Solutions-Class-12-Chemistry-Chapter-8-Transition and Inner transition Elements-Maharashtra Board

Transition and Inner transition Elements

Maharashtra Board-Class-12-Chemistry-Chapter-8

Solutions

Question 1. Choose the most correct option.

(i) Which one of the following is dimagnetic

(a) Cr2+

(b) Fe3+

(c) Cu2+

(d) Sc3+

Answer :

(d) Sc3+

(ii) Most stable oxidation state of Titanium is

(a) +2

(b) +3

(c) +4

(d) +5

Answer :

(c) +4

(iii) Components of Nichrome alloy are are

(a) Ni, Cr, Fe

(b) Ni, Cr, Fe, C

(c) Ni, Cr

(d) Cu, Fe

Answer :

(c) Ni, Cr

(iv) Most stable oxidation state of Ruthenium is

(a) +2

(b) +4

(c) +8

(d) +6

Answer :

(b) +4

(v) Stable oxidation states for chromium are

(a) +2, +3

(b) +3, +4

(c) +4, +5

(d) +3, +6

Answer :

(d) +3, +6

(vi) Electronic configuration of Cu and Cu+1

(a) 3d10, 4s0; 3d9, 4s0

(b) 3d9, 4s1; 3d94s0

(c) 3d10, 4s1; 3d10, 4s0

(d) 3d8, 4s1; 3d10, 4s0

Answer :

(c) 3d10, 4s1; 3d10, 4s0

(vii) Which of the following have d0s0 configuration

(a) Sc3+

(b) Ti4+

(c) V5+

(d) all of the above

Answer :

(d) all of the above

(viii) Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is

(a) 2

(b) 3

(c) 4

(d) 5

Answer :

(d) 5

(ix) In which of the following series all the elements are radioactive in nature

(a) Lanthanides

(b) Actinides

(c) d-block elements

(d) s-block elements

Answer :

(b) Actinides

(x) Which of the following sets of ions contain only paramagnetic ions

(a) Sm3+, Ho3+, Lu3+

(b) La3+, Ce3+, Sm3+

(c) La3+, Eu3+, Gd3+

(d) Ce3+, Eu3+, Yb3+

Answer :

(d) Ce3+, Eu3+, Yb3+

(xi) Which actinoid, other than uranium, occur in significant amount naturally?

(a) Thorium

(b) Actinium

(c) Protactinium

(d) Plutonium

Answer :

(a) Thorium

(xii) The flux added during extraction of Iron from teamatite are its?

(a) Silica

(b) Calcium carbonate

(c) Sodium carbonate

(d) Alumina

Answer :

(b) Calcium carbonate

Question 2. Answer the following

(i) What is the oxidation state of Manganese in (i) MnO42- (ii) MnO4- ?

Answer :

Oxidation state of Manganese in

(i) MnO42- is +6 (ii) MnO4- is +7

(ii) Give uses of KMnO4

Answer :

Potassium permanganate is used :

  • as an antiseptic.
  • as a powerful oxidising agent in laboratory and industry.
  • in detection of unsaturation in organic compounds in laboratory. (Baeyer's reagent, alkaline KMnO4).
  • for detecting halides in qualitative analysis.
  • in volumetric analysis for the estimation of H2O2, FeSO4

(iii) Why salts of Sc3+, Ti4+, V5+ are colourless ?

Answer :

(a) Sc3+ salts are colourless :

  • The electronic configuration of 21Sc [Ar] 3d14s2 and Sc3+ [Ar] d°.
  • Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
  • Therefore, Sc3+ ions do not absorb the radiations in the visible region. Hence salts of Sc3+ are colourless (or white).

(b) Ti4+ salts are colourless :

  • The electronic configuration of 22Ti [Ar] 3d24s2 and Ti4+ : [Ar] d°
  • Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
  • Therefore, Ti4+ ions do not absorb the radiation in visible region. Hence salts of T4+ are colourless.

(c) V5+ salts are colourless :

  • The electronic configuration of 23V : [Ar] 3d34s2 and V5+ : [Ar] 3d°
  • Since there are no unpaired electrons in 3d-subshell, d → d transition is not possible.
  • Therefore, V5+ ions do not absorb the radiations in the visible region. Hence, V5+ salts are colourless.

(iv) Which steps are involved in manufacture of potassium dichromate from chromite ore ?

Answer :

Steps in manufacture of potassium dichromite from chromite ore are :

  • Concentration of chromite ore.
  • Conversion of chromite ore into sodium chromate (Na2CrO4).
  • Conversion of Na2CrO4 into sodium dichromate (Na2Cr2O7).
  • Conversion of Na2Cr2O7 into K2Cr2O7.

