March 17, 2012

d - Block Elements Three Marks

1. Explain why Mn2+ is more stable than Mn3+
Outer Electronic Configurations of
Mn (Z = 25) - 3d54s2
Mn2+ - 3d5 and
Mn3+ - 3d4
Mn2+ has stable half filled d – orbital when compared to Mn3+ which has partially filled d – orbital. Hence Mn2+ is more stable than Mn3+.

Half filled 3d – orbital in Mn2+ have been most stable than Mn3+due to symmetry (Hund’s rule)
Hence Mn2+ is more stable than Mn3+.
2. What is the action zinc on hot NaOH solution?
Zinc dissolves in hot NaOH solution forming soluble zincate ion.
Zn + 2NaOH + 2H2Na2ZnO2 + H2O
                                                                    Sodium zincate
3. Why are transition metal ions coloured?
1. The presence of unpaired electrons in transition metal ions.
2. The energy gap between two energy levels in the same d- subshell being small.
3. The very small amount of energy required for excitation of electrons from one energy level to the other is easily provided by the visible light.
4. The colour observed corresponds to the complementary colour of the light absorbed.
4. Why do d-block elements exhibit variable oxidation states? Or The transition elements show variable oxidation states. Give reason.
1.  These elements have several (n – 1) d and ns electrons.
2.  The energies of (n – 1) d and ns orbitals are fairly close to each other.
5. Why do transition elements form complexes? Or d-block elements form complexes compounds. Why?
1.  Small size and high positive charge density.
2. Presence of vacant (n – 1) d orbitals which are of appropriate energy to accept lone pair and unshared pair of electrons from the ligands for bonding with them.
6.  Write a note on chrome plating. Or How is chrome plating done?
In chrome – plating generally the articles are first plated with nickel and then subjected to chromium plating.
Anode - Lead.
Cathode - The articles to be plated with chromium.
Electrolyte – Chromic acid and Sulphuric acid
During electrolysis chromium deposits on the article (Cathode).
7. Write the action of aqua regia on gold. Or Write the reaction of gold with aqua regia.
Gold dissolves in aquaregia (3 parts of Con.HCl + 1 part of Con.HNO3) to form auric chloride.
                  2Au + 9 HCl + 3HNO3    2AuCl3 + 6H2O + 3NOCl
8. A substance is found to have a magnetic moment of 3.9 BM. How many unpaired electrons does it contain?
μ = √n(n + 2)
3.9 = = √n2 + 2n
Squaring on both sides
15.21 = n2 + 2n
n2 + 2n – 15.21 = 0
Number of unpaired electrons n= 3
9. Explain Chromyl chloride test with equation.
            When salt containing chloride is treated with K2Cr2O7 and Con. H2SO4 reddish brown vapours of chromyl chloride (CrO2Cl2) are obtained.
K2Cr2O7 + 4KCl + 6H2SO4   2CrO2Cl2 + 6 KHSO4 + 3H2O
10. Explain electrolytic refining of copper.
For electrolytic refining of copper,
Anode - Impure copper metal.
Cathode - Pure copper metal.
Electrolyte - Copper sulphate solution acidified with Sulphuric acid.
When electric current is passed through the electrolytic solution pure copper get deposited on the cathode, impurities settle near the anode in the form of sludge called anode mud.
11. Give any two evidences for the oxidising nature of potassium dichromate.
In presence of dil. H2SO4, one mole of K2Cr2O7 gives three atoms of oxygen.
K2Cr2O7 + 4 dil. H2SO4   K2SO4 + Cr2 (SO4)3 + 4H2O+ 3(O)
K2Cr2O7 liberates I2 from KI
K2Cr2O7 + 7H2SO4 + 6KI 4K2SO4 + Cr2 (SO4)3 + 3I2 + 7H2O
K2Cr2O7 oxidises ferrous to ferric salt
K2Cr2O7 + 7H2SO4 + 6 FeSO4 K2SO4 + Cr2 (SO4)3 + 3Fe2 (SO4)3+ 2H2O
K2Cr2O7 oxidises H2S to sulphur
K2Cr2O7 + 4H2SO4 + 3H2S K2SO4 + Cr2 (SO4)3 + 7H2O + 3S
12. How is Purple of Cassius prepared?
Purple of Cassius is prepared by mixing very dilute solution of gold chloride with stannous chloride solution. It is only a form of colloidal gold
2AuCl3 + 3 SnCl2    2Au + 3SnCl4
The gold thus precipitated is adsorbed by stannic hydroxide formed by the hydrolysis of SnCl4.
SnCl4 + 4H2O  Sn(OH)4 + 4HCl
Purple of Cassius is a combination of gold with colloidal stannic hydroxide.
13. What happens when KI solution is added to an aqueous solution of copper sulphate?
                When KI is added to a solution of CuSO4, a white precipitate of Cuprous iodide (Cu2I2) is produced.
CuSO4 + 2KI     CuI2 + K2SO4
2CuI2    Cu2I2 + I2
                                      (White ppt.)
14. What is spitting of silver? How is it prevented?
             Molten silver absorbs about 20 times its volume of oxygen which it again expels on cooling.
      Globules of molten silver are thrown off. This is called “spitting of silver”.
              This can be prevented by covering the molten metal with a layer of charcoal.
15. What is the action of heat on copper sulphate crystals? Write the equation.
               On heating CuSO4.5H2O loses its water of crystallization and decomposes at 720°C to give cupric oxide and sulphur trioxide.
                                    100°C                             230°C                720°C
CuSO4 .5H2O  CuSO4 .H2O CuSO4   SO3 + CuO
(Blue)             – 4H2O                                                                                       – H2O (White)

