However, both precipitatesdissolve in excess iodide solution. What might be the reason for this?

8. A solution containing 2.665g of 3 2CrCl .6H O is passed through a cation exchanger. The chloride ions

obtained in solution react with 3AgNO and give 2.87 g of AgCl. Determine the structure of thecompound.

9. Explain the following

i) A little acid is always added in the preparation of aqueous ferrous sulfate solution.ii) Mercuric chloride and stannous chloride cannot exist as such if present together in an aqueoussolution.

iii) A ferrous salt turns brown in air.

iv) Copper hydroxide is soluble in ammonium hydroxide but not in sodium hydroxide.

10: Nickel can be determined by the precipitation of nickel dimethyl glyoximate.

a) What is the reaction?

b) A 1.502 g sample of steel yields 0.259 g of nickel dimethyl glyoximate. What is the per cent of Niis in the steel ? (Ni =59.0)





LEVEL-IV

1. 2NiCl in the presence of dimethyl glyoxime (DMG) gives a complex which precipitates in the

presence of 4NH OH, giving a bright red colour.

a) Draw its structure and show H-bonding

b) Give oxidation state of Ni and its hybridization

c) Predict whether it is paramagnetic or diamagnetic

2. Write the IPUAC nomenclature of the given complex along with its hybridization and structure.

2 3 4K [Cr(NO)(NH )(CN) ], 1.73BM .

3. Deduce the structures of 2–4[NiCl ] and 2–

4[Ni(CN) ] considering the hybridization of the metal ion.Calculate the magnetic moment (spin only) of the species.

4. A metal complex having composition Cr 3 4 2(NH ) Cl Br has been isolated in two forms (A) and (B).

The form (A) reacts with 3AgNO to give a white precipitate readily soluble in dilute aqueousammonia, whereas (B) givens a pale yellow precipitate soluble in concentrated ammonia. Write theformula of (A) and (B) and state the hybridization of chromium in each. Calcualte their magnetic

moments (spin-only value).5. a) Write the chemical reactions associated with the brown ring test’

b) Draw the structures of 33 6[Co(NH ) ] , 2–

4[Ni(CN) ] and 4[Ni(CO) ]. Write the hybridization of atomic orbitals of the transition metal in each case.

c) An aqueous blue coloured solution of a transition metal sulphate reacts with 2H S in acidicmedium to give a black precipitate A, which in insoluble in warm aqueous solution of KOH. Theblue solution on treatment with KI in weakly acidic medium, turns yellow and produces a whiteprecipitate B. Identify the transition metal ion. Write the chemical reactions involved in theformation of A and B.

6. A compound of vanadium has a magnetic moment of 1.73 BM. Work out the electronic configurationof the vanadium ion in the compound.

15. ANSWERS

ANSWERS TO EXCERCISES

1. Cu is a 10d ion so that there is no d— d transition and thus the ion is colourless. But 2Cu O and

2Cu S are coloured because of charge transfer of electrons from 2–O or 2–S to the vacant orbital of

Cu .

2. Consider the two compounds 22 6[Co(H O) ] and 4–

6[Co(CN) ] each of which contains 2Co ion in

an octahedral field. Obviously the compound 22 6[Co(H O) ] is high spin having a magnetic moment

value of 4.6 BM whereas 4–6[Co(CN) ] is low spin with a magnetic moment value of 1.9 BM. The

complex 4–2 6[Fe(NO ) ] contains 2Fe which belongs to a 6d system. As the magnetic moment of the

complex is zero, it indicates that all the six d-electrons are paired up. Hence the compound is low spin.On the other hand, the magnetic moment of 5.94 BM for the complex 3

2 6[Fe(H O) ] corresponds to

five unpaired electrons in 3Fe . It indicates that there is no pairing of d-electrons of 3Fe ion, hencethe complex is high spin.

3: Here both the positive and negative part has the same metal. Procedure is same as earlier. The IUPACname is Tetrapyridlneplatinum(II) tetrachloroplatinate(II).





