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d- & f- BLOCK ELEMENTS

C1 PHYSICAL PROPERTIES

As discussed in unit-2 (The Periodicity), three series of element are formed by filling the 3d, 4d and 5dshells of electrons. Together these comprise the d-block elements. They are often called transitionelements because their positions in the periodic table is between the s-block and p-block elements. Theelectrons are added to the penultimate shell, expanding it from 8 to 18 electrons.

Elements in the main transition series-the d-block elements- have atoms or ions with partially filled dorbitals. The Zinc group has a d10 configuration hence their compounds show quite difference from othercompounds in many properties (e.g., colour, magnetic behaviour etc.)

Metallic Character :

Nearly all the transition metals have the simple hcp, ccp or bcc lattice characteristics of true metals. Theyhave relatively high densities. They are malleable, ductile and have high tensile strength, thermal andelectrical conductivity and lustre. Melting points of 3d-series (shown in fig.) rise to a maximum and fall asthe atomic number increases except in Mn where m.p. is minimum. High m.p. and high heat of atomisationindicate that the atoms in these elements are held together by strong metallic bonds that are present in themolten states as well. Greater the number of valence shell, stronger is the resultant bonding.

0

Fig. : M.pt. of 3d-series elements (not according to scale)

Oxidation States :

A characteristic property of the d-transition metals is their ability to exhibit several oxidation states. Thestability of a given oxidation state depends on the nature of the element with which the transition metal iscombined. The highest oxidation states are found in the compounds of fluorides and oxides because themost electronegative nature of F and O.

Except Sc, the most common oxidation state of the first row transition elements is II which arises from theloss of two 4s electron. This means that after Sc, d-orbitals become more stable than the s-orbital.

In the II and III oxidation states, bonds formed are mostly ionic. In compounds of the higher oxidationstates, the bonds are formed by the removal or sharing of d-electron hence the bonds formed are essentiallycovalent (as in MnO

4—)

Within a group, the maximum oxidation state increases with atomic numbers.

Fe II, III

Ru, Os II, III, IV, VI, VIII

Transition metals also form compounds in low oxidation states (1, 0). Such compounds are expected to beunstable except in cases where vacant d-orbitals are used for accepting lone-pairs from -bonding ligandse.g., Ni(CO)

4, [Ag(CN)

2]—, [Ag(NH

3)

2]+.

When an element exists in more than one oxidation state, standard electrode potential data provide a clue tothe relative stabilities of different states w.r.t. oxidation. Cr2+ is unstable in water w.r.t. oxidation whereasCr3+ is stable in water; Fe2+ is unstable in aerated water (and changes to Fe3+) w.r.t. oxidation.

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The electronic structure of the atoms in the second and third rows do not always follow the pattern of thefirst row. The structure of the nickel group are :

Ni 3d84s2

Pd 4d105s0

Pt 5d96s1

Since a full shell of electron is a stable arrangement, the place where this occurs is of importance.

The d-levels are complete in

Ni Cu 3d104s1 Zn 3d104s2

Pd 4d105s0 Ag 4d105s1 Cd 4d105s2

Pt Au 5d106s1 Hg 5d106s2

Even though the ground state of the atom has a d10 configuration, Pd and the coinage metals Cu, Ag and Aubehave as typical transition elements. This is because in their most common oxidation states, Cu (II) has ad9 configuration, and Pd (II) and Au (III) have d8 configuration (incomplete d level). However, Zn (II), Cd(II) and Hg (II) have a d10 configuration. Because of this, these elements do not show the propertiescharacteristic of transition elements.

Ionisation Energies and Electrode Potential :

The magnitude of ionisation energies provide an indication of the energy needed to raise the metal to aparticular oxidation state in a compound.

(IE)1 of 5d elements are higher than those of the 3d and 4d elements. This is due to greater effective

nuclear charge acting on outer valence electrons because of the weak shielding of the nucleus by 4felectrons.

It is possible to determine thermodynamic stability of the transition metal compounds based on the valuesof (IE) of the metals. (IE)

1, (IE)

2, (IE)

3 and (IE)

4 of Ni and Pt in MJ mol–1 are compared.

(IE)1 + (IE)

2(IE)

3 + (IE)

4Total

Ni 2.49 8.80 11.29

Pt 2.66 6.70 9.36

Thus, nickel (II) compounds tend to be thermodynamically more stable than platinum (II) while platinum

(IV) compounds are relatively more stable than nickel (IV). IV

62PtClK is well known but such type of

nickel compound is not formed. Following factors also decide the stability of the compounds :

— the energy of the sublimation,

— the lattice energy,

— the solvation energy.

A qualitative treatment of the stability of transition metal ion in different oxidation states in aqueous me-dium is obtained from the electrode potential data. The value of E0 for M2+/M denotes the emf of the cell inwhich the reaction

2H+ (aq) + M(s) M2+ (aq) + H2(g)

occurs under reversible conditions. The total energy H of process

M(s) M2+ (aq)

can be represented as equal to sum of the energy of the following steps :

M(s) M(g)

Hsub

(enthalpy of sublimation)

M(g) M2+(g) (IE)1 + (IE)

2

M2+(g) M2+ (aq)

Hhyd

(enthalpy of hydration)

)aq(M)s(M 2

H = Hsub

+ (IE)1 + (IE)

2 + H

hyd.

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Complexes :

The transition elements have an unparalleled tendency to form coordination compounds with Lewis basesthat is with groups which are able to donate an electron pair (called ligands). A ligand may be a neutralmolecules such as NH

3, or an ion such as Cl— or CN—

Co3+ + 6NH3 [Co(NH

3)

6]3+

Fe2+ + 6CN— [Fe(CN)6]4—

Cr3+ + 6H2O [Cr(H

2O)

6]3+

This ability to form complexes is in marked contrast to the s- and p-block elements which form only a fewcomplexes. Tendency to form complexes by transition metal is due to :

— their smaller size,

— higher nuclear charge,

— presence of low energy vacant orbitals to accept lone pair of electrons donated by ligand.

Complexes where the metal is in the (III) oxidation states are generally more stable than those where themetal is in the (II) state. Thus, [FeIII(CN)

6]3– is more stable than [FeII(CN)

6]4–

Magnetic Properties :

Two types of magnetic behaviour are found in substances :

— diamagnetism

— paramagnetism

Diamagnetic substances are repelled by an applied magnetic field. Such substances have no unpairedelectron.

Paramagnetic substances are attracted by an applied magnet field. Ferromagnetism is a special type ofparamagnetism in which permanent magnetic moment is acquired by the substance due to presence ofunpaired electrons. It is shown by Fe, Co and Ni.

Transition metals and many of their compounds show paramagnetic behaviour where there are unpairedelectrons. The magnetic moment arise from the spin and orbital motions in ion or molecules. Magnetic

moment of N unpaired electron is )2N(Nµ Bohr Magneton

The spin only magnetic moment : )1S(S4Ms B. M

Magnetic moment increases as number of unpaired electron increases, For 3d-series element, variation ofmagnetic moment is given in fig. This clearly shows magnetic moment is maximum for chromium (with sixunpaired electrons.)

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Z 21 22 23 24 25 26 27 28 29 30

N 1 2 3 6 5 4 3 2 1 0

µ 3 8 15 48 35 24 15 8 3 0

Fig. Magnetic Moment of 3d-series elements (on arbitrary scale)

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Colour :

Many ionic and covalent compounds of transition elements (and also inner transition elements) are coloured.In contrast, compounds of the s- and p- block elements are almost always white. Colour may arise from anentirely different cause in ions with incomplete d or f shells. In a free isolated gaseous ions, the five dorbitals are degenerate, that is, they are identical in energy. The surrounding groups, which can be solventmolecules in solution or ligands in a complex or other ions in a crystal lattice, affect the energy of some dorbitals more than others. Thus the d orbitals no longer degenerate and at their simplest they form twogroups of orbitals of different energy. Thus in transition element with a partly filled d shell, it is possible topromote electrons from one d level to another d* level of higher energy (called d-d* transition). Thiscorresponds to a fairly small energy difference, and so light is absobed in the visible region. The colour ofa transition metal complex is dependent on how big the energy difference is between the two d-levels. Thisin turn depends on the nature of the ligand, and the type of the complex formed.

[Ni(NH3)

6]2+ blue

[Ni(H2O)

6]2+ green

[Ni(NO2)

6]2— brown-red

ZnSO4 (Zn2+ with d10 configuration) and TiO

2 (Ti4+ with d0 configuration) are white-in both of them d-d*

spectra are impossible and they are colourless.