16. What is the reaction of CuSO4 with KCN?
          A yellow precipitate of cupric cyanide is first formed with KCN and it decomposes to give cyanogen gas.
CuSO4 + 2KCN     Cu(CN)2 + K2SO4
2Cu(CN)2     Cu2(CN)2 + (CN)2
17. Why are Zn2+ salts colourless while Ni2+ salts are coloured?
Outer Electronic Configurations of  Zn (Z = 30) - 3d104s2 and Ni (Z = 28) - 3d84s2
        Zn2+ - 3d10 

and   Ni2+ - 3d8

Zn2+ salts are coloureless because of the absence of vacant d - orbitals to which electrons can be excited.
Ni2+ salts are coloured because of the presence of 2 unpaired electrons in d - orbitals and the energy gap between two energy levels in the same d-subshell being small, can be excited from one energy level to the other by the visible light leading to green colour. 
18. Give the percentage composition and use of Nichrome

% Composition
Cr = 15%,
Ni = 60%,
Fe = 25%
It is used in resistance wires,
for electrical heating
19. What is the action of heat on heating, K2Cr2O7?
On heating, K2Cr2O7 decomposes to give potassium chromate with evolution of O2 gas.
4K2Cr2O7 4K2CrO4 + 2Cr2O3 + 3O2

20. Most of the transition metals and their compounds have catalytic activity. Why
Most of the transition metals and their compounds are used as catalyst. The catalytic activity of transition metals is due to the following reasons.
1. They show a variety of oxidation states and thereby can form intermediate products with various reactants.
2. They are also capable of forming interstitial compounds which can adsorb and activate the reacting species
21. Why do transition elements form alloys?
Transition metals form alloys with each other. This is because they have almost similar size and the atoms of one metal can easily take up positions in the crystal lattice of the other.
Eg. Alloys of Cr - Ni, Cr – Ni - Fe, Cr – V - Fe, Mn - Fe, etc.
22. How is lunar caustic prepared?
Lunar caustic (Silver nitrate) is prepared by dissolving silver in dilute Nitric acid.
3Ag + 4HNO3 → 3AgNO3 + 2H2O + NO↑

23. What is the action of heat on Silver nitrate?
On heating AgNO3 decomposes in two stages
             723 K
2AgNO3     2AgNO2 + O2
            980 K
AgNO2      Ag + NO2
24. What is Philosopher’s wool? How is it formed?
When Zinc is heated in air at 773 K, it burns to form a white cloud of Zinc oxide which settles to form a wooly flock called philosopher’s wool.
     773 K
2Zn + O2 → 2ZnO
25. Write two alloys of copper and their uses
For making utensils, condenser tubes, wires, etc.
For making cooking utensils, statues, coins, etc.
Gun metal
For making gun barrels, gears, castings, etc.
26. What is called Bordeaux mixture? Mention its use
A mixture of Copper sulphate and Lime, commonly known as Bordeaux mixture.
It is used as Fungicide.
27. Give the reaction of ammoniacal silver nitrate with Formic acid
Ammoniacal silver nitrate (Tollen’s reagent) is reduced to Silver mirror by compounds like Formic acid, Formaldehyde or Glucose
2AgNO3 + 2NH4OH → Ag2O + 2NH4NO3 + H2O
Ag2O + HCOOH → 2Ag↓ + H2O + CO2


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