ELECTROPLATING AND IT’S ADVANTAGES

 

ELECTROPLATING AND IT’S ADVANTAGES

One day Shruti went to the market with her mother for shopping. Her mother wanted to buy some daily articles for their home, so they went to a nearby supermarket. As they arrived to the market, Shruti noticed that some food items and cold drinks were packed inside the metallic utensils. Shruti remembered that her teacher told her about rusting that metals like iron forms rust when they kept in open or comes in contact of moisture. Shruti got confused that why the cans and utensils are not getting rusted and how can we eat or drink these food items while these are stored inside the utensils for a long period. The next day she went to her school and asked her teacher so the teacher told her about the process of “ELECTROPLATING”. The teacher said, “Electroplating is the process of plating one metal over another. It is done for various purposes, most commonly for imparting corrosion resistance and decorative appearance. There are a number of advantages of this process”.

Ques. 14.1 Cans and utensils not getting rusted after keeping in contact with liquid items like cold drinks.

What will happen if these food items will be packed inside metal boxes without any process, in their original form?

A. The food item will remain unchanged.

B. The food item will become long lasting

C. The food item will become spoiled.

D. The food item will become costly.

Ques. 14.2 Can we apply electroplating technique to the following things? Circle ‘Yes’ or ‘No’ for each case.

A copper necklace	Yes/No
A plastic bucket	Yes/No
A Notebook	Yes/No
An iron bucket	Yes/No

 

Ques. 14.3 Can you explain more such examples of daily usage items surrounding you, which are metallic but not getting rusted after in contact of water and air?

……………………………………………………………………………………………………………………………..

……………………………………………………………………………………………………………………………..

Ques. 14.4 How much do you agree with the following statements? Tick only one in each row.

	Agree	Disagree
Electroplating enhances durability of metals
Electroplating can be done anytime & anywhere
Electroplating is beneficial for food industry
Electroplating doesn’t have any disadvantages

 

Answer 14.1 Option C. The food item will become spoiled.

Answer 14.2

A copper necklace	Yes
A plastic bucket	No
A Notebook	No
An iron bucket	Yes

 

Answer 14.3

Bathroom taps, car parts, metallic utensils, metallic buckets, jewellery, handles of doors etc.

Answer 14.4

	Agree	Disagree
Electroplating enhances durability of metals	☑
Electroplating can be done anytime & anywhere		☑
Electroplating is beneficial for food industry	☑
Electroplating doesn’t have any disadvantages		☑

REFRACTION OF LIGHT

Sameer performs an experiment with the help of clear round drinking glass, pencil and tap water. He fills the drinking glass about two-thirds of the way with water. Then he place the pencil inside the glass on an angle, so it’s resting on the rim.

 

                           

He then bends down until he is looking in line with the top of the water. He observes that the pencil is broken in two.

 

Question 1- A ray of light is travelling from a rarer medium to a denser medium. What will happen?

     …………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

 

 

Question 2- A light ray enters from medium A to medium B as shown in figure. The refractive index of medium B relative to A

 

(a)   greater than unity

(b)   less than unity

(c)   equal to unity
(d)   zero

Question 3- The optical phenomena, twinkling of stars, is due to

(a)   Atmospheric reflection

(b)  Total reflection

(c)   Atmospheric refraction

(d)   Total refraction

 

Question 4- Which one is optically denser in between Water and air?

 

(a)  Water

(b) Air

(c)  Both are same

(d) None of these

MARKING SCHEME:

1-    When a ray of light is travelling from a rarer medium to a denser medium, it bends towards the normal.
Correct answer - FULL CREDIT                                    (1 marks)

      Any other response- NO CREDIT

2-    (a)- FULL CREDIT, Any other response- NO CREDIT              (1 marks)

3-    (c)- FULL CREDIT, Any other response- NO CREDIT               (1 marks)

4-    (a)- FULL CREDIT, Any other response- NO CREDIT               (1 marks)

 

  

d-and f- block elements -Important key points & question -answer

• The transition elements may be defined as the elements whose atoms or simple ions contain at least one partially filled  d –orbitals. 