In the series Sc (+III), Ti (+IV), V (+V) Cr (+VI) and Mn (+VII), all have empty d-orbitals, hence d-d*spectra are impossible and they should be colourless. But following ions in aq. solution are :

VO2

2+ (oxidn. state +V) — pale yellow

CrO4

2— (oxidn. state +VI) — deep yellow

MnO4

— (oxidn. state +VII) — intense purple

The colours arise by charge transfer, for example, in MnO4— an electron is momentarily transferred from

O to the metal and thus oxygen changes from O2— to O— and manganese from Mn(+VII) to Mn(+VI).Detailed study is beyond the scope of standard XII/IIT-JEE/REE.

Size of Atoms and lons :

The covalent radii of the element (summarised in table) decrease from left to right across a row in thetransition series, until near the end when the size increases slightly. On passing from left to right, extraprotons are placed in the nucleus and extra orbital electrons are added. The orbital electrons shield theNuclear charge incomplete (order being d < p < s). Thus the nuclear charge attracts all the electrons morestrongly, hence a contration in size occurs.

Th radii of the element from chromium to copper, however, and very close to one another. This is due to thefact that successive addition of d-electrons screen the outer electrons (4s) from the inward pull of thenucleus. As a result of this, the size of the atom does not alter much in moving from chromium to copper.

Atoms of the transition elements are smaller than those of the Group I and II elements in the samehorizontal period. This is partly because of the usual contraction in size across a horizontal period andpartly because the orbital electrons are added to the penultimate d shell rather than to the outer shell of theatom.

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There is a gradual decrease in size of the 14 lanthanide elements from Cerium (58) to Lethetium (71). Thisis called Lanthanide Contraction. The lanthanide contraction cancels almost exactly the normal sizeincrease on decending a group of transition elements. Thus covalent and ionic radii of Nb are thesame as the values for Ta. Same is the case for Zr and Hf.

Table : Effect of Lanthanide Contraction on Ionic Radii (pm)

Ca2+ 100 Sc3+ 74.5 Ti4+ 60.5 V3+ 64.0

Sr2+ 118 Y3+ 90.0 Zr4+ 72.0 Nb3+ 72.0

Br2+ 135 La3+ 103.2 Hf4+ 71.0 Ta3+ 72.0

The effect of lanthanide contraction are less pronounced towards the right of the d-block.

Nonstoichiometry :

Transition elements have the property of forming compounds of indefinite structures and proportions.

These are called Nonstoichiometric compounds. Iron (II) oxide FeO is written as FeO to indicate that

ratio of Fe and O atoms is not exactly 1 : 1. Analysis shows that the formula varies between Fe0.94

O andFe

0.84O. Vanadium and selenium form a series of compounds ranging from VSe

0.98 to VS

e2 :

)VSVSe(VSe 2.198.0

)VSeVSe(SeV 6.12.132

)VSeVSe(SeV 26.142

CuS (insoluble) is actually a mixture of CuS and Cu2S. Nonstoichiometry is shown particularly among

transition metal compound of the Group VI elements (O, S, Se, Te) and is mostly due to variable valency oftransition elements, and also sometimes due to defects in solid structures.

Catalytic Properties

The catalytic activity of the transition metal and their compounds is ascribed to their ability to adopt mul-tiple oxidation states and to form complexes. Some of the most important ones are summarised in table.

Table : Transition Metals and their Compounds as Catalyst

Catalyst Function

TiCl4

Used as Natta catalyst in the production of polythene

FeSO4 + H

2O

2Used as Fenton’s reagent in the oxidation of alcohols to aldehyde

Ni Reney nickel in reduction processes :

— manufacture of hexamethylenediamine,

— production of H2 from NH

3,

— reducing anthraquinone to anthraquinol in the productionof H

2O

2

Ni Complexes Reppe synthesis (polymerisation of alkynes to give benzene, cyclooctatetraene).

Cu Used in manufacturing of (CH3)

2SiCl

2 used to make silicones

Cu/V Used in the oxidation of cyclohexanol/cyclohexanone mixture to acidic acidwhich is used to prepare Nylon-66.

CuCl2

Decon process of making Cl2 and HCl.

MnO2

Used in the decomposition of KClO3 to KCl and O

2.

V2O

5Used in contact process of H

2SO

4.

Fe Promoted Fe used in the Haber-Bosch process of NH3

FeCl3

Used in the production of CCl4 from CS

2 and Cl

2

PdCl2

Wacker process for converging C2H

4 into CH

3CHO.

Pt/PtO Adams catalyst in reduction.

Pt In three stage convertors for cleaning car exhaust fumes.

Pt/Rh In Ostwald process for HNO3 from NH

3

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ZnO/CuO In synthesis of CH3OH from CO and H

2

Co/Fe In synthesis of hydrocarbons by CO and H2 in Fischer-Tropsch process.

Enzymes are catalysts that enhance the rates of specific reactions. Some enzymes require the presence ofmetal ions as cofactors called metalloenzymes which generally contain transition metal, e.g. Mo in xan-thine oxidase (metabolism of purines).

Alloy and Interstitial Compounds :

The atomic size of the transition metals are very similar and hence in the crystal lattice, one metal can bereadily replaced by another metal giving a solid solution and smooth alloys. The alloys so formed are hardand have high m.p. (refer Appendix VI for various types of alloys)

The transition metals form a number of interstitial compounds, in which they take up atoms of small size(like H, C and N) in the vacant spaces in their lattices and also form bonds with them. The productsobtained in this way are also hard and rigid. Steel and cast iron become hard due to the formation of aninterstitial compounds with carbon.

C2 COMPOUNDS

Potassium Dichromate (K2Cr

2O

7)

K2Cr

2O

7 is obtained from chromate ore (FeCr

2O

4) in following steps :

Step I : Chromite ore is fused with 4FeCr2O

4 + 8Na

2CO

3 + 7O

2

molten Na2CO

3 in the

2lelubinsowater

32lelubsowater

42 CO8OFeCrONa8

presence of air when

Na2CrO

4 and Fe

2O

3 are

formed. Fe2O

3 is water

insoluble while Na2CrO

4

is water soluble hence

separation is done.

Step II : Aq. solution of Na2CrO

4 is 2Na

2CrO

4 +

dil42SOH Na

2Cr

2O

7

acidified when Na2Cr

2O

7 is + Na

2SO

4

formed. + H2O

Step III : Sodium dichromate Na2Cr

2O

7 + 2KCl K

2Cr

2O

7

(Na2Cr

2O

7) is more soluble + 2NaCl

and less stable than K2Cr

2O

7

hence Na2Cr

2O

7 changes to

K2Cr

2O

7 on reaction with

KCl. On cooling concent-

rated solution of K2Cr

2O

7

orange crystals are obtained

It exists as orange-red crystals (m.p 3980C) and is moderately soluble in cold water but freely soluble in hotwater.

In alkaline solution (pH > 7), orange colour of Cr2O

72— (dichromate ion) changes to yellow colour due to

formation of CrO42— (chromate ion) and again yellow colour changes to orange colour in acid medium

(pH < 7) :

OHOCrH2CrO2

OHCrO2OH2OCr

2272

24

2yellow

24

redorange

272

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Thus CrO42— and Cr

2O

72— exist in equilibrium and are interconvertible by altering the pH of the solution :

2CrO42— + 2H+ 2HCrO

4— + H

2O

4K2Cr

2O

7

K2CrO

4 + 2Cr

2O

3 + 3O

2

(NH4)

2Cr

2O

7

N2 +

green32OCr + 4H

2O

Once this reaction starts, it will keep going at a resonable rate. Also a small amount of it gives a much largervolume of Cr

2O

3 (green). Both features of the reaction lead to it being used in indoor fireworks.

K2Cr

2O

7 when heated with conc. H

2SO

4 and soluble chloride (like NaCl, KCl), gives deep red vapours of

CrO2Cl

2 (chromyl chloride).

OHCrO2KHSO2SOH2OCrK 234.conc

42722

OHClCrOHCl2CrO

HClNaHSOSOHNaCl

2223

4.conc

42

CrO2Cl

2 when passed into aq. NaOH solution, yellow colour solution of Na

2CrO

4 is obtained. This on

reaction with (CH3COO)

2Pb/CH

3COOH gives yellow ppt. of PbCrO

4 :

COONaCH2PbCrOPb)COOCH(CrONa

HCl2CrONaNaOH2ClCrO

342342

4222

This test called ‘Chromyl-Chloride Test’ is used to identify Cl in inorganic salt analysis.