(v) Balance the following equation

(i) KMnO4 + H2C2O4 + H2SO4 MnSO4 + K2SO4 + H2O + O2

Answer :

2KMnO4 + 5H2C2O4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O

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

Answer :

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

(vi) What are the stable oxidation states of plutonium, cerium, manganese, Europium ?

Answer :

Plutonium : +3 to +7

Cerium : +3, +4

Manganese : +2, +4, +6, +7

Europium : +2, 34

(vii) Write probable electronic configuration of chromium and copper.

Answer :

(a) Chromium (24Cr) has electronic configuration,

24Cr (Expected) : 1s22s22p63s23p63d44s2

(Observed) : 1s22s22p63s23p63d54s1

Explanation :

  • The energy difference between 3d- and 4s-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are half-filled (3d).
  • Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d54s1 and not [Ar] 3d44s2.

(b) Copper (29Cu) has electronic configuration,

29Cu (Expected) : 1s22s22p63s23p63d94s2

(Observed) : 1s22s22p63s23p63d104s1

Explanation :

  • The energy difference between 3d- and 4s-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are completely filled.
  • Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.
  • Hence, the configuration of Cu is [Ar] 3d104s1 and not [Ar] 3d94s2.

(viii) Why nobelium is the only actinoid with +2 oxidation state?

Answer :

Nobelium has the electronic configuration

102No : [Rn] 5f146d07s2

No2+ : [Rn] 5f146d0

Since the 4f subshell is completely filled and 6d0 is empty, +2 oxidation state is the stable oxidation state.

Other actinoids in +2 oxidation state are not as stable due to incomplete 4f subshell.

(ix) Explain with the help of balanced chemical equation, why the solution of Ce(IV) is acidic.

Answer :

Ce4 undergoes hydrolysis as, Ce4+ + 2H2O -> Ce(OH)4 + 4H+.

Due to the presence of H+ in the solution, the solution is acidic.

Hence, the solution of Ce(IV) is acidic in nature.

(x) What is meant by ‘shielding of electrons’ in an atom?

Answer :

  • The shielding effect is a mechanism where the inner shell electrons in an atom screen protect the outermost electron from nuclear attraction,
  • The magnitude of shielding effect depends upon the number of inner electrons.

(xi) The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?

Answer :

The 90th element thorium has electronic configuration, [Rn] 6d27s2. Since it has 2 unpaired electrons it is paramagnetic.

Question 3. Answer the following

(i) Explain the trends in atomic radii of d block elements.

Answer :

  • Atomic radii of the elements of the transition series decrease gradually from left to right.
  • As we move across a transition series from left to right, the nuclear charge increases by one unit at a time.
  • The last filled electron enters a penultimate (n - 1)d subshell. However, d orbitals in an atom are less penetrating or more diffused and, therefore d electrons offer smaller screening effects.
  • The result is that effective nuclear charge also increases as the atomic number increases along with a transition series.
  • Hence, the atomic radii gradually decrease across a transition series from left to right.
  • The decrease in atomic radii is less. There is slight increase in atomic radii of last two elements Cu and Zn.

(ii) Name different zones in the Blast furnace. Write the reactions taking place in them.

Answer :

There are three zones of temperature in the Blast furnace in which three main chemical reactions take place.

(a) Zone of combustion : This is 5 - 10 m from the bottom. The hot air blown through the tuyers, oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.

C + \(\frac{1}{2}\)O2 → CO ΔH = —220 kJ

Some part of CO dissociates to give finely divided carbon and O2.

2CO → 2C + O2

The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(b) Zone of reduction (22-25 m near the top) : At about 900 K, CO reduces Fe2O3 to spongy (or porous) iron.

Fe2O3 + 3CO → 2Fe + 3CO2

Carbon also reduces partially Fe2O3 to Fe.

Fe2O3 + 3C → 2Fe + 3CO

 (c) Zone of slag formation (20 m unit) : At 1200 K limestone, CaCO3 in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO2, Al2O3) and forms a slag of CaSiO3 and Ca3AlO3.

CaCO3 → CaO + CO2

CaO + SiO2 → CaSiO3

12CaO + 2Al2O3 → 4Ca3AlO3 + 3O2

The slag is removed from the bottom of the furnace through an outlet.

(d) Zone of fusion (15 m ht): The impurities in ore like MnO2 and Ca3(PO4)2 are reduced to Mn and P while SiO2 is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4 % carbon.

