Chemistry d and f block notes and questions

d AND  f- BLOCK ELEMENTS

THE SQUARE THING SHOWS THE SIGN OF(---->) "GIVES."

d-BLOCK ELEMENTS (TRANSITION ELEMENTS)

GENERAL ELECTRONIC CONFIGURATION- ns^1-2(n-1)d^1-10

There are 4 transition series in d-block each having 10 elements as a d- sub-shell can have a maximum of 10 electrons. d-block elements represents the change or transition in properties from more electropositive s-block elements to less electropositive p-block elements and thus they are known as transition elements.

Transition elements are those elements which have incompletely filled d-sub shell in their elemental form or the most common oxidation state.

PROPERTIES OF d-BLOCK ELEMENTS

1. ATOMIC RADII- The atomic radii of d-block elements in a series decrease with increase in atomic number but the decrease in atomic size is small after midway. At the end of the period, there is a slight increase in the atomic radii.

2. METALLLIC CHARACTER- The metallic character of transition elements is due to their low ionization enthalpy.

3. ENTHALPY OF ATOMISATION- Due to presence of the strong metallic bonds the transition metals are hard and have high enthalpies of atomization.

4. IONISATION ENTHALPY- The increase in ionization enthalpy is due to increase in nuclear charge which tends to attract the electron cloud with greater force of attraction.

5. OXIDATION STATE- Transition elements exhibits a number of oxidation states. The highest oxidation state is found in the compounds of Fluorine and Oxygen due to the high electro negativity and small size. In the first transition series the highest oxidation state is exhibited in Manganese and it is +7. The variable oxidation state of transition elements is due to the participation of ns and (n-1)d electrons in bonding.The highest oxidation state is exhibited by Osmium and is +8.

MAGNETIC PROPERTIES

Transition elements generally contain one or more unpaired electron in them and their compounds are generally paramagnetic.

          The magnetic behaviour is expressed in terms of Magnetic Moment (μ):

                

μ = √n (n+2)    B.M.

                       

where n number of unpaired electron and B.M. stands for Bohr Magneton.

The compounds of Zn are diamagnetic in nature because Zn²⁺ do not contain any unpaired electron and all its orbitals are completely filled.

COMPLEX FORMATION

The tendency of transition elements to form complex is due to following reasons:

(i)                 Small size and high charge density of transition metal ions.

(ii)               Presence of vacant orbital of appropriate energy which can accept lone pair of electron donated by other groups.

CATALYTIC PROPERTIES

Many transition elements and their compounds act as good catalyst for various reactions due to their tendency to form variable oxidation state which leads to formation of unstable intermediate compounds. The transition metals also provide a larger surface area for the reactants to be adsorbed.

ALLOY FORMATION

Transition metals have tendency to form a number of alloys as these are quite similar in size.

INTERSTITIAL COMPOUNDS

The transition metal can form interstitial compounds due to the presence of vacant spaces (voids) between the metal atoms which arise due to their tendency to exhibit oxidation state.

COLOUR

Transition elements have partially filled d-orbital and thus the transition of electron can take place from one of the lower d-orbital to higher d-orbital within the same sub shell. Thus they appear colored.

POTASSIUM PERMAGNATE

PREPARATION-

It can be prepared from Pyrolusite Ore (MnO₂).

KMnO₄ can be prepared from pyrolusite ore by the following steps:

(i)                 Conversion of MnO₂ into potassium magnate. KMnO₄ is fused with potassium hydroxide/K₂CO₃ in the presence of air or strong oxidizing agent like nitric acid to form a green mass due to formation of potassium magnate.

      

               MnO₂ + KOH + O₂        K₂MnO₄           +    H₂O

                                                                         potassium magnate

                                                                               (green)

               MnO₂   +   K₂CO₃   +O₂         K₂MnO₄       +  CO₂

                                                                                                                                      (green)

 Conversion of  K₂ MnO₄  to KMnO₄

Potassium magnate when treated with chlorine/O₃ gets oxidized to potassium permagnate.

K₂ MnO₄  +    Cl₂          KMnO₄  +  KCl

                                                                (purple)

PROPERTIES OF KMnO₄

(i)                 COLOUR – It exists as deep purple black prisms with a greenish lustre.