Cr2O

72– is a good oxidising agent in acidic medium :

OH7Cr2e6H14OCr 2green

3272

E0 = 1.33 V (A)

The reaction is spontaneous (E0 > 0) and Cr2O

72— is used as a primary standard in volumetric (titrimetric)

analysis.

6

.wt.mol)OCr(.wt.eq;e6OCr 2

72272

Following reactions are affected by Cr2O

72— / H+ and solution turns green due to formation of

Cr3+ (green) :

e6H12SO3OH6SO3

e6Fe6Fe6

e6S3H6SH3

e6I3I6

2422

32

2

2

When H2O

2 is added to an acidified solution of a dichromate [or any other Cr(VI) species], a complicated

reaction occurs. The products depend on the pH and the concentration of Cr :

Cr2O

72— + 2H+ + 4H

2O

2 2CrO(O

2)

2 + 5H

2O

A deep blue-violet coloured perexo compound CrO(O2)

2 called chromic peroxide is formed. This

decomposes rapidly in aqueous solution into Cr3+ and oxygen. The peroxo compound can be extracted intoether, where it is stable; it reacts with pyridine forming the adduct : Py. CrO(O

2)

2

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(chromic peroxide)

In less acidic solution, K2Cr

2O

7 and H

2O

2 give salts which are violet coloured and diamagnetic due to

formation of [CrO(O2)(OH)]—

In alkaline solution with 30% H2O

2, a red-brown compound K

3CrO

8 is formed. It is tetraperoxo species

[Cr(O2)

4]3— and thus contains Cr (V) ion.

In ammonical solution (NH3)

3CrO

4, a dark red-brown compound with Cr(IV) ion is formed.

PbCrO4

— Chrome yellow

PbCrO4 . PbO — Chrome red

PbCrO4 + [FeFe(CN)

6]— — Chrome green

prussian blue

Zn(OH)2.ZnCrO

4 . H

2O — Yellow

Cr2O

3 . 2H

2O — Guigret’s green

CrO3

— Orange (called chromic acid)

K3[CrO

8] — Red-brown

Chrome alum is obtained when acidified K2Cr

2O

7 solution is saturated with SO

2 :

Potassium Permanganate (KMnO4) :

KMnO4 is prepared from MnO

2 (pyrolusite) in following steps :

Step I : MnO2 is fused with KOH and KClO

33MnO

2 + 6KOH + KClO

3

of KNO3 when potassium manganate OH3KClMnOK3 2

green42

(K2MnO

4) is formed as indicated

by its green colour. KClO3/KNO

3

acts as oxidising agent.

Step II : The fused mass is extracted with

water and solution is green

(MnO42—)

(a) when Cl2 gas is passed, K

2MnO

4 2

green42 ClMnOK2

is oxidised to KMnO4 (purple)

purple4KMnO2KCl2

(b) MnO42— can be oxidised to at anode : eMnOMnO 4

24

MnO4— electrochemically at anode. at cathode : 2H+ + 2e H

2

Strong oxidising agents such as PbO2 or sodium bismuthate (NaBiO

3) can also oxidise Mn2+ to MnO

4—.

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MnO42— is unstable in acidic medium and disproportionates :

3MnO42— + 4H+ MnO

2 + 2MnO

4— + 2H

2O

KMnO4 crystallises as deep purple rhombic prisms.

MnO4— (VII) has 3d0 electronic configuration in Mn and thus MnO

4— would be expected to be colourless,

but colour arises due to charge transfer by oxygen to Mn and thus one of the oxygen is O— and Mn changesfrom +VII to +VI (with one unpaired electron in 3d) hence coloured.

It is isomorphous with KClO4, potassium per chlorate; highly soluble in water :

22424 OMnOMnOKKMnO2

KMnO4 is good oxidising agent; it finds use in volumeric analysis as secondary standard in the estimation

of Fe2+, C2O

42—, H

2O

2 etc. in acidic medium :

MnO4— + 8H+ + 5e— Mn2+ + 4H

2O

The purple colour of MnO4— acts as its own indicator (self) in titrations

MnO4— 5e—

Thus, equivalent weight of 5

.wt.molMnO4

in acidic medium.

Some oxidation reactions (using MnO4—/H+) are :

Fe2+ Fe3+ + e—

C2O

42— 2CO

2 + 2e—

2I— I2 + 2e—

2Cl— Cl2 + 2e—

H2O

2 O

2 + 2H+ + 2e—

H2O + SO

32— SO

42— + 2H+ + 2e—

H2S S + 2H+ + 2e—

CH3CH

2OH O

CH3CHO + H

2O

In alkaline solution, MnO2 is formed.

MnO4— + 2H

2O + 3e— MnO

2 + 4OH—

Eq. wt. (in basic medium) =

3

.wt.mol

Thus equivalent weight of KMnO4 varies with medium.

Alkaline KMnO4 is called Baeyer’s reagent and is used to test unsaturation is an organic compound,

when it decolorises KMnO4.

2

OH|

2

OH|KMnOalkaline

222 HCHCOOHCHCH4

However, in very strong alkali solution, MnO4— changes to MnO

42—

4MnO4— + 4OH— 4MnO

4— + MnO

2 + 2H

2O

In dilute alkali, water or acidic solutions, MnO42— again disproportionates :

3MnO42— + 4H+ 2MnO

4— + MnO

2 + 2H

2O

3MnO42— + 2H

2O 2MnO

4— + MnO

2 + 4OH—

Permanganate solution are intrinsically unstable is acidic solution, and decompose slowly. Decompositionis catalysed by sunlight, so KMnO

4 solution should be stored in dark bottles and they must be standarised

at the moment of its use in volumetric analysis.

4MnO4— + 4H+ 4MnO

2 + 3O

2 + 2H

2O

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If a small amount of KMnO4 is added to concentrated H

2SO

4, a green solution containing MnO

3+ ions is

formed :

OHHSO3MnOKSOH3KMnO 34green

3424

If KMnO4 is in excess, an explosive oil, Mn

2O

7, is formed :

OHSOKOMnSOHKMnO2 24272424

Zinc Oxide (ZnO) :

ZnO is formed when ZnS is oxidised :

2ZnS + 3O2 2ZnO + 2SO

2

Zn(OH)2 on strongly heating gives ZnO :

OHZnO)OH(Zn 22

ZnO is white when it is cold, a property that has given it a use as a pigment in paints. However, it changescolour, when hot, to a pale yellow. This is due to change in the structure of lattice.

ZnO is soluble both in acid and alkali and is thus amphoteric in nature :

ZnO + 2H+ Zn2+ + H2O

22

ionzincate

242 ZnOor])OH(Zn[OHOH2ZnO

Or ZnO + 2HCl ZnCl2 + H

2O

ZnO + 2NaOH OHZnONa 2zincatesodium

22

ZnO + C

C10000 Zn + CO

ZnO + CO Zn + CO

2

Zinc Sulphate (ZnSO4) :

ZnSO4 . 7H

2O (also called white vitriol) is formed :

— by decomposing ZnCO3 with dil H

2SO

4

ZnCO3 + H

2SO

4 ZnSO

4 + H

2O + CO

2

— by heating ZnS (zinc blende) in air at lower temperature and dissolving the product in dil. H2SO

4

OHZnSOSOHZnO

SOZnSOZnOO5.3ZnS2

2442

242

Highly soluble in water and solution is acidic in nature due to hydrolysis

ZnSO4 + 2H

2O Zn(OH)

2 + H

2SO

4

4C280

24C100

24 ZnSOOH6.ZnSOOH7.ZnSO00

It is isomorphous with Epsom salt and used in the manufacture of lithophone (which is a mixture of BaS +ZnSO

4 and is used as white pigment).

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Silver Nitrate (Lunar Caustic) AgNO3 :

When Ag is heated with dil. HNO3, AgNO

3 is formed. Crystals separate out on cooling the concentrated

solution of AgNO3.

OH2NOAgNO3HNO4Ag3 233

Colourless crystalline compound soluble in H2O and alcohol; M.P. 2120C

When exposed to light, it decomposes hence stored in a brown coloured bottle :

2 Ag + 2NO2 + O

2 hotred,

2AgNO3 C212T, 0

2AgNO2 + O

2

It is reduced to metallic Ag by more electropositive metals like Cu, Zn, Mg and also by PH3.

2AgNO3 + Cu Cu(NO

3)

2 + 2Ag

6agNO3 + PH

3 + 3H

2O 6Ag + 6HNO

3 + H

3PO

3

It dissolves in excess of KCN :

eentocyanidargpotassium

lelubso2

KCN

.pptwhite

KCN3 ])CN(Ag[KAgCNAgNO

AgNO3 gives white ppt. with Na

2S

2O

3; white ppt. changes to black.