OBJECTIVE ASSIGNMENTS

LEVEL - I

1. (a) 2. (a) 3. (a) 4. (a) 5. (c)

6. (d) 7. (a) 8. (a) 9. (d) 10. (c)

11. (a) 12. (a) 13. (d) 14. (b) 15. (c)16. (b) 17. (d) 18. (b) 19. (c) 20. (d)

LEVEL - II

1. (c) 2. (a) 3. (c) 4. (c) 5. (d)

6. (a) 7. (c) 8. (d) 9. (c) 10. (d)

11. (b) 12. (c) 13. (a) 14. (a) 15. (a)16. (d) 17. (b) 18. (d) 19. (c) 20. (a)

SUBJECTIVE ASSIGNMENTSLEVEL - I

1. 2Zn has completely filled d-orbitals 10(3d ) while 2Ni has incompletely filled d-orbitals 8(3d )2. 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 increase and size decreases.

3. Zn and Cd have electronic configuration 10 2[Ar]3d 4s and 10 2[Kr]4d 5s , respectively. Therefore, thereis no unpaired electron for metallic bonding. Thus, these metals are soft.

4. The oxidation number of osmium in 4OsO is +8. The stability of higher oxidation states increases aswe go down the group of the transition metals. Osmium being in third transition series is, therefore,stable in oxidation state +8 and exists as 4OsO

5. a) No metallic bond formation

b) Ti(IV) had 0d configuration6. The coordination number and oxidation state of the metal ion in (a) to (d) are given below:

a) +6, +2 b) +6, +3c) +4, +2 d) +4, +2

7. Stability of higher oxidation state increase as we go down a group.8. Ion Electronic Number of Magnetic

Configuration unpaired electrons moment

a) 2Fe 6 0[Ar]3d 4s 4 4.90

b) 7Mn 0 0[Ar]3d 4s 0 0 BM

c) Cu 10 0[Ar]3d 4s 0 0 BM

d) 3 Ti 1 0[Ar]3d 4s 1 1.73 BM

9. In 4[Ni(CO) ] Ni is in zero oxidation state whereas in 2–4[NiCl ] , 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 topair up the unpaired electrons.

10: 3 Ti has some electron in the d-orbital 1(3d ) which can absorb energy corresponding to yellow

wavelength and jump from lower energy level to higher energy level. But 3Sc has no electron in d-orbital.





LEVEL – II

1. i) 4 3 7(NH ) [ZrF ]

ii) 2 2 4FeCl (H O) ]

iii) 4 2 2 4[Cr(en) Cl ] [PdCl ]

iv) 2 2 3[Co(en) (CN) ]ClO

v) 23 4 2[Pt(NH ) Cl ]

vi) 3 3 2[Pt(NH ) Br]NO

vii) –4 2 2[CrCl (H O) ]

viii) 5 5 2[Fe(C H ) ]

ix) 2 6 4[Fe(H O) ]SO

x) 3 6K [Fe(CN) ]

xi) 2 3 4[CoCl (NH ) ]

xii) 3 2[lr(Ph P) (CO)Cl]

xiii) 3 4 2 6[Co(NH ) Cl ][Cr(CN) ]

xiv) 53 5 3 5[(NH ) Cr— OH — Cr(NH ) ]

xv) 3 3 3[(Co) Fe(CO) Fe(CO) ]

2. i) Bromoaquatetraamminecobalt (III)nitrateii) Tris(ethylenediamine)cobalt(III)chlorideiii) -hydroxobis(pentamminechromium(III))chloride

iv) Dicarbonylbis(triphenylphosphine)nickel(0)v) Tetrachlorobis(diethylether)titanium(IV)vi) Hexaamminechromium(III)ion

vii) Tetrapyridine platinium (II) tetrachloroplatinate(II)viii) Tetrathiocyanato-N-zinc(II )ix) Tetrathiocyanato-S-cadmium(I)x) Diamminebromochloroethylenediaminecobalt(III) nitrate

3. 2Cu in 4CuSO has 9 0[Ar]3d 4s configuration and its electron can be promoted to the half filled dorbital. Thus it can undergo d-d transition which absorbs mainlyin the red region of the visible lightand 4CuSO appears blue in colour (blue is complementary colour of red). Because 2Zn in 4ZnSO

has the configuration [A 10 0r]3d 4s , the transition of electron from one d orbital to another is not

possible and no light is absorbed in the visible region of spectrum by 4ZnSO and therefore, it appearswhite.