(iii) What are the differences between cast iron, wrought iron and steel.

Answer :

Cast iron Wrought iron Steel
Hard and brittle Very soft Neither too hard nor too soft.
Contains 4% carbon. Contains less than 0.2% carbon. Contains 0.2 to 2% carbon
Used for making pipes,

Manufacturing automotive parts, pots, pans, utensils

Used for making pipes, bars for stay bolts, engine bolts and rivetts. Used in buildings infrastructure, tools, ships, automobiles, weapons etc.

(iv) Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable ? Why ?

Answer :

The electronic configuration of Fe2+ and Fe3+ :

Fe2+: 1s22s22p63s23p63d6

Fe3+: 1s22s22p63s23p63d5

Due to loss of two electrons from the 4s-orbital and one electron from 3d-orbital, iron attains 3+ oxidation state.

Since in Fe3+, the 3d-orbital is half filled, it gets extra stability, hence Fe3+ is more stable than Fe2+.

(v) Give the similarities and differences in elements of 3d, 4d and 5d series.

Answer :

Similarity :

  • They are placed between s-block and p-block of the periodic table.
  • All elements are metals showing metallic characters.
  • Some are paramagnetic.
  • Most of them give coloured compounds.
  • They have catalytic properties.
  • They form complexes.

Differences :

  • In 4d and 5d series lanthanide and actinoid contraction is observed. In 3d series atomic size changes are less marked.
  • 4d and 5d elements have high coordination numbers compared to 3d elements.
  • 4d and 5d series have similar properties whereas 3d series have different properties.
  • Although most properties exhibited by d block elements are similar, the elements of the first row differ from second and third rows in the stabilization of higher oxidation states in their compounds.

(vi) Explain trends in ionisation enthalpies of d block elements.

Answer :

  • The ionisation enthalpies of transition elements are quite high and lie between those of s-block and p-block elements. This is because the nuclear charge and atomic radii of transition elements lie between those of s-block and p-block elements.
  • As atomic number of transition elements increases along the period and along the group, first ionisation enthalpy increases even though the increase is not regular.
  • If IE1, IE2 and IE3 are the first, second and third ionisation enthalpies of the transition elements, then IE1 < IE2 < IE3.
  • In the transition elements, the added last differentiating electron enters into (n - 1) d-orbital and shields the valence electrons from the nuclear attraction. This gives rise to the screening effect of (n -1) d electrons.
  • Due to this screening effect of (n - 1) d electrons, the ionisation enthalpy increases slowly and the increase is not very regular.

(vii) What is meant by diamagnetic and paramagnetic? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.

Answer :

Diamagnetic substances : When a magnetic field is applied, substances which are repelled by the magnetic fields are called diamagnetic substances.

Paramagnetic substances : When a magnetic field is applied, substances which are attracted towards the applied magnetic field are called paramagnetic substances.

Examples :

  Transition metal Lanthanoid metal
Diamagnetic Zinc Ytterbium
Paramagnetic Titanium Cerium

(viii) Why the ground-state electronic configurations of gadolinium and lawrentium are different than expected?

Answer :

  • Gadolinium and lutetium show different ground-state electronic configurations because the 5d and 4f-orbitals are near to the same energy due to which the next electron goes in 5d-orbital instead of 4f-orbital.
  • This retains the extra stability achieved when the 4f-orbital is half-filled and completely filled respectively. Thus, the 5d-orbital contains one electron in the case of gadolinium and lutetium.

(ix) Write steps involved in metallurgical process.

Answer :

The various steps and principles involved in the extraction of pure metals from their ores are as follows :

  • Concentration of ores in which impurities (gangue) are removed.
  • Conversion of ores into oxides or other reducible compounds of metals.
  • Reduction of ores to obtain crude metals.
  • Refining of metals giving pure metals.

(x) Cerium and Terbium behaves as good oxidising agents in +4 oxidation state. Explain.

Answer :

  • The most stable oxidation state of lanthanoids is +3.
  • Hence, Ce4+ (cerium) and Tb4+ (terbium) tend to get +3 oxidation state which is more stable.
  • Since they get reduced by accepting electron, they are good oxidising agents in +4 oxidation state.

(xi) Europium and xtterbium behave as good reducing agents in +2 oxidation state explain.

Answer :

  • The most stable oxidation state of lanthanoids is +3.
  • Hence, Eu2+ and Yb2+ tend to get +3 oxidation states by losing one electron.
  • Since they get oxidised, they are good reducing agents in + 2 oxidation state.

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