42black

22322

3.pptwhite

3223223

SOHSAgOHOSAg

NaNO2OSAgOSNaAgNO2

Ammonical AgNO3 is called Tollen’s reagent and is used to identify reducing sugars (including aldehydes)

:

OH2Ag2RCOOOH3Ag2RCHO 2

It is called ‘silver mirror test’ of aldehydes and reducing sugar (like gluocose, fructose).

Ferric Chloride (FeCl3) :

32 FeCl2Cl3Fe2 (anhydrous)

When Fe is dissolved in aqua-regia or Fe2O

3 is dissolved in HCl, hydrated FeCl

3 . 6H

2O is formed.

Fe + 4HCl + 2Cl 2FeCl3 + 2H

2

Fe2O

3 + 6HCl FeCl

3 + 3H

2O

Anhydrous FeCl3 is black but hydrate FeCl

3.6H

2O is yellowish brown, highly deliquescent crystalline solid.

It sublimes at 3000C giving a dimeric gas :

C3003

0

FeCl2

FeCl3 dissolves in both ether and water giving solvated monomeric species :

3FeClhydrated

22423 OH2.Cl]Cl)OH(Fe[FeCl

solvated FeCl3

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It normally consists of trans [Fe(H2O)

4Cl

2] . 2H

2O but in strong HCl it forms tetrahedral [FeCl

4]— ion.

Aq. solution is acidic due to hydrolysis :

FeCl3 + 3H

2O Fe(OH)

3 + 3HCl

Oracid3

base

252

base2

acid

362 OH)]OH()OH(Fe[OH])OH(Fe[

FeCl3 is also an oxidising agents :

2FeCl3 + H

2S 2FeCl

2 + 2HCl + S

2FeCl3 + SnCl

2 2FeCl

2 + SnCl

4

Yellow colour of aqueous Fe (III) changes to light green aqueous Fe (II)

Fe3+ solution gives blood red colour with SCN— ions :

Fe3+ + SCN— [Fe(SCN)]2+

This red colour also contains [Fe(SCN)(H2O)

5]2+, [Fe(SCN)

3] and [Fe(SCN)

4]—. However, when F— is

added, colour fades due to formation of [FeF6]3—.

Fe3+ solution also gives deep blue ppt. of Pressian blue with K4[Fe(CN)

6], potassium ferrocyanide :

KCl12])CN(Fe[Fe])CN(Fe[K3ClFe4blueussianPr

36

IIIII

46

II

43

III

or KCl3]6)CN(Fe[FeK])CN(Fe[KFeClblueussianPr

IIIII

643

A deep blue colour is also produced when Fe2+ reacts with K3[Fe(CN)

6], potassium ferricyanide :

blues'Turnbull6

IIIII

6

III

34

II

])CN(Fe[FeK])CN(Fe[KSOFe

Recent X-ray work, IR and other spectroscopic methods have proved that Turnbull’s blue is identical toPrussian blue. The intense colour arises from electron transfer between Fe(+II) and Fe(+III).

Fe3+ partly oxidised 3

6

III4

6

II

])CN(Fe[to])CN(Fe[ and itself is reduced to Fe2+, and also Fe2+ partly

reduces 4

6

II3

6

III

])CN(Fe[to])CN(Fe[ and itself is oxidised to Fe3+. Thus,

])CN(Fe[Feand])CNFe(Fe[ 6

IIIII

6

IIIII are identical.

Ferric Oxide (Fe2O

3) :

Hydrolysis of FeCl3 actually gives a red-brown gelatinous ppt. of the hydrous oxide Fe

2O

3(H

2O)n which on

heating at 2000C gives red-brown a-Fe2O

3 (which occurs as the mineral hematite).

On oxidation of Fe3O4, -Fe

2O

3 is formed.

-Fe2O

3 has hexagonally close-packed lattice of O2— ions with Fe3+ ion in two thirds of the octahedral holes

while -Fe2O

3 has cubic close-packed arrangement of O2— ions with Fe3+ ions randomly distributed in both

the octahedral and tetrahedral sites.

243

,C140032 OOFe4OFe6

0

Fe3O

4 is a mixed oxide

III

32

II

OFe.FeO .

FeO, Fe2O

3 and Fe

3O

4 all tend to be nonstoichiometric. Fe

3O

4 occurs in nature as magnetite.

Fe2O

3 is predominantly basic. Freshly precipitated Fe

2O

3 . (H

2O)

n dissolves in acid giving pale violet

[Fe(H2O)

6]3+ ion. This [Fe

2O

3.(H

2O)

n] also dissolves in concentrated NaOH forming [Fe(OH)

6]3–. This

shows that Fe2O

3 is slightly amphoteric.

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Fusion of Fe2O

3 with Na

2CO

3 gives NaFeO

2 (sodium ferrites) which is hydrolysed to Fe

2O

3 and NaOH :

Na2CO

3 + Fe

2O

3 2NaFeO

2 + CO

2

2NaFeO2 + H

2O 2NaOH + Fe

2O

3

(In earlier days this method, called Lowing Process, was used to manufacture NaOH).

If Cl2 gas is passed into an alkaline solution of hydrated ferric oxide, a red purple solution is formed

containing the ferrate ion

VI2

4 ]FeO[

Fe2O

3 + 2NaOH 2NaFeO

2 + H

2O

2NaFeO2 + Cl

2 + 4NaOH 2Na

2FeO

4 + 2NaCl + 2H

2

Na2FeO

4 can also be obtained by oxidation of Fe

2O

3 with NaOCl (solution hypochlorite). Na

2FeO

4 has

Fe(+VI) and is a strong oxidising agent (like KMnO4).

Ferrous Sulphate (FeSO4.7H

2O – Green Vitriol) :

2442

2442

HFeSOSOHFe

SHFeSOSOHFeS

Na2FeO

4 can also be obtained by oxidation of Fe

2O

3 with NaOCl (sodium hypochlorite). Na

2FeO

4 has

Fe(+VI) and is a strong oxidising agent (like KMnO4).

Ferrous Sulphate (FeSO4 . 7H

2O – Green Vitriol) :

24.dil

42

2442

HFeSOSOHFe

SHFeSOSOHFeS

FeSO4 solution is concentrated when pale green crystals are obtained.

32324 SOSOOFeFeSO2

Anhydrous FeSO4 is coloruless/white. On exposure to atmospheric it turns brownish yellow due to

formation of basic ferric sulphate :

4FeSO4 + 2H

2O + O

2 4Fe(OH)SO

4

FeSO4 . 7H

2O is isomorphous with ZnSO

4 . 7H

2O (white vitriol) and MgSO

4 . 7H

2O (Epsom salt).

Aq. solution of FeSO4 is acidic due to hydrolysis of Fe2+ :

Fe2+ + 2H2O Fe(OH)

2 + 2H+

Acidified MnO4— and CrO

72— oxidise Fe2+ to Fe3+ :

MnO4— + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H

2O

Cr2O

72— + 14H+ + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H

2O

Acidified MnO4— and Cr

2O

72— oxidise Fe2+ to Fe3+ :

MnO4— + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H

2O

Cr2O

72— + 14H+ + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H

2O

Fe2+ salts form blue ppt. Turnbull’s blue with K3[Fe(CN)

6]. Turnbull’s blue is used as pigment in ink and

paint :

blues'Turnbull6

IIIII36

III2 ])CN(Fe[Fe])CN(Fe[Fe

The complex [Fe(H2O)

5NO]2+ is formed in the brown ring test for nitrates when freshly prepared FeSO

4

solution is added to aq. solution of NO3— followed by addition of conc. H

2SO

4. The colour is due to charge

transfer. Its magnetic moment is approximately 3.87 BM confirming the presence of three unpairedelectrons in iron. Thus iron exists as Fe(+I) and nitrosyl as NO+.

FeSO4 and H

2O

2 is used as Fenton’s reagent for producing hydroxyl radicals and for oxidising alcohols to

aldehydes.

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Saturated solution of FeSO4 and (NH

4)

2SO

4 on concentration and cooling gives crystal of Mohr’s salt,

FeSO4 . (NH

4)

2SO

4 . 6H

2O which is widely used as primary standard for standarisation of KMnO

4 and

K2Cr

2O

7 solution.