4. The observed magnetic moment differs from that of the calculated spin only magnetic moment due tothe contribution of orbital motion of the electrons. The observed magnetic moment for 22 6[Co(H O) ]

has contribution from the spin as well as orbital angular momentum and thusthe observed magneticmoment is higher than the calculated spin-only magnetic moment.

5. The compound 2 2 2[Co(en) Cl ]NO may have 2 2[Co(en) (NO )Cl]Cl as an ionization isomer. Thisionization isomer in turn may give a linkage isomer

2 2[Co(en) (NO )Cl]Cl . These three isomers would thus constitute what are known as structural

isomers. All three structural isomers can exist as cis- and trans- isomers, one such pair is given below: The cis- form of each of the above three compounds would have an optical isomer or an enantiomer. This would make a total of nine possible isomers for the given compound.





6. Since 2 3 2[PdCl (PPh ) ] is a four coordinated complex and gives geometrical isomers, it must have

square planar geometry. Though 2 3 2[NiCl (PPh ) ] is also four coordinated and belongs to a 8d system just like the first compound, yet it does not give geometrical isomers. This indicates a tetrahedralgeometry for the compound which will also be paramagnetic.

7. In aq. solution (0.001 M) the complexes will dissociate to give the ions.4 –

3 6 4 3 6[Pt(NH ) ]Cl [Pt(NH ) ] 4Cl (5 ions)

3 –3 6 3 3 6[Cr(NH ) ]Cl [Cr(NH ) ] 3Cl (4 ions)

–3 4 3 3 4 2Co(NH ) .Cl [Co(NH ) Cl ] Cl (2 ions)

2–2 6 6K [PtCl ] [PtCl ] 2K (3 ions)

Thus, their conductivities (in solution) will increase with increasing number of ions liberatedso,

3 4 3 2 6 3 6 3 3 6 4(leastconducting) (mostconducting)

Co(NH ) .Cl K PtCl Cr(NH ) Cl Pt(NH ) Cl

8. i) 11Cu is reduced to 1Cu by –I but not by –Cl .

ii) Theoxidation is aided by the complexation of the product gold (III) as –4AuCl .

iii) The stability of 2–4Cu(CN) ion is so great that the E value for oxidation is reduced to a point

where 2H O can oxidise the copper.

iv) When zinc is exposed to moist air, the surface is affected with the formation of a film of basiczinc carbonate on it. Due to this zinc becomes dull.

2 2 2 3 24Zn 3H O CO 2O ZnCO .3Zn(OH)

9. For complex 3–6[Mn(CN) ] , the number of unpaired electrons is calculated as,

2.8 n(n 2) n 2. 3–

6[Mn(CN) ] has two unpaired electrons. Hence the geometry is octahedral with 2 3d sp

hybridization. For complex 2–4[MnBr ] , the number of unpaired electrons is calculated as

5.9 n(n 2) n 5 2–

4[MnBr ] , has 5 unpaired electrons. Hence the geometry is tetrahedral with 3sp hybridization.

10. a) Isomer (i) gives white ppt of 4BaSO with 2BaCl whereas isomer (ii) does not form a ppt. At the

same time isomer (ii) gives a yellow precipitate of silver bromide with 3AgNO but (i) does not.

b) The water molecule in Isomer (ii) lost easily on heating whereas the water molecule in isomer (i)are not lost easily, being coordinated to the central atom.

LEVEL – III

1. Remembering that co-ordination number of CO is 6 the formula of the complex will be

2 3 4[CoCl (NH ) ]Cl . The name will betetraminedichlorocobalt (III) chloride.

2. Dissolution : 22AgO 2H Ag H O

Redox : 22 24Ag 2H O 4Ag 4H O

3. As no ammonium ionis are detected in the solutions of these compounds, it indicates that ammoniamolecules are in the first coordination sphere of the metal ion in both the cases. The compound

3 3CoCl .6NH can, therefore, be formulated as 3 6 3[Co(NH ) ]Cl which may give four ions per mole in

solution as indicated by the conductivity of the solution. Since, the solution of 3 3CoCl .4NH is less

conducting than that of 3 3CoCl .6NH , it means it gives less than four ions per mole in solution. It will





be possible only if two of the chlorines are also present in the first coordination sphereand thecompound has the formula 3 4 2[Co(NH ) Cl ]Cl .

