2nd PUC Chemistry Question Bank Chapter 8 The d and f Block Elements

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Karnataka 2nd PUC Chemistry Question Bank Chapter 8 The d and f Block Elements

Question 1.
Define the term transition element.
Answer:
Transition element is defined as the one which has incompletely filled d-orbitals in its ground state or in any one of its oxidation states.

Question 2.
Give the general electronic configurations of ‘d’ block elements.
Answer:
[rare gas] (n – 1)d1-10. ns1-2

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 3.
Why scadium (z = 21) is a transition element but zinc Is not?
Answer:
Scadium has incompletely filled ‘d’ orbitais. Zinc has completely filled d-orbitais.

Question 4.
Write electronic configuration, calculate magnetic moment and predict magnetic properties of some ‘3d’ clement ions.
Answer:
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 1

Question 5.
Write electronic configuration, calculate magnetic moment and predict magnetic properties of some 3d element ions.
Answer:
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 2

Question 6.
Write the formula to calculate the spin only magnetic moment.
Answer:
µ = \(\sqrt{n(n+2)}\) where ‘n’ is no. of unpaired electrons.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 7.
Name the metal of the 1st row transition series that
(a) has maximum no. of unpaired electrons in its ground state.
(b) has zero spin only magnetic moment in its +2 oxidation state.
(c) Exhibits maximum number of oxidation states.
Answer:
(a) Chromium
(b) Zinc
(c) Manganese.

Question 8.
With reference to the first row transition metals.
(a) Name a metal which shows maximum number of oxidation states.
(b) Among Zn2+ and Cu2+ which is colourless.
(c) Between Ti2+ which conatins more number of unpaired electrons.
Answer:
(a) Manganese
(b) Zn2+
(c) V2+

Question 9.
(a) Calculate the spin only magnetic moment of Fe2+.
[Atomic no. of iron 26].
(b) Which element of 3d series exhibit maximum oxidation state.
Answer:
(a)
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 2
(b) Manganese (Mn).

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 10.
Between Cu2+ (aq) and Cu+ (aq) which is more stable?
Answer:
Cu2+(aq)

Question 11.
Transition metals are very hard and have high melting and boiling points. Give reason.
Answer:
Involvement of greater number of electrons from (n – 1)d in addition to the ns electrons in the interatomic metallic bonding makes the metal hard and have high m.p and b.p.

Question 12.
‘d’ block elements shows variable oxidation states. Why.
Answer:
‘d’ block elements shows variable oxidation states, because the energy gap between 4s and 3d subshells is very less and 3d electrons also loses easily in addition 4s electrons.

Question 13.
Name a transition element which does not exhibit variable oxidation states.
Answer:
Scandium does not exhibit variable oxidation states.

Question 14.
Sc3+ ions are colourless where as V3+ ions are coloured. Give reason.
Answer:
Sc3+ ions does not contain any unpaired electrons hence d-d transition does not take place.
V3+ ion contain two unpaired electrons, hence d-d transition takes place. Therefore V3+ ions are coloured.

Question 15.
Why Sc3+ salts are colourless whereas Cr3+ salts are coloured.
Answer:
Sc3+ ions does not contain any unpaired electrons hence d-d transition does not take place.
Cr3+ ions contain 3 unpaired electrons, hence d-d transition take place. Therefore, Cr3+ ions are coloured.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 16.
Why Ti4+ is colourless where as Cr3+ salts are coloured.
Answer:
Sc3+ ions does not contain any unpaired electrons hence d-d transition does not take place.
Cr3+ ions contain 3 unpaired electrons, hence d-d transition take place. Therefore, Cr3+ ions are coloured.

Question 17.
Name the electrode system which have positive E0 value in 3d series. Give reason.
Answer:
Copper electrode system have +ve E0 value. This is because high energy to transform Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy.

Question 18.
Give any two reasons for the formation of large number of complex compounds by transition metals.
Answer:
(a) The transition metals form a large number of complex compounds. This is because :

  • Smaller size of metal ions.
  • High ionic charges.
  • Availability of empty d-orbital for bond formation.

(b) Manganese exhibits highest oxidation states.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 19.
Cu2+ ions are coloured and Zn2+ ions are colourless. Give reason
Answer:
Cu2+ ions contain one unpaired electron in its d-subshell and can undergo d-d transition. Therefore it is coloured.

In Zn2+, there is no unpaired electrons, hence it does not undergo d-d transition. Hence it is colourless.