Copper Sulphate (CuSO4.5H

2O – Blue Vitiol)

CuO + H2SO

4 CuSO

4 + H

2O

2Cu + 2H2SO

4 + O

2

2CuSO4 + 2H

2O

On concentration and cooling, blue crystal of CuSO4 . 5H

2O separate.

The Cu(II) ion or cupric ion is written more precisely as [Cu(H2O)

4]2+ when in aq. solution. It has a bright

blue colour. In CuSO4 . 5H

2O four of the water molecules are associated with Cu2+, and the fifth is

hydrogen – bonded to the sulphate ion as well as to the water molecules on the copper ion, hence this it iswritten as [Cu(H

2O)

4]SO

4 . H

2O

When this is heated, it loses its water of hydration in stages :

22C750

white4

C230

whitebluish24

C100

blue24 OO5CuOCuSOOH.CuSOOH5.CuSO

000

It is the complexing of water molecules to the copper ion is responsible for the blue colour of the pentahydrate.

When the copper (II) ion (complexed with tetrate ion) is heated in basic solution with a reducing sugar,such as glucose), the ion is reduced to Cu

2O (brick red precipitate). (Mixture of alkaline CuSO

4 and sodium

potassium tetrate is called Fehling’s solution and this test called Fehling’s Test of aldehydes and reducingsugars) :

RCHO + 2Cu2+ + 5OH— RCOO— + Cu2O + 3H

2O

If excess of KI is added to an acidified solution of CuSO4(Cu2+ ions), I

2 is produced by oxidation of I—

which reduces Cu2+ to CuI (or Cu2I

2) :

2.pptwhite

4 ICuI2KI4CuSO2

I2 + KI KI

3

I2 produced can be determined by titration with hypo solution :

I2 + 2Na

2S

2O

3 2NaI + Na

2S

4O

6

Thus CuSO4 (Cu2+ salts) can be determined iodometrically :

3224

32224

OSNaICuSO

OSNa2ICuSO2

Eq. wt. of CuSO4 = mol. wt.

Eq. wt. of Na2S

2O

3 = mol. wt.

Addition of KCN to CuSO4 solution first causes reduction and then precipitates CuCN (cuprous

cyanide). This reacts with excess CN—, forming a soluble four co-ordinate complex [Cu(CN)4]3— which is

tetrahedral in shape :

34

I

cyanogen2

I2

])CN(Cu[CN3CuCN

)CN(CNCu2CN4Cu2

[Cu(CN)4]3— Is a stable complex. Cd2+ also forms complex [Cd(CN)

4]2— but unstable, which ionises to

form Cd2+ and CN—. Thus in Cd2+ and Cu2+ both are present, Cu2+ and Cd2+ both are complexed byCN—. If H

2S gas is passed, Cd2+ in unstable [Cd(CN)

4]2— gives yellow ppt. of CdS.

Addition of aq. NH3 to Cu2+ salts give blue ppt. of Cu(OH)

2 which dissolves in excess of aq. NH

3 due to the

formation of the deep blue complex [Cu(NH3)

4]2+ :

OH2])NH(Cu[NH4)OH(Cu

SO)NH()OH(CuOHNH2CuSO

)lelubso(bluedeep

24332

424244

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CuSO4 solution on reaction with Na

2S

2O

3 gives colurless (white) NaCuS

2O

3 due to reduction of Cu(II) to

Cu(I).

CuSO4 + Na

2S

2O

3 CuS

2O

3 + Na

2SO

4

2CuS2O

3 + 2Na

2S

2O

3 2NaCuIS

2O

3 + Na

2S

4O

6

On boiling or in acid solution, NaCuS2O

3 changes to Cu

2S.

2NaCuS2O

3 + H

2O Na

2S

2O

3 + Cu

2S + H

2SO

4

Mercurous Chloride (Hg2Cl

2 – Calomel) :

Hg2Cl

2 (calomel) is formed as white precipitate when soluble chloride (say NaCl) is added to soluble

mercurous salt, say Hg2(NO

3)

2 :

42222

322232

SnClClHgSnClHgCl

NaNO2ClHgNaCl2)NO(Hg

Excess of SnCl2 would reduce Hg

2Cl

2 to Hg :

222

422

ClHgHgHgCl

SnClHg2SnClHgCl

It treated with aq. NH3, it turns black due to formation of Hg(NH

2)Cl . Hg

ClNHHgClHgNHNH2ClHg 4black

2322

In Hg2Cl

2 two Hg atoms are bonded together using the 6s orbitals.

Mercurous ion is diamagnetic; this indicates that three is no unpaired electron, hence mercurous ion is notHg+ but Hg

22+ :

HgHgClHgClClHg

HgCl2ClClHg

2222

2222

C3 Lanthanoids : Properties

1. Electronic Configuration : (n – 2)f1-14(n – 1)d0-1ns2. Atoms of these elements have configuration with 6s2

common but with variable occupancy of 4f level.

However the electronic configuration of all the tripositive ions are same having the general form of 4fn

(n = 1 to 14) Pr (Praseodymium) [Xe]4f36s2

CeZ = 58 [Xe] 2s6d5f4 Z = 59

Gd2

stablemoreorbitalfilledHalf

7 s6d5f4]Xe[ inspite of [Xe]4f86s2

Atomic number Z = 64

Terbium (Tb) [Xe]4f96s2 (Z = 65)

Lutetium (Lu) [Xe]4f145d16s2 (Z = 71)

(Last element of 4f inner transition elements)

Atomic and Ionic Size :

The overall decrease in size (i.e., atomic and ionic) from Lanthanum to Leutetium (the LanthanoidContraction). Thus contraction is of course similar to observe in an ordinary transition series. i.e. imperfectshielding of one electron by another in the same sub-shell. However the shielding of one 4f electron byanother is less than one d-electron by another with the increase in nuclear charge along the series.

of contraction of Lanthanoid series known as Lanthanoid contraction, causes the radii of the members ofthe third transition series to be very similar to those of corresponding members of the second series.

The almost identical radii of Zr(160 pm) and Hf (159 pm), a consequence of the Lanthanoid contraction.

Oxidation States : +3 is the predominant O.S. occasionally they do exist in +2, +4 O.S. e.g. Ce+4 formationis favoured by its noble gas configuration, but it is strong oxidant reverting the common +3 state.

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d- and f- BLOCK ELEMENTS (Practice Problems)

1. Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is atransition element ?

[The outer electronic configuration of Ag (Z = 47) is 4d105s1. In addition to +1, it shows an oxidationstate of +2 also. In +2 oxidation state, the configuration is d9, i.e., the d-subshell is incompletely filled.Hence, it is a transition element]

2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e.,126 kJ mol–1. Why ?

[In the series, Sc to Zn, all elements have one ore more unpaired electrons except zinc which has nounpaired electron, its outer electronic configuration being 3d104s2. Lower the number of unpairedelectrons, lower is the metal-metal bonding. Hence, metal-metal bonding is weakest in zinc.Therefore, enthalpy of atomisation is lowest]

3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states andwhy ?

[Manganese (Z = 25) shows maximum number of oxidation states because of its electronicconfiguration 3d54s2. As 3d and 4s are close in energy, it has maximum number of electrons to lose orshare (as all the 3d electrons are unpaired). Hence, it shows maximum oxidation states from +2 to +7(+2, +3, +4, +5, +6 and +7)]

4. The E0(M2+/M) value for copper is positive (+0.34 V). What is possibly the reason for this ?

[E0 (M2+/M) for any metal depends upon the sum of the enthalpy changes taking place in thefollowing steps : M(s) +

aH M(g), (

aH = enthalpy of atomisation), M(g) +

iH M2+(g),

(iH = ionisation enthalpy), M2+(g) + aq M2+(aq) +

hydH(

hydH = hydration enthalpy)] Copper

possesses a high enthalpy of atomisation (i.e., energy absorbed) and low enthalpy of hydration, (i.e.,energy released). Hence; E0(Cu2+/Cu) is positive]

5. How would you account for the irregular variation of ionisation enthalpies (first and second) in thefirst series of the transition elements ?

[First ionisation enthalpy : As we move from left to right, it is expected in general that the firstionisation enthalpy should show an increasing trend. However, the trend is irregular becauseremoval of the electron alters the relative energies of 4s and 3d orbitals. Thus, there is a reorganisationenergy accompanying ionisation. The results into the release of exchange energy which increases asthe number of electrons increases in the d-orbitals and also from the transference of s-electrons intod-orbitals. Cr has low first ionisation energy because loss of one electron gives stable electronicconfiguration (3d5). Zn has very high ionisation energy because electron is removed from 4s orbitalof the stable configuration (3d10 4s2). Half-filled and fully filled configuration are more stable thanothers. Second ionisation enthalpy : After the loss of one electron, the removal of second electronbecomes much more difficult. Hence, second ionisation enthalpies are much higher and in generalincrease from left to right. However, Cr and Cu show much higher values because the secondelectron has to be removed from the stable configuration of Cr+(3d5) and Cu+(3d10)]

6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only ?

[Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidise the metal tothe highest oxidation state by prompting all the valence electrons to participate in bonding]

7. Which is a stronger reducing agent Cr2+ or Fe2+ and why ?

[Cr2+ is a stronger reducing agnet than Fe2+. Reason : 0

Cr/Cr 23E is –ve (–0.41 V) whereas 0

Fe/Fe 23E

is +ve (+0.77 V). Thus, Cr2+ is easily oxidised to Cr3+ but Fe2+ cannot be easily oxidised to Fe3+. Hence,Cr2+ is stronger reducing agent than Fe2+]

8. Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

[Spin only magnetic moment (µ) = BM87.3BM15)23(3BM)2n(n ]

9. Explain why Cu+ ion is not stable in aqueous solution.

[Cu2+(aq) is much more stable than Cu+(aq). Although second ionisation enthalpy of copper is largebut

hyd H for Cu2+(aq) is much more negative than that for Cu+(aq) and hence it more than

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compensates for the second ionisation enthalpy of copper. Therefore, many copper (I) compoundsare unstable in aqueous solution and undergo disproportionation]

10. Actionoid constraction is greater from element to element than lanthanoid contraction. Why ?

[This is due to poor shielding by 5f electrons in the actinoids than that by 4f electrons in thelanthanoids]

11. Write down the electronic configuration of :

(i) Cr3+ (ii) Cu+ (iii) Co2+ (iv) Mn2+

(v) Pm3+ (vi) Ce4+

[(i) Cr3+ = [Ar]3d3 (ii) Cu+ = [Ar]3d10 (iii) Co2+ = [Ar]3d7 (iv) Mn2+ = [Ar]3d5 (v) Pm3+ = [Xe]4f4

(vi) Ce4+ = [Xe] (vii) Lu2+ = [Xe]4f145d1 (viii) Th4+ = [Rn]]

12. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state ?

[Electronic configuration of Mn2+ is 3d5 which is half filled and hence stable. Therefore, third ionisationenthalpy is very high, i.e., third electron cannot be lost easily at it would affect and stability of 3d5

configuration. In case of Fe2+, electronic configuration is 3d6. Hence, it can lose one electron easily togive the stable configuration 3d5]

13. Explain briefly how +2 state becomes more and more stable in the first half of the first rowtransition elements with increasing atomic number ?

[Except scandium (which shows an oxidation state of +3), all other first row transition elementsshow an oxidation state of +2. This is due to loss of two 4s electrons. In the first half, as we move fromTi2+ to Mn2+, the electronic configuration changes from 3d2 to 3d5, i.e., number of half-filledd-orbitals increases imparting greater and greater stability to +2 state. In the second half, i.e., Fe2+ toZn2+, the electronic configuration changes from 3d6 to 3d10, i.e., the number of half-filled orbitaldecreases. Hence, the stability of +2 state decreases]

14. To what extent do the electronic configuration decide the stability of oxidation states in the firstseries of the transition elements ? Illustrate your answer with example.

[In a transition series, the oxidation state which lead to noble gas or exactly half-filled or completelyfilled d-orbitals are more stable. For example, electronic configuration of Mn(Z = 25) is [Ar] 3d54s2.It shows oxidation states from +2 to +7 but Mn (II) is most stable because it has the half-filledconfiguration [Ar]3d5. Similarly, Sc3+ and Zn2+ are more stable because of noble gas configurationand completely filled configuration. Sc = [Ar]3d14s2, Sc3+ = [Ar], i.e., noble gas configuration,Zn = [Ar] 3d10 4s2, Zn2+ = [Ar]3d10, i.e., completely filled configuration]

15. What may be the stable oxidation state of the transition element with the following d-electronconfiguration in the ground state of their atoms : 3d3, 3d5, 3d8 and 3d4 ?

16. Name the oxometal anions of the first series of the transition metals in which the metal exhibits theoxidation state equal to its group number.

[Cr2O

72– and CrO

42– (Group Number = Oxidation state of Cr = 6) MnO

4– (Group = Oxidation state of

Mn = 7)]

17. What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ?

18. What are the characteristics of the transition elements and why are they called transition elements ?Which of the d-block elements may not be regarded as the transition elements ?

[Zn, Cd and Hg are not regarded as transition elements]

19. In what way is the electronic configuration of the transition elements different from that of the non-transition elements ?

[Transition elements contain incompletely filled d-subshell, i.e., their electronic configuration is(n – 1)d1 – 10ns0 – 2, whereas non-transition elements have either no d-subshell or their d-subshell iscompletely filled and have ns1-2 or ns2np1-6 in their outermost shell]

20. What are the different oxidation states exhibited by the lanthanoids ?

[The most common oxidation state of lanthanoids is +3. However, some lanthanoids also show anoxidation state of +2 and +4. For example, Eu shows an oxidation state of +2 and Ce shows anoxidation state of +4]

21. What are interstitial compounds ? Why are such compounds well known for transition metals ?

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22. How is the variability in oxidation states of transition metals different from that of thenon-transition metals ? Illustrate with examples.

23. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect ofincreasing pH on a solution of potassium dichromate ?

[On increasing pH of potassium dichromate solution(i.e., on adding alkali), it changes to potassiumchromate. 2K

2Cr

2O

7 + 2KOH 2K

2CrO

4 + H

2O]

24. Describe the oxidising action of potassium dichromate and write the ionic equations for its reactionwith : (i) iodide (ii) iron (II) solution and (iii) H

2S.

[(i) 2232

72 I3OH7Cr2I6H14OCr

(ii) OH7Fe6Cr2H14Fe6OCr 23322

72

(iii) S3OH7Cr2SH3H8OCr 23

2272 ]

25. Describethe preparation of potassium permanganate. How does the acidified permanganatesolution react with (i) iron (II) ions (ii) SO

2 and (iii) oxalic acid ? Write the ionic equations for the

reactions.

[(i) Iron (II) ions (ii) SO2 and (ii) SO

2 and (iii) Oxalic acid.

32

224 Fe10OH8Mn2Fe10H16MnO2

H4Mn2SO5OH2SO5MnO2 224224

2222

424 CO10OH8Mn2OC5H16MnO2 ]

26. For M2+/M and M3+/M2+ systems, the E0 values for some metals are as follows :

Cr2+/Cr = –0.9 V Cr3+/Cr2+ = – 0.4 V

Mn2+/Mn = –1.2 V Mn3+/Mn2+ = + 1.5 V

Fe2+/Fe = –0.4 V Fe3+/Fe2+ = + 0.8 V

Use this data to comment upon :

(a) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and

(b) the ease with which iron can be oxidising as compared to the similar process for eitherchromium or manganese metals.

[(a) Thus, the stability follows the order : Cr3+ > Fr3+ > Mn3+ (b) Thus, the order of getting oxidisedwill be : Mn > Cr > Fe]

27. Predict which of the following will be coloured in aqueous solution ?

Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, and Co2+

Give reasons for each.