(ii)               SOLUBILITY- It is moderately soluble in water at room temperature and is more soluble in cold water.

(iii)             ACTION OF HEAT- On heating it forms potassium magnate along with the formation of MnO₂ and evolution of Oxygen.

                                 Δ

2KMnO₄       K₂MnO₄  +  MnO₂  +  O₂↑

(iv) OXIDISING NATURE

KMnO₄ acts as a strong oxidizing agent in acidic, alkaline as well as neutral medium.

     2KMnO₄ + 3H₂SO₄                             K₂SO₄  +  2MnSO₄ +3 H₂O + 5[O]

To provide acidic medium dilute sulphuric acid is used and not diluted nitric acid or hydrochloric acid. Nitric acid is not used to acidify the medium as it itself acts as a stronger oxidizing agent. HCl is also not used to acidify the medium as it gets oxidized to Cl₂ by KMnO₄.

MnO^-₄   +  8H⁺  + 5e^-                       Mn⁺²  +   4H₂O

A.   ACIDIC MEDIUM

(i)                 Oxidation of Acidified Ferrous salt to Ferric Salt

     2KMnO₄  +  3H₂SO₄     K₂SO₄   +    2MnSO₄    +    3H₂O   +   5[O]

2Fe( SO₄)  +  H₂SO₄   + [O]            Fe₂(SO₄)₃ +  H₂O   }  x 5

     2KMnO₄ + 10FeSO₄ + 8 H₂SO₄   K₂SO₄+ 2MnSO₄ +5Fe2(SO₃)₃  +  8H₂O

     2MnO^-₄ + 5Fe²⁺ + 8H⁺     Mn²⁺ + 5Fe³⁺  + 4H₂O

(ii)      Oxidation of Oxylic Acid to CO₂

  

2KMnO₄ + 3H₂SO₄     K₂SO₄ + 2MnSO₄  + 3H₂O

H₂C₂O₄  +  [O]    2CO₂  + H₂O     }  x  5

2KmnO4 + 3H₂SO₄  + 5H₂C₂O₄    K₂SO₄  +2 MnSO₄  +  8H₂O +10CO₂

2MnO^-₄ + 16H⁺   +  5C₂O^2-₄   2Mn²⁺ + 8 H₂O + 10CO₂

B.         NEUTRAL MEDIUM  

 KMnO₄ in neutral medium forms MnO₂ along with KOH. Although the reaction starts in neutral medium, in due course of time it becomes alkaline in nature due to formation of KOH or release of hydroxide ions in the solution.

2KMnO₄ + H₂O  2KOH + 2MnO₂ +  3[O]

(i) Oxidation of Na₂S₂O₃ to Na₂SO₄

KMnO₄ + H₂O + 3Na₂S₂O₃   2KOH + 2MnO₂ +  3[O]

Na₂S₂O₃ + [O]          Na₂SO₄ + S  }  x  3

2KMnO₄ + H₂O + 3Na₂S₂O₃  3KOH + 2MnO₂ +  3S

C.  ALKALINE MEDIUM

In alkaline medium KMnO₄ first reduced to potassium magnate and then it is further reduced to MnO_2. Thus, the solution first changes from purple to green and then finally becomes colourless due to formation of MnO_2.

(i)Oxidation of KI to KIO₃

  

 2MnO₄ + H₂O  2MnO₂ +  2KOH

 KI  +  [O]         KIO₃     } x 3

2MnO₄ + H₂O + KI   2MnO₂ +  4KOH + KIO₃

2MnO^-₄ + I^-    2MnO₂  + 2OH^- + 2IO^-₃

POTASSIUM DICHROMATE (K₂Cr₂O₇)

PREPARATION: K₂Cr₂O₇ can be prepared from chromite ore (FeCr₂O₄) by following steps:

FeCr₂O₄  + NaOH + O₂    Na₂CrO₄ + Fe₂CO₃ + H₂O↑

FeCr₂O₄  +  Na₂CO₃ + O₂  Na₂CrO₄ + Fe₂CO₃ + CO₂↑

Na₂CrO₄ + H₂SO₄   Na₂Cr₂O₇  +  Na₂SO₄ +  H₂O

Na₂Cr₂O₇  + KCl    K₂Cr₂O₇ + NaCl

PROPERTIES OF K₂Cr₂O₇

ü  ACTION OF HEAT- It decomposes to give potassium chromate along with chromic oxide.