Question 20.
Transition elements shows catalytic property. Give two reasons.
Answer:
Transition metals and their compounds are known for their catalytic activity. This is because transition metals.

  • Adopt multiple oxidation states.
  • Form intermediate complexes,
  • Lowers the activation energy of the reaction.

Question 21.
What are interstitial compounds?
Answer:
Interstitial compounds are those which are formed by transitional elements, when small atoms like H, C and N are trapped inside the crystal lattices of metals. Example: VH0.5

Question 22.
How is potassium dichromate manufactured from chromite ore?
Answer:
Step – 1: Potassium dichromate is prepared by fusion of chromite ore [FeOCr2O3l with sodium carbonate with excess of air.
4 FeO.Cr2 O3 + 8Na2CO3 + 7O2 → 2Fe2O3 + 8Na2CrO4 + 8CO2

Step – 2: The yellow solution of sodium chromate is filtered and acidified with sulphuric acid. Orange solution of sodium dichromate is obtained.
2Na2CrO4 + 2H+ → Na2Cr2O7 + H2O + 2Na+

Step – 3: The sodium dichromate solution is treated with potassium chloride solution to get potassium dichromare.
Na2Cr2O7 + 2KCI → K2Cr2O7 + 2NaCl
Orange crystals of potassium dichromate crystalises out.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 23.
Show that chromates and dichromates are interconvertible.
Answer:
The chromate and dichromates are interconvertible in aqueous solution depending upon pH of the solution.
At low pH (acidic medium) :
Chromate converts into dichromate
2CrO2-4 + 2H+ → Cr2O72- + H2O
At high pH (alkaline medium) :
dichromate is converted into chromate
Cr2O2-7 + 2OH → 2CrO2O42- + H2O

Question 24.
Give the structure of chromate ion?
Answer:
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 3

Question 25.
Give two examples to show that potassium dichromate is an oxidising agent.
Answer:
(i) Acidified potassium diehrumale exidises iodides lo iodine.
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 4

(ii) Acidified potassium dichromate oxidises Fe2+ (ferrous) to Fe3+ [ferric].
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 5

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 26.
How is potassium permanganate manufactured from MnO2 (pyrolusite)?
Answer:
Potassium permanganate is prepared by fusion of MnO2 with potassium hydroxide in presence of oxidising agent KNO3 to get potassium manganate [K2MnO4]
2 MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O.

potassium manganate on acidification gives potassium permanganate.
3 MnO42- + 4H+ → 2MnO4 + MnO2 + 2H2O.

Question 27.
Name a compound which is isostructural with KMnO4.
Answer:
Potassium perchiorate.

Question 28.
What happens, when. potassium permanganate crystals are heated?
Answer:
When potassium permanganate crystals are heated at 513 K it decomposes to give oxygen gas.
2KMnO4 → K2MnO4 + MnO2 + O2

Question 29.
What is the shape of manganate and permanganate ions? Write structures.
Answer:
Both manganate and permanganates are tetrahedral in structure.
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 6

Question 30.
What is the magnetic nature of manganate and permanganate ions.
Answer:
Mananate ion is paramagnetic in nature with one unpaired electron and permanganate is diamagnetic in nature.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 31.
Give four examples to show that acidified KMnO4 is an oxidising agent.
Answer:
(i) Acidified KMnO4 oxidises Iodide ion to Iodine
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 7

(ii) Acidified KMnO4 Oxidises Fe2+ to Fe3+
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 8

(iii) Acidified KMnO44 oxidises H2S to S
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 9

(iv) Acidified KMnO4 oxidises oxalate ion or oxalic acid to CO2 at 333K
2nd PUC chemistry Question Bank Chapter 8 The ‘d’- and ‘f’- Block Elements 10

Question 32.
What is lanthanoid contraction? Mention the cause for it.
Answer:
Steady decrease in the size of lanthanoids with increase in atomic number is known ns lanthanoid contraction.
Lanthanoid contraction is due- to the imperfect shielding of 4f electrons.

Question 33.
What is lanthanoid contraction? Write the general oxidation state of actinoids.
Answer:
Steady decrease in the size of lanthanoids with increase in atomic number is known ns lanthanoid contraction.
Lanthanoid contraction is due- to the imperfect shielding of 4f electrons.
General oxidatation stale of actinoids is +3.

Question 34.
Give two consequences of lanthanoid contraction.
Answer:
Consequences:

  1. Radii of members of 3rd transition series are very much similar to corresponding members of 2nd series.
  2. Very difficult to separate lanthanoids of 2nd series elements from 3rd series elements in pure state during extraction.