[Only those ions which have incompletely filled d-orbitals are coloured. Those with fully-filled orempty d-orbitals are colourless. Thus, Ti3+, V3+, Mn2+, Fe3+ and Co2+ are all coloured due to d-dtransition. MnO

4– is also coloured but due to charge transfer. Only Sc3+ (3d0) is colourless]

28. Compare the stability of +2 oxidation state for the elements of the first transition series.

[The decreasing negative electrode potentials of M2+/M in the first transition series shows that ingeneral, the stability of +2 oxidation state decreases from left to right (the exceptions being Mn andZn). The decrease in the negative electrode potentials is due to increase in the sum IE

1 + IE

2. The

greater stability of +2 state for Mn is due to half-filled d-subshell (d5) in Mn2+ and that of Zn is due tocompletely filled of subshell (d10) in Zn2+]

29. Compare the chemistry of actinoids with that of the lanthanoids with special reference to :

(a) electronic configuration (b) oxidation state

(c) atomic and ionic sizes, and (d) chemical reactivity

[(a) The general electronic configuration of lanthanoids is [Xe]4f1-145d0-16s2 whereas that of actionoids

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is [Rn]5f1-146d0-17s2. Thus, lanthanoids belong to 4f-series whereas actionoids belong to 5f-series.(b) Lanthanoids show limited oxidation states (+2, +3, +4) out of which +3 is most common. This isbecause of large energy gap between 4f, 5d and 6s subshells. On the other hand, actionoids show alarge number of oxidation states because of small energy gap between 5f, 6d and 7s subshells. (c)Both lanthanoids and actinoids show decrease in size of their atoms or ions in +3 oxidation state. Inlanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoidcontraction. However, the contraction is greater from element to element in actinoids due to poorershielding by 5 f electrons. (d) Important reactions of lanthanoids and actinoids are given below :Lanthanoids : (i) They combine with H

2 on gentle heating with carbon, then form carbides. When

burnt in presence of halogen, they form halides. (ii) They react with dilute acids to liberate hydrogengas. (iii) They form oxides and hydroxides of the type M

2O

3 and M(OH)

3 which are basic in

character. Actinoids : (i) They combine with most of the non-metals at moderate temperature withcarbon, they form carbide, with sulphur, sulphides are formed and with halogens, halides are formed.(ii) These metals are attached by hydrochloric acid, but the effect of nitric acid is small due to theformation of protective oxide layer on the surface. (iii) They react with boiling water to give amixture of oxides and hydroxides. (iv) Alkalis have no action on them]

30. How would you account for the following :

(i) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidising ?](ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is

easily oxidised ?

(iii) The d1 configuration is very unstable in ions.

[(ii) Co(III) has greater tendency to form coordination complexes than Co(II). Hence, in thepresence of ligands, Co(II) changes to Co(III), i.e., is easily oxidized. (iii) The ions with d1

configuration have the tendency to lose the only electron present in d-subshell to acquire stable d0

configuration. Hence, they are unstable and undergo oxidation or disproportionation]

31. What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction inaqueous solution.

[ OH2MnOMnO2H4MnO3 22424 , OH4CrCrO2H8CrO 2

324

34 ]

32. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently andwhy ?

[Cu has the electronic configuration 3d10 4s1. It can easily lose 4s1 electron to give the stable3d10 configuration. Hence, it shows +1 oxidation state]

33. Calculate the number of unpaired electrons in the following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+.Which one of these is the most stable in aqueous solution ?

[Mn3+ : 3d4 has 4 unpaired electrons, Cr3+ : 3d3 has 3 unpaired electrons, V3+ : 3d2 has 2 unpairedelectrons, Ti3+ : 3d1 has 1 unpaired electron. Cr3+ is most stable out of these in aqueous solutionbecause it has half-filled t

2g level (i.e., t3

2g).

34. Give examples and suggest reasons for the following features of the transition metal chemistry :

(i) The lowest oxide of transition metal is basic, the highest is amphoteric acidic.

(ii) A transition meal exhibits highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

35. Indicate the steps in the preparation of

(i) K2Cr

2O

7 from chromite ore.

(ii) KMnO4 from pyrolusite ore.

36. What are alloys ? Name an important alloy which contains some of the lanthanoid metals. Mentionits uses.

[An alloy is a homogeneous mixture of two or more metals or metals and non-metals. An importantalloy containing lanthanoid metals is mischmetal which contains 95% lanthanoid metals and 5%iron alongwith traces of S, C, Ca and Al. it is used in Mg-based alloy to produce bullets, shells andlighter flints]

37. What are inner transition elements ? Decide which of the following atomic numbers are the atomicnumbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

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[The f-block elements i.e., in which the last electron enters the f-subshell are called innertransitionelements. These include lanthanoids (58-71) and actinoids (90-103). Accordingly, elements with atomicnumbers 59, 95 and 102 are inner transition elements]

38. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify thisstatement by giving some examples from the oxidation state of these elements.[Lanthanoids show limited number of oxidation state, viz. +2, +3 and +4 (out of which +3 is mostcommon). This is because of large energy gap between 4f, 5d and 6s subshells. The dominantoxidation state of actinoids is also +3 but they show a number of other oxidation states also, e.g.,uranium (Z = 92) and plutonium (Z = 94) show 3, +4, +5 and +6, neptunium (Z = 94) shows +3, +4, +5and +7, etc. This is due to small energy difference between 5f, 6d and 7s subshells of the actionoids.Depending upon the reaction conditions, any number of electrons from 5f, 6d and 7s canparticipate]

39. Which is the last element in the series of the actinoids ? Write the electronic configuration of thiselement. Comment on the possible oxidation state of this element.[Last actinoid is Lawrencium (Z = 103). Electronic configuration = [Rn]5f14d17s2. Possible oxidationstate = +3]

40. Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magneticmoment of the basis of ‘spin-only’ formula.[Cerium has the configuration : [Xe] 4f15d16s2. Configuration of Ce3+ may be written as : [Xe]4f1, i.e.,there is only one unpaired electron, i.e., n = 1 (1.73 BM)]

41. Name the numbers of the lanthanoid series which exhibit +4 oxidation states and those whichexhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurationof these elements.[+4 oxidation state is given by :

58Ce,

59Pr,

60Nd,

65Tb,

66Dy. +2 oxidation state is given by : 60Nd, 62Sm,

63Du, 69Tm, 70Yb. +2 oxidation state is exhibited when the lanthanoid has the configuration 5d06s2 sothat 2 electrons are easily lost. +4 oxidation state is exhibit when the configuration left is close to4f0 (e.g. 4f0, 4f1, 4f2) or close to 4f1 (e.g. 4f7 or 4f8)]

42. Compare the chemistry of the actionoids with that of lanthanoids with reference to :(i) electronic configuration(ii) oxidation states, and(iii) chemical reactivity

43. Write the electronic configuration of the elements with atomic numbers 61, 91, 101 and 109.44. Compare the general characteristics of the first series of transition metals with those of the second

and third series metals in the respective vertical columns. Give special emphasis on the followingpoints : (i) electronic configurations, (ii) oxidation states, (iii) ionisation enthalpies and (iv) atomicsizes.

45. Write down the number of 3d electrons in each of the following ions :Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+

Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions(octahedral).

46. Comment on the statement that elements of the first transition series possess many propertiesdifferent from those of heavier transition elements.

47. What can be inferred from the magnetic moment values of the following complex species ?Example Magnetic moment (BM)K

4[Mn(CN)

6] 2.2

[Fe(H2O)

6]2+ 5.3

K2[MnCl

4] 5.9

48. Explain giving reasons :(i) Transition metals and many of their compounds show paramagnetic behaviour.(ii) The enthalpies of atomisation of the transition metals are high.(iii) The transition metals generally form coloured compounds.(iv) Transition metals and their may compounds act as good catalyst.

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E X E R C I S E (OBJECTIVE)

1. Out of SiCl4, TiCl

4, PO

43—, SO

42—, CrO

42—, CCl

4

isostructural are

(a) SiCl4, TiCl

4(b) SO

42—, CrO

42—

(c) both (d) none

2. Maximum oxidation state is shown by

(a) Os (b) Mn

(c) Cr (d) Co

3. Following elements do not show the propertiescharacteristics of d-block elements

(a) Cu, Ag, Au (b) Zn, Hg, Cd

(c) Sc, Ti, V (d) Fe, Co, Ni

4. AgCl and NaCl are colourless NaBr and NaI arealso colourless but AgBr and AgI are coloured. Thisis due to

(a) Ag+ polarises Br— and I—

(b) Ag+ has unpaired d-orbital

(c) Ag+ depolarised Br— and I—

(d) none is correct

5. Which is not true statements ?

(a) ions of d-block elements are coloured dueto d—d transition

(b) ions of f-block elements are coloured dueto f—f transition

(c) [Sc(H2O)

6]3+, [Ti(H

2O)

6]4+ are coloured

complexes

(d) Cu+ is colourless ion.

6. Match the catalyst in column X with their uses incolumn Y

Column X Column Y

A : TiCl4

I : Adams catalystin reduction

B : PdCl2

II : in preparationof (CH

3)

2SiCl

2

C : Pt/ III : Reppesynthesis

D : Cu IV : used as thecatalyst inpolytheneproduction

E : Ni V : Wake processfor convergingC

2H

4 to

CH3CHO

A B C D E

(a) IV V I II III

(b) IV V II I II

(c) V IV I III II

(d) II I III V IV

7. When (A) NH4VO

3 is heated.