                                Δ

4 K₂Cr₂O₇   4K₂CrO₄ + 2Cr₂O₃ + 3O₂

ü  REACTION WITH ALKALIES

 K₂Cr₂O₇ + KOH  K₂CrO₄ + H₂O

K₂CrO₄ + H₂SO₄  K₂Cr₂O₇ + K₂SO₄ + H₂O

                                                          _alkali

                 K₂Cr₂O₇   K₂CrO₄

                          ^{ORANGE               acid                     YELLOW}

                                                          _alkali

                 Cr₂O^2-₇      CrO^2-₄

                                                           ^acid

OXIDISING PROPERTIES

Potassium dichromate acts as an oxidising agent in acidic medium.

(i) Oxidation of acidified potassium iodide to iodine

 K₂Cr₂O₇  +7H₂SO₄ + 6KI   3I₂+ 4K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O

Cr₂O^2-₇ + 14H⁺  +16I^-  3I₂+ 7H₂O +2Cr³⁺

(ii) Oxidation of acidified ferrous salt to ferric salt

K₂Cr₂O₇  + 7H₂SO₄ + 6FeSO₄   3Fe₂(SO₄)₃ + K₂SO₄  + Cr₂(SO₄)₃ + 7H₂O

Cr₂O^2-₇ + 14H⁺  + 6Fe²⁺     6Fe³⁺+ 7H₂O +2Cr³⁺

f- BLOCK ELEMENTS

f- block elements are also known as inner transition elements and they are present at the bottom in periodic table. It comprises of two series, lanthanoids and actinoids.

  

GENERAL ELECTRONIC CONFIGURATION: 4f^1-145d^0-16s^1-2

OXIDATION STATE: The common oxidation state of lanthanoids is +3 but in some cases  they exhibit the oxidation state of +2 or +4.This is so because in doing so they attain the stable configuration of 4f⁷/4f¹⁴.

Ø  Cerium and Terbium exhibits oxidation state of +4.

Ø  Europium and Ytterbium exhibits the oxidation state of +2.

LANTHANOID CONTRACTION

The steady decrease in the atomic sizes of lanthanoids with increasing atomic number is called lanthanoid contraction.

ð  CONSEQUENCES

1. Similarity in atomic sizes of elements of second and third transition series present in same group.

2. Separation of lanthanoids was made possible.

©	LANTHANOIDS	ACTINOIDS
1. 2. 3. 4.	Besides +3 oxidation state they show +4 and +2 oxidation state only in few cases. They have less tendency towards complex formation. Except promethium they are non-radioactive. Their magnetic properties can be explained easily.	Besides +3 oxidation state they show higher oxidation state of +4,+5,+6 and +7. They have greater tendency towards complex formation. They are radioactive. Their magnetic properties cannot be explained easily as they are more complex.

Chemical elements in d-block
Group →	3	4	5	6	7	8	9	10	11	12
↓ Period
4	21 Sc	22 Ti	23 V	24 Cr	25 Mn	26 Fe	27 Co	28 Ni	29 Cu	30 Zn
5	39 Y	40 Zr	41 Nb	42 Mo	43 Tc	44 Ru	45 Rh	46 Pd	47 Ag	48 Cd
6	71 Lu	72 Hf	73 Ta	74 W	75 Re	76 Os	77 Ir	78 Pt	79 Au	80 Hg
7	103 Lr	104 Rf	105 Db	106 Sg	107 Bh	108 Hs	109 Mt	110 Ds	111 Rg	112 Uub

f-block

Period

f1

f2

f3

f4

f5

f6

f7

f8

f9

f10

f11

f12

f13

f14

d1

6

57 La

58 Ce

59 Pr

60 Nd

61 Pm

62 Sm

63 Eu

64 Gd

65 Tb

66 Dy

67 Ho

68 Er

69 Tm

70 Yb

71 Lu

7

89 Ac

90 Th

91 Pa

92 U

93 Np

94 Pu

95 Am

96 Cm

97 Bk

98 Cf

99 Es

100 Fm

101 Md

102 No

103 Lr

VERY SHORT ANSWER TYPE QUESTIONS:   ( 1  Mark each)

VERY SHORT ANSWER QUESTIONS ( 01 MARKS ).