Question 35.
What happens when lanthanoids are heated in
(a) Oxygen
(b) Nitrogen
(c) Halogen
(d) Carbon?
Answer:
(a) 4Ln + 3O2 → 2Ln2 O3 Lanthanoid oxide
(b) 2Ln + N2 → 2 LnN Lanthanoid nitride
(c) 2Ln + 3X2 → 2LnX3 Lanthanoid halide
(d) Ln + 2C → LnC2 Lanthanoid carbide

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 36.
How lanthanoids reacts with water and dil. acids?
Answer:
With water lanthanoids gives hydroxides with the liberation of hydrogen.
6Ln + 6H2O → 2 Ln (OH)3 + 3H2.
With dil acid they liberate hydrogen
Ln + dil acids → H2 gas.

Question 37.
What is mischmetall? Give its one use.
Answer:
Mischmetall is an alloy of lanthanoids with iron (Ln – 95% and Fe 5%).
It is used in bullets.

Question 38.
What, is actinoid contraction? Why actinoid contraction is greater than lanthanoid contraction?
Answer:
Steady decrease in the size of actinoids with increase in atomic number is known as actinoid contraction.
Actinoid contraction is greater from element to element in the series due to poor shielding by 5f electrons compared to 4f electrons.

Question 39.
Give reason: Actinoid contraction is greater from element to element than lanthanoid contraction.
Answer:
Steady decrease in the size of actinoids with increase in atomic number is known as actinoid contraction.
Actinoid contraction is greater from element to element in the series due to poor shielding by 5f electrons compared to 4f electrons.

Question 40.
Give general oxidation state of actinoids. Why actinoids shows variable oxidation states?
Answer:
General oxidation state of actinoids is +3.
Actinoids shows variable oxidation states due to the fact that 5f, 6d and 7s levels are of comparable energies.

Question 41.
Zr and Hf have almost identical atomic radii. Give reason.
Answer:
Poor shielding of 4f electrons decrease. the size of Hf due to lanthanoid contraction. Hence it maintain the size equivalent to Zr.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 42.
Give reasons:
1. Actinoids show variable oxidation states.
2. Zr and Hf have almost identical radii.
Answer:
1. This is due to that 5f, 6d and 7s orbitals have comparable energies.
2. Due to lanthanoid contraction.

Question 43.
Give reason: Why Lanthanoids are less reactive than actinoids.
Answer:
4f electrons are firmly attracted by nuclear charge. Hence 4f electrons are not easily available for bonding compared to 5f electrons of the actinoids.

Question 44.
Out of the following elements identify the element which does not exhibit variable oxidation state Cr, Co, Zn.
Answer:
Zn.

Question 45.
Give reason.
1. Most of the transition metal have high melting point and boiling point.
2. 2nd ionisation enthalpy of Cu is exceptionally high.
3. Atomic size of 4d and 5d series elements are almost the same.
Answer:
1. Electrons of (n – 1)d orbitals along with ns electrons are also involved in metallic bonding.
2. Cu+ ion has completely filled orbitals more energy is needed to remove second electron from stable configuration.
3. Due to lanthanoid contraction.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 46.
Study of actinoid elements is difficult. Give two reasons.
Answer:
Actinoids:

  1. are prepared in nanograms.
  2. have very less half lifes.

Question 47.
Give reasons: Element cerium (Ce) exhibits +4 oxidation state.
Answer:
It has noble gas configuration in +4 oxidation state.

Question 48.
Among Fe2+ and Fe3+ which is more stable? Give reason.
Answer:
Fe3+ is more stable. Fe3+ has stable half filled configuration.

Question 49.
Give reason:
Mn exhibits the higher oxidation state of +7 among 3d transition elements.
(b) Cu2+ Isparamampøtic and Cu+ is diamagnetic
Answer:
(a) Due to the participation of all the electrons of 3d and 4s orbitais. in bonding.
(b) Cu+ has no unpaired electrons and Cu2+ has one unpaired electron in 3rd orbitais.

2nd PUC Chemistry Question Bank Chapter 8 The 'd' and 'f' Block Elements

Question 50.
Give reason “transition metals generally form coloured compounds”.
Answer:
When a visible light falls on the metal ion, the unpaired electron of d subshell jumped from lower energy d-orbitals to higher energy d-orbitals by absorbing light of particular wavelength. Therefore the unabsorbed light is transmitted as complimentary colour.