(B) (NH4)

2Cr

2O

7 is heated :

(a) in both cases N2 is formed

(b) in both cases NH3 is formed

(c) in (A) NH3 and in (B) N

2 are formed

(d) in (A) N2 and in (B) NH

3 are formed

8. FeCr2O

4 (chromite) is converted to Cr following

steps :

CrOCrCrONaChromiteIII

32II

42I

I, II and III are

I II III

(a) H2SO

4/air, C C

(b) NaOH/air, C, Al,

(c) HCl/air, C, Mg,

(d) conc. H2SO

4, NH

4Cl, C,

9. When H2O

2 is added to an acidified solution of

K2Cr

2O

7

(a) solution turns green due to formation ofCr

2O

3

(b) solution turns yellow due to formation ofK

2CrO

4

(c) a deep blue-violet coloured compoundCrO(O

2)

2 is formed

(d) solution gives green ppt of Cr(OH)3

10. Match the compounds of column X with oxidationstate of column Y

Column X Column Y

I [Cr(H2O)

6Cl

3] 5

II CrO5

4

III K3CrO

86

IV (NH3)

3CrO

43

I II III IV

(a) 3 6 5 4

(b) 3 4 5 6

(c) 4 5 6 3

(d) 6 5 4 3

11. For CrO3 following is not true statement

(a) it is called chromic acid

(b) it is colourless due to 3d0 configuration

(c) it is bright orange solid and colour arisesdue to charge transfer

(d) it is toxic and corrosive

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12. 272

ypH24

xpH272 OCrCrOOCr pH

values x and y can be

(a) 4 and 5 (b) 4 and 8

(c) 8 and 4 (d) 8 and 9

13. Cl2 gas is obtained by various reactions but not by

(a) KMnO4 + con. HCl

(b) KCl + K2Cr

2O

7 + conc. H

2SO

4

(c) MnO2 + conc. HCl

(d) KCl + F2

14. Which is not true statement about KMnO4

(a) its solution in unstable in acidic medium

(b) its small quantity added to conc. H2SO

4,

a green coloured solution containingMnO

3+ ions is formed

(c) MnO4— changes to Mn2+ in basic solution

(d) it is self-indicator in Fe2+ or C2O

42—

titration

15. Pyrolusite is MnO2 used to prepare KMnO

4. Steps

are

4II2

4I

2 MnOMnOMnO

I and II are

(a) fuse with KOH/air, electrolytic oxidation

(b) fuse with KOH/air, electrolytic reduction

(c) fuse with conc. HNO3/air, electrolytic

reduction

(d) all correct

16. Fe is made passive by

(a) dil. H2SO

4(b) dil HCl

(c) aqua regia (d) conc. H2SO

4

17. Which is not ture statement about FeO ?

(a) it is non-stoichiometric and is metaldeficient

(b) it is basic oxide

(c) its aqueous solution changes to Fe(OH)3

and then to Fe2O

3.(H

2O)n by

atmospheric oxygen

(d) it gives red colour with KCNS

18. In Na2[Fe(CN)

5NO], sodium nitrophrusside :

(a) oxidation state of Fe is +1

(b) this has NO+ as ligand

(c) both correct

(d) none is correct

19. FeSO4 solution gives brown colour ring in testing

nitrates or nitrites. This is

(a) [Fe(H2O)

5NO]2+

(b) [Fe(H2O)

5NO

2]2+

(c) [Fe(H2O)

4(NO)

2]2+

(d) [Fe(H2O)

4NO]2+

20. Ni2+, in traces, can be tested using

(a) sodium nitroprusside

(b) dimethyl glyoxime

(c) ammonium sulphocyanide

(d) potassium ferocyanide

21. CuSO4 can be estimated volumetrically

(a) by reaction with KI following by reactionwith Na

2S

2O

3

(b) by reaction with BaCl2

(c) by reaction with K4Fe(CN)

6

(d) none is correct

22. Which is correct statement ?

(a) ammoniacal CuCl is used to measure theamount of CO in gas samples

(b) ammoniacal CuCl gives red ppt. withCH CH

(c) both correct

(d) none is correct

23. When KCN is added to CuSO4 solution

(a) KCN acts a reducing agent

(b) KCN acts as complexing agent

(c) K3[Cu(CN)

4] is formed

(d) all correct

24. Out of AgNO3, AgF and AgClO

4, water soluble salts

are

(a) AgF

(b) AgF, AgNO3

(c) AgF, AgNO3, AgClO

4

(d) none

25. Stability of Cu+ and Ag+ halide complexes is in or-der

(a) I > Br > Cl > F (b) F > Cl > Br > I

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

26. Which of them exists as white salt is in anhydrousstate ?

(a) CuF2

(b) CuSO4

(c) both (d) none

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27. When KI (excess) is added to

I : CuSO4

II : HgCl2

III : Pb(NO3)

2

(a) a white ppt of CuI in I, a orange ppt HgI2

in II and a yellow ppt PbI2 in III

(b) a white ppt of CuI in I, an orange pptdissolving to HgI

42— in II, and a yellow

ppt of PbI2 in III

(c) a white ppt of CuI, HgI2 and PbI

2 in each

case

(d) none is correct

28. Magnetic moment of Cr(Z = 24), Mn+ (Z = 25) andFe2+ (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

29. Interstitial compound is formed by

(a) Fe, Co (b) Co, Ni

(c) Fe, Ni (d) all

30. A jeweller is selling 22-carat gold articles with 95%purity, it is approximately

(a) exact (b) 3.5% higher

(c) 3.5% lower (d) 5% lower

31. Paramagnetism is given by the relation

)1s(s2µ magnetons where ‘s’ is the total

spin. On this basis, the paramagnetism of Cu+ ionis :

(a) 3.88 magnetones

(b) 2.83 magnetones

(c) 1.41 magnetones

(d) zero

32. The oxygen-carrying pigment, oxyhaemocyanin,containing two copper ions is diamagnetic, because

(a) the two copper ions are in +1 oxidationstate

(b) one of the copper ions is in +1 oxidationstate and the other is in +2 oxidationstate

(c) there are strong anti-ferromagnetic interactions between the two copper ions

(d) there are ferromagnetic interactionsbetween the two copper ions

33. Match List I with List II and select the correct an-swer using the codes given below the lists :

List I List II

(Alloys) (Constituents)

A. Gun metal 1. Lead + Tin

B. German silver 2. Copper + Tin +Zinc

C. Brass 3. Copper + Zinc

D. Solder 4. Copper + Zinc+ Nickel

Codes :

A B C D

(a) 1 3 2 4

(b) 4 2 1 3

(c) 2 4 3 1

(d) 3 1 2 4

34. An acidic solution contains Cu2+, Pb2+ and Zn2+. Ifhydrogen sulphide gas is passed through thissolution, the preipitate will contain

(a) CuS and ZnS

(b) PbS and ZnS

(c) CuS and PbS

(d) CuS, PbS and ZnS

35. Ti2+ is purple while Ti4+ is colourless, because :

(a) there is no crystal field effect in Ti4+

(b) Ti2+ has 3d2 configuration

(c) Ti4+ has 3d2 configuration

(d) Ti4+ is a very small cation when comparedto Ti2+ and hence does not absorb anyradiation

36. Match List I with List II and select the correctanswer using the codes given below the lists

List I List II

(Metals) (Ores)

A. Zinc 1. Azurite

B. Tin 2. Carnallite

C. Copper 3. Calamine

D. Magnesium 4. Cassiterite

Codes :

A B C D

(a) 3 4 2 1

(b) 3 4 1 2

(c) 4 1 3 2

(d) 4 3 2 1

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37. Haemoglobin and Chlorophyll contain

(a) Fe, Co (b) Fe, Mn

(c) Mg, Fe (d) Fe, Mg

38. In Cr2O

72— every Cr is linked to

(a) two O atoms (b) three O atoms

(c) four O atoms (d) five O atoms

39. For Ni and Pt different I.P. values are givenbelow :

21 )IP()IP( 43 )IP()IP(

Ni 2.49 8.80

Pt 2.60 6.70

hence :

(a) nickel (II) compounds tend to be thermodynamically more stable than platinum(II)

(b) platinum (IV) compounds tend to bemore stable than nickel (IV)

(c) both correct

(d) none is correct

40. Which one is green vitriol ?

(a) CuSO4

(b) FeSO4

(c) MgSO4

(d) ZnSO4

A N S W E R S

1. c

2. a

3. b

4. a

5. c

6. a

7. c

8. b

9. c

10. a

11. b

12. c

13. b

14. c

15. a

16. c

17. d

18. c

19. a

20. b

21. a

22. c

23. d

24. c

25. a

26. c

27. b

28. c

29. d

30. b

31. d

32. a

33. c

34. c

35. b

36. b

37. d

38. c

39. c

40. b

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