01)  Why are Zn, Cd and Hg not regarded as transition metals ?

        Ans = Because neither these metals nor their ions have incompletely

        filled d- orbitals.

02)    In the transition series, the atomic radius does not increase with increase in

         atomic number. Why ?

         Ans =  As the atomic number increases, the nuclear charge also increases

         which tends to decrease the size but there is simultaneous  increase in the no.

         of d-e^-s which increases the shielding effect. These two opposing effects

         counter balance each other and thus there is not much increase in atomic

         radii with increase in at. no.

03)  Name the transition metal in first transition series that exhibits the highest

       oxidation state.

       Ans = Manganese with +7  oxidation state.

04)  K₂PtCl₆ is a well known compound  while corresponding  Ni  compound  is

       not known. Why ?

       Ans = As the sum of first four IE of Pt is less than that of Ni  :. Pt⁴⁺  ion is

        more  stable than Ni⁴⁺.

05)   Of Fe²⁺ and Fe³⁺,  which is more stable and Why ?

        Ans = Fe²⁺  -   [ Ar ] 3d₆

                   Fe³⁺  -   [ Ar ]  3d₅ .

                   Fe³⁺  is more stable than Fe²⁺ due to its stable configuration.

06)  Why do Zr and Hf  exhibit similar properties ?

       Ans = Due to lanthanoid  contraction.

07)  Why is HCl not used to acidify a permanganate solution in volum etric

       estimations of Fe²⁺ or C2O₄^2-  ?

       Ans =  As HCl will be oxidized to Cl₂ by KMnO₄.

08)  E⁰ (M³⁺ / M²⁺ ) for Sc is low.

Ans = Because Sc³⁺  is stable due to its noble gas configuration.

09)  Why is KMnO₄  kept in dark bottles ?

        Ans = Because it gets decomposed in Sunlight to green coloured K₂MnO₄.

10)   Why does Orange coloured solution of K₂Cr₂O₇ turn yellow on addition of

         NaOH to it ?

        Ans = This is due to formation of potassium chromate which is yellow in

        colour.

                    Cr₂O₇^2-   + 2OH^-                    2CrO₄^-2    +  H2O

                            (Orange)                            (Yellow)

11)  Of Cu⁺ and Cu²⁺, which is more stable and why ?

        Ans = Cu²⁺ is more stable than Cu⁺ as Cu²⁺ has higher hydration energy due

         to smaller size and higher charge. It is for this reason that Cu⁺

        disproportionates  in aqueous solution to form Cu²⁺.

12)    IE₃ of Mn is unexpectedly higher. Why ?

        Ans = This is so because after the loss of 2 electrons Mn²⁺ acquires a stable

         electronic configuration of exactly half  filled d- orbitals and to remove

         further an electron from it requires very high amount of energy.

13)  Write the ionic equation  representing reaction between potassium oxalate and

        KMnO₄  in acidic medium.

        Ans = 2MnO^-₄     +     5 COO^-    + 16H⁺                    2Mn²⁺  +  10CO₂  + 8H₂ .

                                    

                                             COO^- .

14)   Zn²⁺ Salts are white while Cu²⁺ salts are Blue. Why ?

        Ans =   Zn²⁺   :  [ Ar]  3d¹⁰

                    Cu²⁺   :  [ Ar]  3d⁹

        This is so because Zn²⁺ salts do not have  any  unpaired  electrons  while Cu²⁺

         salts have one unpaired electron that undergoes d-d  transition and hence are

         blue in colour.

15)   Name a compound that is isostructural with KMnO₄.

        Ans =  KClO₄.

16)  Why is K₂Cr₂O₇ generally preferred over Na₂CrO₇ in volumetric analysis

        although both are oxidizing agents ?

        Ans =  Because Na₂Cr₂O₇  is  hygroscopic in nature.