4 i) 3Fe is more paramagnetic than 2Fe as 3Fe consists five unpaired electrons while 2Fe

possesses four unpaired electrons.

ii) Any ion of transition elements which possesses unpaired d-electrons, i.e. d–d transition is possibleshows a characteristic colour 0(n–1)d (or) 10(n–1)d configuration does not involve d–d

transition and hence, is colourless. 2Fe , 2Mn and 3Cr are coloured, while 3Cu ,Se and4 Ti are colourless.

iii) Chromium and copper. chromium attains 5 13d 4s configuration in which all the d-orbitals are

unpaired in order to get extra stability. Copper attains 10 13d 4s configuration in which all the d-orbitals are paired in order to get extra-stability.

5. This is because a part of the oxygen produced from 4KMnO +HCl is used in oxidizing HCl to 2Cl

2 24HCl 2(O) 2H O 2Cl

6. 2– 2 3 32 7 22 3612

Cr O 14H 6Fe 2Cr 6Fe 7H O

2– –2 7Cr O 6e ] gain of

–

6e by two Cr atoms.0.1 M 2– 2–

2 7 2 7Cr O 0.6N Cr O

– 2 2 34 222 37

MnO 8H 5Fe Mn 5Fe 4H O

– –4MnO 5e

0.1 M – –4 4MnO 0.5NMnO

2 –4Fe 0.5N MnO

24.5 mL of 0.1 M 2– –2 7 4Cr O V mL od 0.1M MnO

24.5 mL 0.6 N2–

2 7Cr O V mL of 0.5 N–

4MnO24.5 0.6

V 29.4mL5

7. a) The black colour is due to the manganese (IV) oxide, 2MnO . It is made by the 2Mn(OH) being

oxidized by oxygen in the air:

2 2Mn(OH) MnO H O

2 2air black

1MnO O MnO

2

b) It is due to formation of 2–4HgI (a soluble complex) in both the cases with 2HgI :

– 2–2 4HgI 2I HgI But in 2 2Hg I , first there is oxidation of Hg(I) to Hg(II) and then complex formation takes place;

it is by following disproportionation reaction:2 – 2–2 4

01 2

Hg 4I HgI Hg

8. 143.45g of AgCl contains 35.45 g –Cl ions

2.87 g of AgCl will contain35.45 2.87

143.45

=0.709 g –Cl ions





266.35 g 3 2CrCl .6H O contains n 35.45 g of ionisable –Cl ions (where n =no. of –Cl ions outside

the coordination sphere).

Thus, 2.665 g 3 2CrCl .6H O will containn 35.45 2.665

266.35

g

Alson 35.45 2.665

0.709 n 2

266.35

Keeping in view the octahedral geometry of the complex, its structure may be written as

2 5 2 2[CrCl(H O) ]Cl .H O

9. i) Ferrous sulfate is a salt of a weak base and a strong acid. Thus its hydrolysis occurs when it isdissolved in water and solution becomes turbid due to formation of ferrous hydroxide

4 2 2 2 4FeSO 2H O Fe(OH) H SO

Addition of a small amount of acid shifts the equilibrium towards left and prevents hydrolysis.

ii) 2HgCl is an oxidizing agent while 2SnCl is a reducing agent. When both are present, a redoxreaction occurs forming Hg and stannic chloride as final products.

2 2 2 2 4SnCl 2HgCl Hg Cl SnCl

2 2 2 4Hg Cl SnCl 2Hg SnCl

iii) A ferrous salt turns brown in air due to oxidation to ferric salt

iv) 2Cu(OH) dissolves in 4NH OH by forming a complex.

2 2 3 4 2 2Cu(OH) 4NH OH [cu(NH ) ](OH) 4H O

2Cu(OH) is insoluble in NaOH as no such complex is formed

10. a) 2Ni salts (in basic medium) on reaction with dimethyl glyoxime give cherry red ppt of nickeldimethyl glyoximate:

2NiCl 24

2NH OH CH3 C

CCH3 NOH

NOH

(in alcohol)

4 22NH Cl 2H O

N

CH3

CH3

N

O

Ni

O H

N

H O

O

NC

C

CH3

CH3

(Cherry red ppt.)

b) 28 4 14 4

59 289

xg 0.259g

Ni (insteel) NiC N H O

Stioichiometrically:x 0.25959 289

x =0.05288 g pure Ni in 1.502 of steel sample.