• Properties of these elements that are transitional between s & p block elements, there fore they are called as transition elements.

• General electronic configuration of these elements as ns^1-2 (n-1)d^1-10 .

• Lanthanides and actinides are called inner transition elements.  Their general electronic configuration is ns²(n-1)d^0-1 (n-2)f^1-14

• Transition metals are having strong metallic bonding due to presence of large no. of valence electrons and greater nuclear charge.  Hence these are hard, possess high densities and high enthalpy of atomization.
• Transition elements are having high melting and boiling points due to presence of large no. of half filled d – orbitals which causes strong inter particle attractions.

• Ionization energy increases from left to right due to increase in the nuclear charge, as the increased nuclear charge is partly cancelled by screening effect there is no much increase.

• The reduction potential values of transition elements varies irregularly because it depend up on Enthalpy sublimation, Ionisation energy and hydration energy.  All these values are irregularly changing hence their reduction potential values are also varies irregularly.

• Atomic and ionic radii decreases to middle and become constant in middle and at the end it increases.  The variation is not much because increased nuclear charge is partly cancelled by screening effect.

• Transition elements possess variable oxidation states(o.s.) due to involvement of both ns electrons and (n-1)d electrons in bonding.   Lower o.s is zero and highest is +8, +2 is the most common o.s.

• Transition metals forms complex compounds due to small size, high charge density and presence of empty d- orbitals.

• Transition elements forms colored complexes due to d – d transition.

• The transition metals ions generally contain one or more unpaired electrons in them and hence their complexes are generally paramagnetic.  If a metal ion does not have at least one electron it behave as diamagnetic.

• Transition metal ions and their compounds known to act as catalysts.  The catalytic activity of compounds is due to variable oxidation states, and providing surface for adsorption.

• Transition metals are almost similar size, there fore these elements can mutually substitute their positions in their crystal lattices and forms alloys.

• Transition elements are capable of entrapping smaller atoms of other elements such as H, C & N in the interstitial position and the trapped atoms get bonded with transition elements.  And these compounds are known as interstitial compounds.  Example Steel.

• Preparation of potassium dichromate from chromite involves the following steps( in current syllabus it has been removed) 

      The chromite ore is finely ground and heated strongly with  molten alkali in  the  presence of air.

       2FeCr₂O₄ + 8NaOH + 7 ½ O₂ à 4Na₂CrO₄ + Fe₂O₃ + 4H₂O

       The solution of sodium chromate is filtered and acidified with dilute sulphuric acid so that sodiumdichromate is  obtained.

       2NaCrO₄ + H₂SO₄ à Na₂Cr₂O₇ + Na₂SO₄ + H₂O

       A calculated quatity of potassium chloride is added to a hot  oncentrated

       solution of sodium dichromate. Potassium dichromate is less   soluble therefore  it crystallizes out first.

       Na₂Cr₂O₇+ 2KCl à K₂Cr₂O₇ + 2NaCl

§     In alkaline solution dichromate which is orange in colour is converted to chromate which is yellow. The acidification reverses the reaction.( in current syllabus it has been removed) 

2CrO₄ ^2-   + 2H⁺ à  2HCrO4^- à Cr₂O₇ ^2-    +  H₂O        (reversible)

 a)It oxidizes potassium iodide to iodine

  Cr₂O₇ ^2-    + 14H⁺ + 6I^-   à 2Cr³⁺  +  7H₂O  + 3I₂

  b)It oxidizes iron (ii) solution to iron(III) solution

  Cr₂O₇ ^2-    + 14H⁺ + 6Fe²⁺  à 2Cr³⁺  +  7H₂O  + 6Fe³⁺

   c) It oxidizes H₂S to S

     Cr₂O₇ ^2-    + 8H⁺ + 3H₂S   à 2Cr³⁺  +  7H₂O  + 3S

 §     Pyrolusite ore is fused with alkali in the presence of air when potassium manganate is formed.( in current syllabus it has been removed) 

2MnO₂ + 4KOH + O₂ à 2K₂MnO₄  + 2H₂O

Potassium manganate is oxidized by using either CO₂, ozone or chlorine to potassium permanganate.