17)  In MnO₄^-  ion, all the bonds formed between Mn and  oxygen are covalent.

       Give reasons.

        Ans =  In MnO₄^-, Mn is in  +7 oxidation state Mn cannot lose 7 electrons

        because very high energy is required to remove 7 electrons. Thus it formed

        covalent bonds.

18)   Ce⁴⁺ is a good oxidizing agent. Why?

        Ans= because after gaining an electron, it acquires a stable oxidation

        state of +3.

19)    V₂O₅  acts as a catalyst. Why?

        Ans= Because Vanadium exhibits variable oxidation state.

20)   Transition metals form a number of complex compounds.

  Ans= this is due to the presence of vacant d- orbitals in them as well as due to

  comparative smaller size of metal ions and high ionic charge.

21)   Transition metals readily form alloys. Why?

        Ans= Because they have almost similar radii.

22)   What is Lanthanoid contraction?

  Ans= The steady decreases in the atomic and ionic radii along the

  Lanthanoid   series   is called Lanthanoid contraction.

23)    What are transuranic elements?

  Ans=  the elements beyond _a2U are called transuranic elements. They are

   also called synthetic or man made elements.

24)   Name a Lanthanoid element that exhibits a +4 oxidation state.

        Ans= Terbium.

25)    Actinoid  contraction  is greater from element to element than Lanthanoid

          contraction, Why ?

         Ans = This is due to poor shielding by 5f  electrons.

26)         Actinoid exhibit greater range of  oxidation states than lanthanoid. Why ?

       Ans =   This is due to comparable energies of 5f, 6d and 7s levels.

27)   Calculate magnetic moment of Cr.

        Ans = Cr  : [Ar]4s¹     3d⁵

        No.of unpaired electrons = 6

         µ  =    n  ( n+2)   =     6  ( 6+2)   =     48   =  6.9 BM.

28)    What is a disproportionation rection. Give examples.

         Ans = Disproportionate reaction is a reaction in which the  same species

         undergoes oxidation as well as reduction.       

                   (+6)                                                    (+7)                (+4)

     eg:   3MnO₄^2-     +    4H⁺                           2MnO₄ ^-        +   MnO₂   +  2H₂O.

             Cu⁺(ag)                      Cu²⁺(ag)    +  Cu

29)    Calculate the number of unpaired electrons in a species if its magnetic    moment is 5.5 B.M.

Ans =     µ =  n(n+2)   squaring both the sides, we get

              (5.5) = n (n+2)

             30.25   =   n² + 2n

             n² + 2n  - 30.25  = 0

30)     State the  constituents of alloy ‘mischmetal’.

       Ans =   Lanthanoid metal        ~      95%

                    Iron                             ~        5%

               Traces of   S,  C,  Ca    and  Al.

SHORT ANSWER TYPE  QUESTIONS ( 2 MARKS )

01)           a) Of  Co²⁺,   Ti⁴⁺  and   Cu⁺,  which one will be coloured in aqueous

              solutions?.

            b) If each one of the above ionic species is in turn kept in a magnetic

                field, how will it respond and why ?

Ans  =  a)  Co²⁺  will be  coloured in aqueous solution as only it has unpaired

             electrons while Ti⁴⁺  and Cu⁺  do not contain any unpaired electrons.

            b)  Co²⁺ will be attracted towards magnetic field as it is paramagnetic in

                 nature while Ti⁴⁺  and  Cu  will be repelled by magnetic field as it is

                diamagnetic in nature.

02)    Why is Cr²⁺  reducing while Mn³⁺ oxidizing when both have d⁴

          configuration ?

          Ans =  Cr²⁺ is reducing as it loses an e- to change its configuration from d⁴

          to d³ with half filled t_2g level. Mn³⁺ is oxidizing as by gain of an electron it

         changes to d⁵ configuration which has extra stability.

03)    How  can chromate and dichromate be interconverted ? Give chemical

         equation.

     Ans =  Chromates and dichromates are interconverteble  depending upon the

     pH of the solution, by increasing the pH of the Orange dichromate ions

     dichromate solution, the dichromate ions get converted into yellow coloured

     chromate ions.