Percentage of Ni =3.52%





LEVEL-IV

1. a) Structure of the complex is

CH3 C N

CCH3 N

O H O

O H O

N C CH3

N C CH3

Ni

Hydrogen bonding

Hydrogen bonding

b) The oxidation state of Ni in this compound is +2 and it shows 2dsp hybridization.

c) Since the coordination number of Ni in this complex is 4, the configuration of 2Ni i.e.

3d 4s 4p

At first sight, gives the idea that the complex is paramagnetic with 3sp hybridization as it has twounpaired electrons. However the experiments show that the complex is diamagnetic i.e., it doesnot have any unpaired electron. This is only possible when the 3d electrons rearrange themselvesagainst the Hund’s rule i.e.,

3d 4s 4p

2dsp hybridisation

This isalso in accordance with the fact that the ligand involve here is very strong.

2. IUPAC name of 2 3 4K [Cr(NO)(NH )(CN) ] is potassium ammine tetra cyano nitroso chromate(I)

In it chromium is present as Cr(I), so24 2 2 6 2 6 5 1Cr 1s ,2s 2p ,3s 3p 3d ,4s

2 2 6 2 6 5Cr(I) Cr 1 ,2s 2p ,3s 3p 3d

3d 4s 4p

For it 1.73BM

n(n 2)

1.73 n(n 2) or 21.73 1.73 n 2n

Hence, n =1On the basis of value of n =1, an unpaired electron is present is chromium (I) of this complex ion Thus in excited state of Cr(I) of this complex ion Thus in excited state of Cr(I) (i.e. in complete salt)





3d 4s 4p

2 3d sp hybridisation

hence, in this complex Cr(I) shows 2 3d sp -hybridisation, so structure of this complex is octahedral.

Cr

NO

NH3

C

CC

C

N

N

N

N

2K +

2–

3. In 2–4(NiCl ) –Cl is a weak ligand which is unable to give pair of electron to 2Ni .

In 2–4(NiCl ) oxidation state of Ni is +2

So, 2 2 2 6 2 6 8 0Ni 1s ,2s 2p ,3s 3p 3d ,4s

Thus

3d 4s 4p

Co-ordination number of Ni is 4, so four vacant orbitals are required for the accommodation of fourpair of electrons of –Cl ions and there are one s and three p– orbitals of valence sell.Hence.

3d 4s 4p

3sp hybridisation

The structure of

2–

4(NiCl ) is tetrahedral

Ni

Cl

ClCl Cl

2In it magnetic moment

n(n 2) 2(2 2)

(n 2 (unpaired electron)) =2.82 B.M.

In 2–4(Ni (CN) ) , –CN is a strong ligand which is able to donate the pair of electron to 2Ni O. N. of

Ni in [(Ni 2–4(CN) )] is also+2.

In 2–4[Ni(CN) ) ], 2 2 2 6 2 6 8 0Ni 1s ,2s 2p ,3s 3p 3d ,4s

3d 4s 4p

2dsp hybridisation

3d 4s 4p

In ground state

In excitation

Hence the structure of 2–4(Ni(CN) ) is square planer





Ni

C

CC

C

N

N

N

N

2+

2+

In 2–4(Ni(CN) ) , no. of unpaired electron (n) =0

So, magnetic moment = n(n 2) 0

4. A metal complex having composition Cr 3 4 2(NH ) Cl Br have two forms A and B.

Form `A’ gives white ppt. with 3AgNO , hence it must have chloride ion in form of non-complex ioni.e. outside the complex sphere as

3 4[Cr(NH ) ClBr]Cl–

3 4 3 3 4 3whitepptForm A '

[Cr(NH ) ClBr]Cl AgNO AgCl [Cr(NH ) ClBr] NO

These precipitates of AgCl are soluble in 4NH OH due to formation of complex salt.

4 3 2 2whiteppts ComplexsaltAgCl 2NH OH [Ag(NH ) Cl] 2H O Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in 4NH OH .

Hence form B’ is 3 4 2[Cr(NH ) Cl ]Br.

–3 3 4 2 3 4 2 3

Paleyellow ppt.AgNO [(Cr(NH ) Cl )]Br AgBr [(Cr(NH ) Cl )] NO

4 3 2 2Paleyellowppt.