2K₂MnO₄ + Cl₂ à 2KMnO₄ + 2KCl

Potassium permanganate is crystallized from the solution

a)     It oxidizes iron(II)salts to iron(III) salts.

MnO^4-  + 8H⁺    + 5Fe²⁺   àMn²⁺   + 4H₂O  + 5Fe³⁺

b)It oxidizes SO₂ to sulphuric acid

2MnO₄^-  + 5SO₂ + 2H₂Oà 5SO₄ ^2- + 2Mn²⁺  +    4H⁺

b)    It oxidizes oxalic acid to CO₂ and H₂O

2MnO₄^-  + 16H+  + 5C₂O₄^2-  à 2Mn²⁺  +  8H₂O  + 10CO₂

§  Elements which have partly filled f-orbitals. There are two series of inner transition elements. In the first series of inner transition elements the 4f orbitals are incomplete and electrons are progressively filled in these orbitals as atomic number increases. These elements are called lanthanoids. In the second series of transition elements the electrons  are progressively filled in 5f- orbitals as atomic no. increases. These elements are called actinoids.

§  As we move from La to Lu there is a gradual decrease in size .The steady decrease in size from La to Lu is called lanthanide contraction. This is due to poor shielding effect of    4f  orbitals.

§  The important consequences are

1.     Similarities in the properties of Y and heavier lanthanides .

2.     Similar atomic radii of second and third transition series.

3.     Separation becomes difficult

4.   Causes small differences in the properties like basicity,  solubility of salts, formation of complexes, etc.

§  Colour of the salts and ions in solution Most of the lanthanide trivalent ions are coloured in solid as well as in the solution phase. The ions containing x and (14 – x) electrons show the same colour. The colour of the salts or ions is due to the f – f transition of electrons.

 REASONING QUESTIONs

1. Give two use of lanthanide compounds.

Ans. (i) Misch metal is pyrophoric and used in gas lightness, tracer bullets and shell.

         (ii) Oxides of neodymium and praseodymium are used for making colour glasses.

2. why the melting points of transition elements are high ?

Ans. The melting points of transition elements are high due to the presence of strong   Intermetallic bonds and covalent bonds.

3. Why Zn, Cd and Hg are not regarded as transition elements ?

Ans. Because they have the completely filled d-subshell with outer electronic  configuration (n –1)d¹⁰ ns².

4. Why is K₂Cr₂O₇ generally preferred over Na₂Cr₂O₇ in volumetric analysis   although both are oxidising agents ?

Ans. Because Na₂Cr₂O₇ is hygroscopic, hence it is difficult to prepare its standard  solution for volumetric analysis, but because of non hygroscopic nature of K₂Cr₂O₇ its standard solution can be prepared.

 5. Write any two uses of pyrophoric alloys.

Ans. Pyrophoric alloys contain rare earth metals and are used in the preparation  of ignition and shells and flints for lightness.

  6. Why do Zr and Hf exhibit similar properties ?

Ans. Because of the lanthanide contraction Hf has similar size to Zr, therefore both  Zr and Hf exhibit similar properties.

 7. In the transition servies, with an increase in atomic number      the atomic radius   does not change very much, why is it so ?

    Ans. In the transition series, the effect of increasing nuclear     change is partly   cancelled by the increased screening effect of     the d- electrons of penultimate   shell. Because of this reason, the     atomic radius does not change very much in the   transition series.

8. Why is the third ionization energy of manganese unexpectedly      high ?

Ans. Third electron of the manganese is removed from 3d-orbitals     which have half  filled configuration, thus have extra stability.     Due to this reason high enegy is  required to remove third electron   from Mn.