             Cr₂O₇^2-  +  2OH^-                     2CrO₄^2-   +  H₂O.

      By decreasing the pH ie by adding acid the yellow coloured solution changes

      to orange colour due to formation of dichromate ions.

                2CrO₄^2-  +  2H⁺                       Cr₂O₇^2-   +  H₂O.

04)  Write chemical equations for the reactions involved in the manufacture of

KmnO₄ from pyrolusite ore.

  

    Ans =  2MnO₂  + 4KOH  +  O₂                      2K₂MnO₄   +  2H₂O

              2K₂MnO₄     +    Cl₂                           2KmnO₄     +  2KCl.  

05)  Write chemical equations for the reactions involved in the manufacture of

        K₂Cr₂O₇ from chromite ore.

     Ans =   4FeCr₂O₄   +  8Na₂CO₃   +  7O₂              8Na₂CrO₄  + 2Fe₂O₃  + 8CO₂.

                 2Na₂CrO₄   +  2H⁺                     Na₂CrO₇    + 2Na+     +  H₂O.

                  Na₂Cr₂O₇   +  2KCl                   K₂Cr₂O₇      +  2NaCl.

06)  Write ionic equation representing oxidizing property of acidified KmnO₄ 

        solution.                                               

        Ans =    MnO₄ ^– + 8H⁺   +  5e^-               Mn²⁺  + 4H₂O.

                 

07)   What is Lanthanoid contraction ? State its two consequences.

       Ans =  The  stready decrease in atomic and ionic  radii along

       Lanthanoid series from lanthanum to lutetium is called lanthonoid

      contraction.

      Consequences :    (i)  Separation of lanthanoids became easier.

(ii) The radii of members of second and third transition

    series was almost similar due to Lanthanoid contraction.

08)  State any three differences between lanthanides and actinides.

       Ans =

Sl.No	Lanthanides	Actinides.
01	They are non-radioactive e4xcept promethium	They are all  radioactive
02	They do not from complexes.	They form complexes
03	They  mainly show +3 oxidation state but a few elements exhibits +2 and +4 also.	They show other higher oxidation states of +4.+5 , +6  and +7 apart from +3 oxidation state.

SHORT ANSWER TYPE QUESTIONS (03 MRAKS)

01) Write ionic equation for conversion of

a)     Manganate to permanganate  ion.

b)    Permanganate  ion to manganese ion.

c)     Chromate ion to dichromate ion.

                                      electrolytic  oxidation

       Ans =    a)   MnO₄^2-                                                     MnO₄^-

                                               in alkaline solution.

                    b)   MnO₄^-    +  8H⁺   +  5e^-                           Mn²⁺   +  4H₂O. 

                    c)   2CrO₄^2-    +  2H⁺                   Cr²O₇^2-    +  H²O.

02)  Write  Chemical equations for the conversion of :s

a)     Chromite ore to sodium chromate.

b)     Pyrolusite to potassium manganate.

c)     Potassion permanganate to manganese oxide.

 Ans =   a) 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂     8Na₂CrO₄ + 2Fe₂O₃  + 8CO₂.

              b)  2MnO₂  +  4KOH  + O₂       K₂MnO₄   +  2H₂O. 

                               _Heat     

 c)  2KmnO₄           K₂MnO₄   +  MnO₂    +  O₂.

03)   Of the ions Co²⁺, Sc³⁺ and Cr ³⁺ which   one will give coloured aqueous

        solution and how  will each of them respond to a magnetic field and why ?.

       Ans =         Co2⁺     :    [Ar]  3d⁷

                          Sc³⁺      :    [Ar]  3d^o

                          Cr ³⁺     :    [Ar]  3d³.  Since Co²⁺ and Cr³⁺ contain unpaired

       electrons they will give coloured aqueous solutions and they will be attracted

        to the magnetic field due to presence of unpaired electrons. Sc³⁺ will be

        colourless and will be repelled by the magnetic field as all its electrons are

        paired.

04)   Why do transition metals form complexes ? What type of bond is present in

         these compounds ?

         Ans =  Transition metals form complexes due to presence of vacant

         d- orbitals  in them which can accommodate the lone pair of electrons

         donated by the ligands. Also  they have higher nuclear charge and smaller

         size. The coordinate bond is present in these compounds.