AgBr 2NH OH [Ag(NH ) Br] 2H O

In both complexes, chromium is present as central ion and its oxidation number is +3. So in these2 2 6 2 5 5 1

24Cr 1s ,2s 2p ,3s 3p 3d ,4s3 2 2 6 2 5 3Cr 1s ,2s 2p ,3s 3p 3d

Number of ligands are six and 3Cr shows 2 3d sp hybridization in both complexes A and B

3d 4s 4p


Hence in it, number of unpaired electrons are 3. So magnetic moment ( ) n(n 2)

(where n no. of unpaired electrons)

3(3 2) 15 3.872B.M.

5. a) Aqueous extract or soda extract is acidified with dil. 2 4H SO , add freshly prepared 4FeSO

solution and a few drops of conc. 2 4H SO , appearance of brown ring confirms the presence of nitrate radical.

3 2 4 4 32NaNO H SO NaHSO HNO

3 4 2 4 2 4 3 22HNO 6FeSO 3H SO 3Fe (SO ) 2NO 4H O

4 4Brown ring (Ferrous nitro sulphate)

FeSO NO [Fe(NO)SO ]

b) In 33 6[Co(NH ) ] cobalt is present as 3Co and its coordination number is six.

2 2 6 2 6 7 227Coatom 1s , 2s 2p ,3s 3p 3d ,4s





3 2 2 6 2 6 6Co ion 1s ,2s 2p ,3s 3p 3d

3d 4s 4p


3d 4s 4p

Co3+ion incomplex ion

Hence,

Co

NH3

NH3

NH3

NH3H3N

H3N

3+

or Co

NH3

NH3

NH3

NH3H3N

H3N

3+

Structure of 33[Co(NH )] is octahedral.

In 2–4[Ni(CN) ] nickel is present as 2Ni ion and its coordination number is four.

2 2 6 2 6 8 228Ni atom 1s ,2s 2p ,3s 3p 3d ,4s

2 2 2 6 2 6 8Ni ion 1s ,2s 2p ,3s 3p 3d

3d 4s 4p

2dsp hybridisation

3d 4s 4p

Ni2+ion

Ni2+ion inComplex ion

So structure of 2–4[Ni(CN) ion is square planar which is represented as follows.

Ni

C

CC

C

N

N

N

N

2+

In 4Ni(CO) , nickel is present as Ni atom i.e. its oxidation number is zero and coordinationnumber is four.

Ni atom = 2 2 6 2 6 8 21s ,2s 2p ,3s 3p 3d ,4s

3d 4s 4p

3sp hybridisation

3d 4s 4p

Ni2+ion

Ni in complex

So structure of 4Ni(CO) is tetrahedral which is represented as shown:




Ni

OC

CO

OC CO

c) The transitional metal is 2Cu .

The compound is 4 2CuSO .5H O . It dissolves in water to give blue coloured solution due to

presence of 2 9Cu (d -configuration). On passing 2H S in acidic medium in such solution, the

black precipitate of CuS is obtained which is not soluble in aq. KOH (warm) solution.Acidic

4 2 2 4medium Blackppt.(Insoluble in aq. KOH)

CuSO H S CuS H SO

On addition of KI solution in aqueous solution of 4CuSO , it produces yellow coloured solution

of 2CuI which is decomposed in white ppt. of 2 2Cu I and liberate 2I .

4 2 2 4CuSO 2KI CuI K SO

2 2 2 2whiteppt.

2CuI Cu I I

6. As we show that Magnetic moment ( ) n(n 2) BM

where n number of unpaired electrons 1.73 BM for vanadium ion

1.73 BM n(n 2) . So 2(1.73) n(n 2)

3.0 = 2n 2n or 2n 2n– 3 0 2n 3n– n– 3 0

n(n 3) –1(n 3) 0

(n –1)(n 3) 0

Correct value of n =1. Thus no. of unpaired electrons in vanadium ion =12 2 6 2 6 3 2

23V 1s ,2s 2p ,3s 3p 3d ,4s

So in vanadium (IV) an unpaired electron is present.4 2 2 6 2 6 1V 1s ,2s 2p ,3s 3p 3d

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Transition Elements IItjee