9. All scandium salts are white. Why ?

Ans. Because they have no electron in d-orbital, thus no d - d      transition is possible.Due to this reason all salts of Sc³⁺ are white.

10. The first ionisations energies of the 5d-transition elements are higher then   those of 3d and 4 d transition elements.  Why?

Ans:- Due to  Lanthanoid contraction

11. Why do the d-block elements exhibit a longer number of oxidation states than  f-block elements ?

Ans. Because the energy of ns-electron and (n – 1) d-electrons are nearly same, therefore, ns electrons as well as (n – 1) d- electrons can take part in bond formation in transition elements.In fblock elements last electron goes to the   f-orbitals of second order outer most shell, thus the difference between the energy of ns-electron and (n – 2) f-electrons increases. Due to this reasons   all the (n – 2) f-electrons cannot take point in bond formation.

12. Explain why the first ionisation energies of the elements of  the first transition series do not vary much with increasing  atomic numbers.

Ans. With the increasing atomic number, d-electrons add one by      one in (n – 1) shell   or penultimate shell. The screening effect of these d- electrons shield the outer  s-electrons from inward nuclear pull. The effect of the increase in nuclear charge with the increase in atomic number is opposed by the shielding effect of   d-electrons . thus due to these counter effect there is a very little variations in the values of ionization energies of first transition series.

13. Explain why transition metals are paramagenetic ?

 Ans. The ions of transition metals generally contain one or more      unpaired electrons  hence the compounds of transition metal are      paramagnetic i.e., they are weakly  attracted by magnetic field.The paramagnetic character is directly related to the  value of       magnetic moment, m which intern depends upon the number of unpaired electrons (n) i.e., B.M. = Ö n + (n + 2)

 14. Give plausible reason for the fact that transition metals have       high enthalpy of atomization.?

Ans. Transition elements are metallic in nature and form strong       metallic bonds. Moreever, they have incomplete d-orbitals,       hencey they can form covalent bonds also. Due to these two       reasons transition metals have strong force of attraction.      Therefore, they have high enthalpy of atomization.

15. What is the effect of pH on the color of the solution of  potassium dichromate ?

 Ans. A lower pH, the colour of the solution is orange due to the       solution is orange due To the presence of dichromate ions       (Cr₂O₇ ^2–). But in alkaline P^H, the colour of  the solution changes        to yellow due to the conversion of dichromate ions to  chromate ions.

2CrO₄ ^2-   + 2H⁺ ↔  2HCrO4^- ↔ Cr₂O₇ ^2-    +  H₂O     (reversible)

16. Why the compounds of transitions elements are coloured ?

Ans. The colour of compounds of transition elements depend upon the unpaired  electrons present in d-orbitals fo transition element. If d-orbitals are completely  vacant or completely filled the compounds will be colourless, but if any  unpaired electron      is present in d-orbital, the compound will be coloured due to

d - d transition. The unpaired electron is excited from one energy level to another energy level with in the same d-sub- shell. For this purpose, the energy  is absorbed from visible region of radiation. The complementary part of the absorbed light i.e., reflected light will decide the colour of the  compound.

17. Why the transition elements act as catalyst? Give two  examples.?

Ans. (a) Transition metal show variable oxidation states, therefore, they can from intermediate products of difficult  reactant molecules.

(b) Transition elements are capable to form interstitial compounds due to which they can absorb and activates the  reacting molecules.

Example:

(a) V₂O₅ is used for the oxidation of SO₂ to SO₃ in contact process of H₂SO₄.

(b) Ni is used as a catalyst in the hydrogenation of alkanes and alkynes

18. Why transition elements form

(a) interstitial compounds and       

 (b) Alloys?

Ans. (a) Interstitial compounds: Transition elements form large number of interstitial compounds.

In these compounds small size atoms like hydrogen, carbon, nitrogen, nitrogen, boron etc. occupy the empty space of metal lattice. The small entrapped atom in the interstices forms the bonds with metals due to which malleability and ductility of the metals decrease, whereas tensile strength increases.