05)    Complete the following reactions :

        a)     Cr₂O₇^2-   +   14H⁺  + 6e^-                           ______  +  7H₂O.

        b)    2CrO₄^2-   +  2H⁺     ⇆    _____   +  H₂O.

                                                               

                                                                  alkaline

        c)    MnO₄^-    +   2H₂O    +  3e^-                               ____  + 4OH^- .                             

                                                                  Medium

    Ans =    a)      2Cr³⁺                          (b)   Cr₂O₇^2-                     (c)  MnO₂. 

06)  Complete the following reactions:

           a)   Na₂Cr₂O₇  +   KCl                                                  _____

           b)  2MnO₄^-      +  5C₂O₄^2-    + 16H⁺                              _____

           c)   Cr₂O₇^2-      +  14H⁺       +  6Fe²⁺                             _____

    Ans =    a)   K₂Cr₂O₇   +  2NaCl.

                  b)   Mn²⁺       +  10CO₂   +  8H₂O.

                  c)   2Cr³⁺       +  6Fe³⁺     +  7H₂O. 

LONG ANSWER QUESTIONS ( 05 MARKS ).

01) Give reasons.

a)     Cr²⁺  is a strong reducing agent while Mn²⁺ is not.

b)    The transtion metal ions like Cu⁺,  Ag⁺  and Sc³⁺  are colourless.

c)     The radius of Fe²⁺  is less than that of Mn²⁺.

d)    Chemistry of actinoids is much more complicated than that of lanthanoids.

e)     Transition metals have high boiling points.

Ans =  a) E^o value for Cr³⁺/Cr²⁺ is negative ( -0.41V) while it is positive for

       Mn³⁺/Mn²⁺ ( +1.57V). Thus Cr²⁺ can readily undergo oxidation due  to

       low value of  reduction potential  and hence  is a  reducing agent while

       Mn³⁺ can undergo reduction easily and thus is an oxidizing agent.

     b) As  these s  ions do not have any unpaired electrons.

  c) This is so because the no. of d  electron in Fe²⁺ is more than Mn²⁺ and

        they exert poor  shielding effects and thus electrostatic attraction

        increases and ionic radii decreases.

d)     This is so because actinoids can exist in a- number of oxidation states  as well as they are radioactive in nature, which makes their study  complicated.

e)     This is due to presence of strong metallic bonds.

02)  a) The outer electronics configurations of two members of the lanthanoid

            series are as follows:

           4f¹ 5d¹ 6s²  and  4f⁷  5d^o  6s² .  What are their atomic numbers ?  Predict

           the oxidation states exhibited by these elements in their compounds.

           Ans =  The configuration 4f¹ 5d¹ 6s²  has atomic number of 58 and the

           element is cerium. Its oxidation states are    +2,        +3   and    +4.  The

           configuration  4f⁷  5d^o  6s²  has atomic number of 63 and the element is

           Europium. Its common oxidation state are  +2  and  +3. 

     b)  Complete the following reactions :

            i)  CrO₇^2-   + 3Sn²⁺   + 14H⁺                      2Cr³⁺    + 43Sn⁴⁺    +  7H²O.

           ii)  MnO₄^-    +  5Fe²⁺      +  8H⁺                   Mn²⁺    + 5Fe³⁺   + 4H₂O.

        

  (3)    Use if necessary the following electrode potential data to answer questions

           below::

           i)   Cr₂O₇^2-   + 14H⁺   +  6e^-                 2Cr³⁺   +  7H₂O            E^o = + 1.31V.

          ii)    MnO₄^-    + e^-                      MnO₄^2-                                      E^o = + 0.54V

         iii)    CrO₄^2-  +  4H₂O  + 3e^-                 Cr (OH)₃  +  5OH^-         E^o = - 0.12 V

         iv)    MnO₄^-  + 4H⁺     + 3e^-                  MnO₂      +  2H₂O         E^o = + 1.69 V.

         v)     MnO₄^-   + 8H⁺     +  5e^-                  Mn²⁺      +  4H₂O        E^o = 1.52 V

        vi)    ½ Cl₂    +  e^-                    Cl^-                                                E^o = 1.36 V.