 (b) Alloys: Transition element forms alloys with each other                   

because they have almost similar sizes. Due to similar sizes atoms of one metal in the crystal lattice can easily take up the position of the atom of transition elements. Alloys are more resistant to corrosion than the

constituent elements, and usually harder with higher melting point.

19. Why are Ni²⁺ compounds thermodynamically more stable than Pt²⁺ compounds, whilst Pt⁴⁺ compounds are relatively more stable than Ni⁴⁺ compounds?

Ans. Thermodynamic stability of the compounds can be judged on the basis of the magnitude of ionization energies. The sum of the two first ionization energies of   Ni²⁺ is lower than the sum of first two ionization energies of Pt^2+, therefore, Ni²⁺

 compounds are more stable than Pt²⁺. On the other hand, sum of first four ionization energies of Pt⁴⁺ lower than the sum of the first four ionization energies of Ni^4+, therefore, Pt⁴⁺ compounds are more stable than Ni^4+.

20. Name a transition metal which does not exhibit variation in   oxidation state in its compounds?

Ans. Zinc in its compounds shows an oxidation state of + 2 only.

21.  Why is + 2 oxidation state of manganese (Z = 25) more stable than its + 3 oxidation state, while the same is not true for iron (Z = 26)?

Ans. The electronic configurations of Mn^2+, Mn³⁺, Fe²⁺ and   Fe³⁺ are,

Mn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d ⁵ 4s⁰

Mn³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d ⁴

Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d ⁶ 4s⁰

 Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d ⁵

Due to the extra stability of the half-filled orbitals d ⁵ configuration is more stable than d 4 or d 6. Mn²⁺ and Fe³⁺ both have d ⁵ configuration. Therefore, Mn²⁺ is more stable than Mn3+ and Fe³⁺ is more stable than Fe²⁺. 

22. Of the ions Co²⁺, Sc³⁺ and Cr³⁺ which ones will give coloured aqueous  solutions and how will each of them respond to a       magnetic field and why?

  (Atomic numbers: Co = 27, Sc = 21, Cr = 24)

Ans. The electronic configurations of the given ions are,

Co²⁺ : 1s2 2s2 2p6 3s2 3p6 3d7 No. of unpaired electrons = 3

Sc³⁺ : 1s2 2s2 2p6 3s2 3p6 3d0 No. of unpaired electrons = 0

Cr³⁺ : 1s2 2s2 2p6 3s2 3p6 3d3 No. of unpaired electrons = 3

Co²⁺ and Cr3+ ions will give coloured aqueous solutions. Co²⁺  and Cr³⁺ are paramagnetic, and Sc³⁺ is diamagnetic. Therefore, Co²⁺ and Cr³⁺ ions will get attracted to the magnetic field, whereas Sc³⁺ ion will be repelled by the magnetic field.

23. Why is it that anhydrous copper (II) chloride is a covalent while anhydrous copper (II) fluoride is ionic in nature?

Ans. This is because in halides of transition metals, the ionic  character decreases with      increase in atomic mass of the halogen. Fluorine being the most electronegative, forms ionic        salt with copper. Cupric chloride consists of infinite chains consisting of bridging Cl atoms and has a covalent character.

24. (a) Assign reason for each of the following:

(i) Ce³⁺ can be easily oxidised to Ce^4+.

(ii) E° for Mn^3+/Mn²⁺ couple is more positive than for                   Fe³⁺/Fe²⁺.

(iii) Transition metals exhibit higher enthalpies of                      atomization.

(iv) The transition elements form interstitial compounds.

(b) Mention two uses of potassium permanganate in the  laboratory.  (Atomic number: Mn = 25, Fe = 26, Ce = 58)

Ans. (i) Ce³⁺ has only one electron in its 4f orbitals. Due to extra stability of  completely  empty orbitals belonging to an energy level as compared to having only one   electron in it, Ce³⁺ tends to lose its only electron from  4f orbital and get oxidised to  Ce⁴⁺.

(ii) Mn³⁺ has a 4d  configurtion, so it has greater tendency to accept one  electron to acquire d ⁵ configuration. On the other hand, Fe³⁺ has a d⁵   configuration which is more stable than the6d`6 configuration of Fe²⁺. As a result, reduction of Fe³⁺ to Fe²⁺ is not favoured. Since,  E° values reflect the reduction tendency, therefore, E° value for   Mn³⁺/Mn²⁺ couple is more positive than Fe³⁺/Fe²⁺.

(iii) Transition metals exhibit high enthalpies of atomisation. This is because  the atoms in these elements are held together by strong metallic bonds. The metallic bond is formed as a result of interaction of electrons in the  outermost shells. In general, greater the number of valence electrons  stronger is the metallic bond.

(iv) The transition metals form a number of interstitial compounds in which    small atoms of light elements such  as,  H, C and N occupy the voids in their   lattices. The products   obtained in this way are hard and rigid. For example, steel   and cast iron become hard due to the  formation of an   interstitial compound with  carbon.

(b) The uses of potassium permanganate in the laboratory are

(i) As an oxidising agent,

 (ii) In volumetric estimation of reducing agents such as Fe2+ salts,   oxalic acid etc.

25. Name a transition element which does not exhibit variable  oxidation state?

Ans:- Scandium ( Z = 21) does not exhibit variable oxidation state.

26. Why is Cr⁺²reducing and Mn⁺³ oxidising when both have d⁴  configuration?

Ans:- Cr⁺² is reducing as its configuration changes from d⁴ to d⁵, the latter having a half filled t_2g level. On the other hand the change from Mn⁺² to Mn⁺³ results in the half – filled (d⁵) configuration which has extra stability. 

27. How would you account for the increasing oxidizing power  in  the series VO₂⁺  < Cr₂O₇^2-‑  <  MnO₄^-?

Ans:- This is due to the increasing stability of the lower species  to which they are reduced.

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

Ans:- Irregular variation of ionization enthalpies is mainly       attributed to varying degree of  stability of different 3d   configurations (e.g. d⁰,  d⁵,  d¹⁰ are exceptionally stable.

 29. In the series Sc(Z=21) to Zn (Z=30) the enthalpy of  atomization of zinc is the lowest i.e. 126kJ/mol, why?

Ans:- In the formation of metallic bonds, no electrons from 3d-  orbitals are involved in case of   zinc, while in all other metals        of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds.

30. Which is a stronger reducing agent Cr⁺² or Fe⁺²?

      Ans:- Cr⁺² is stronger reducing agent than Fe⁺² Because d⁴ à d³       occurs in case of Cr⁺² to Cr⁺³   but d⁶ à d⁵ occurs in   case of       Fe⁺² to Fe³⁺. In a medium (like water) d³ is more stable  as  compared to d⁵.

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

 Ans:- Because of small size and high electronegativity oxygen or fluorine can oxidize the metal  to its highest oxidation state.

32. What is meant by disproportionation of an oxidation state?  Give an example?      Ans:- When a particular oxidation state becomes less stable relative to other oxidation states,  one lower, one higher, it is said to undergo disportionation. For example, manganese(VI) becomes unstable relative to Manganese(VII) and manganese (IV) in acidic solution.

  Mn^VIO₄^2-    +   4H⁺    à    2Mn^VIIO₄^-   + Mn^IVO₂   +   2H₂O

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

 Ans:-   Cu⁺ in aqueous solution undergoes disproportionation   i.e.,

 2Cu⁺(aq.)  à Cu⁺²(aq.)   +  Cu(s), The E⁰ value for this is favourable.

34. Name a member of the lanthanoid series which is well known  to exhibit +4 oxidation state?

Ans:-   Cerium (Z = 58)

35. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans:- The 5f electrons are more effectively  shielded from nuclear charge, In other words the 5f electrons themselves  provide poor shielding from element to element in the series.