a)     In alkaline medium  both chromate and dichromate ions contain Cr in +6  oxidation state. Can both be used as oxidation agents for chlorine ion ? Justify your answer.

b)    The conversion of CrO₄^2- to Cr₂O₄^2-  can be brought about by adding a  mineral acid to the former but MnO₄^2- is usually converted to MnO₄^- by electrochemical process. Explain why.

c)     Aqueous solutions of both K₂Cr₂O₇ and KMnO₄ act as oxidizing agents in acidic medium and are used to carry our redox titrations. H₂SO₄  is generally used to acidify the titrant. On the basis of the redox potential data given above, suggest whether HCl or an organic acid can be used in place of H₂SO₄.

                                                         

 Ans = a) only dichromate ion can oxidize Cl^- to Cl₂ as it has higher value of

                   reduction potential.

             b) This is so because the difference in reduction potential of CrO₄^2- and

                   Cr₂O₇^2- is larger while that between MnO₄^2- and MnO₄^_ is lesser.

            c) HCl cannot be used in place of H₂SO₄ as HCl will be oxidized to Cl₂ by

                 Both KmnO₄ as well as K₂Cr₂O₇ as both are stronger oxidizing agents.

HOTS OF d – f  Block,

01) In quantitative analysis of organic alt, the salt of sodium (A) heated with

        K₂Cr₂O₇ and Concentrate H₂SO₄. The salt gives the red vapours of (B)

        Which is passed through NaOH solution gives a yellow solution (C). The

       Yellow solution is acidified with CH₃COOH and lead acetate is added. A

       Yellow ppt of (D) is formed. Identify A, B, C and D. Also write balanced

        equation for each step.

                                        3H₂SO₄

Ans = 4Nacl + K₂ Cr₂O₇                            2CrO₂ Cl₂ + 2Na₂SO₄ + K₂SO₄ + 3H₂O

                 (A)                                          (B)

                  

          Cro₂ Cl₂ + 4Na₂SO₄                          Na₂CrO₄  + 2Ncl +

                   (B)                                               (C)    

                                                                 CH₃COOH                                                                                                     Na₂CrO₄ + (CH₃ COO) ₂ Pb                           PbCrO₄ + 2CH₃COONa

                                                                                   (D) Yellow ppt.

2)    A serious accident occurred in chemistry lab when a student tried to dissolve

        KMnO₄ in conc. H₂SO₄. Why?

       Ans= due to the formation of Mn₂O₇ which is an explosive compound.

3)    A green chromium compound (A) on fusion with alkali gives a yellow compound

       (B) Which on acidification gives orange compound (C). When (C) is treated with

       NH4Cl, another orange compound (D) is formed, which on strong heating

      decomposes to give compound (A) back. Identify A, B, C, and D. 

    

  Ans= Cr₂O₃ + 2KOH    ----------->    K₂CrO₄ + H₂O.

(A)                                                                                                         (B)                                                  

                     

 2K₂CrO₄ + H₂SO_{4  ------------>}      K₂CrO₇ + K₂SO₄ + H₂O

   (B)        (Conc.)                          (C)

 (Yellow)                                    (Orange)

 K₂Cr₂O₇ + 2NH4Cl--------> (NH₄)₂Cr₂O₇ + 2KCl             

                                                                 (D)

                            

         (NH₄)₂Cr₂O_7----------> CrO₃ + N₂+ 4H₂O

                                                (A)

                                               ^(Green)    

4)    An aqueous blue coloured solution of a transition metal sulphate reacts with H2s   in Acidic medium to give a black ppt (A) which is insoluble in warm aqueous

       solution of KOH. The blue solution on treatment with KI in weakly acidic

       medium turns yellow and produces a white ppt (B). Identify the transition

       metal   ion. Write the chemical rxⁿ involved in the formation of A & B.

   

      Ans=                             H⁺

CuSO₄ + H₂S                      CuS + H₂SO₄

                                                                          (A)

                                           Black

2CuSO₄ + 2KI                                 Cu₂I₂ + K₂SO₄ + I₂

                                                   (B)                        ^(Yellow)

                                              ^(White)