CBSE Class 12 Chemistry D And F Block Elements Board Exam Notes

Chemistry Concept Notes for Class 12: Unit 4 The d- and f-Block Elements

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• d‐block consists of elements belonging to group 3‐12.
• The general electronic configuration of these elements in (n‐1) d1–10 ns1–2. They contain unpaired d - electron either in elementary or most commonly occurring oxidation state.
• Transition elements exhibits metallic character, high densities, high melting and boiling point due to strong interatomic interaction between partially filled d - orbits.'
• Transition elements show variable oxidation state due to small energy difference between (n‐1)d & ns orbitals.
• Transition metals show colour due to presence of unpaired electron in (n‐1)d orbital of penultimate shell which undergoes d‐d transition generally
• Transition metal elements are paramagnetic due to presence of unpaired e .
• Most of transition form complex compounds due to
(I) small size
(ii) high charge
(iii) presence of vacant d‐orbital of suitable energy.
 
• Transition elements have lower value of reduction potential due to high ionization potential, high heat of sublimation and low enthalpy of hydration.
• Transition elements form interstitial compounds because as small atoms like C, N, O, H. occupy the voids and get bonded to the atom of transition metals.
• Transition elements form alloys due to similar ionic radii as they can mutually substitute their position in the crystal lattice.
• The oxides of transition metals in lower oxidation states are basic, intermediate oxidation states are amphoteric, highest oxidation state are acidic.
• The highest oxidation state of an element is equal to no. of unpaired electrons present in (n‐1)d & ns orbitals.

LANTHANOIDS
• In lanthanoids filling of electrons take place in 4f orbital which is 3rd antipenultimate shell.
• General electronic configuration is [Xe]4f1–145d0–16s2. last shell or
• Elements after lanthanum having atomic number 58 to 71 are called lanthanoids .
• Principal oxidation state is +3 although Ce shows +4 and Eu +2 due to stable configuration. So Ce (IV) is good oxidizing and Sm (II) is good reducing agent.
• Due to poor shielding effect of electrons. There is a steady decrease in lanthanoids and its trivalent ions known as lanthanoid contraction.

ACTINOIDS
• The 14 elements after actinium having atomic number 90 to 113 are collectively known as Actinoids
• The general electronic configuration of these elements is [Rn]5f1–14, 6d0–1, 7s2.
• The size of actinoids and its trivalent ion decreases from Ac to Lw due to poor shielding of 5f electrons. It is known as actinoid contraction.
• Chemical of actinoids is very complex due to their ability that exist is different oxidation state. More so they are radioactive.

POTASSIUM DICHROMATE
Preparation: ‐ from chromite ore FeCr2O4
I. Conversion of chromite ore to sodium chromate.
II. Conversion of sodium chromate to sodium dichromate.
III. Conversion of sodium dichromate to potassium dichromate.
IV. Following reaction takes place:‐

EFFECT OF pH
Na2Cr2O7 + 2KCl →  K2Cr2O7 + 2NaCl

Chemical properties
Sodium and potassium dichromates acts as strong oxidising agents in acidic medium. Cr2O7 2– + 14H+ + 6e– →2Cr3+ + 7H2O
Oxidation Reaction 6I– → 3I2 + 6e– , 3H2S→6H+ + 3S+ 6e– , 6Fe2+ →6Fe3+ + 6e– , 3Sn2+→3Sn4+ + 6e–

POTASSIUM PERMANGANATE
Preparation: ‐ From pyrolusite ore
I. Conversion of pyrolusite ore into potassium manganate
II. Conversion of potassium manganate to potassium permanganate Following reactions take place:‐
2MnO2 + 4KOH + O2 → 2 K2MnO4 + 2H2O
3MnO4 2– + 4H+ → 2MnO4– + MnO2 + 2H2O

Chemical Properties
Potassium permanganate is a powerful oxidising agent. Neutral solution
2KMnO4 + H2O →2 KOH + 2MnO2 + 3 [ O]
3MnO4 – + 2H2 O+ 3e– →MnO2 + 4 OH–
Alkaline solution
2KMnO4+ 2KOH → 2 K2MnO4 + H2O + 5 [ O]
MnO4 – + e– →MnO4 2–
Acidic solutions
2KMnO4 + 3 H2SO4 →K2SO4 + 2MnSO4 + 3H2O MnO4– + 8H+ + 5e– →Mn2+ +4H2O
FREQUENTLY ASKED QUESTIONS
(1 mark)
Q.1. Cu+ is not stable in aqueous solution. Why?
Ans. In aqueous solution Cu+ undergoes disproportionation to form a more stable Cu2+ion.
2Cu+ (aq) → Cu2+ (aq) + Cu(s)
The higher stability of Cu2+ in aqueous solution maybe attributed to its greater negative ΔhydH
◦ than that of Cu+. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu2+
ions.
Q.2. Which is a stronger reducing agents – Cr2+ or Fe2+ and why?
Ans. Cr2+ is a stronger reducing agent than Fe2+ because after the loss of one electron Cr2+
becomes Cr3+ which has more stable t2g
3 (half filled) configuration in medium like water.
Q.3. Arrange the following increasing order of acidic character: CrO3, CrO , Cr2O3
Ans. CrO <Cr2O3<CrO3. Higher the oxidation state, more will be acidic character.
Q.4. Calculate the 'spin only' magnetic moment of M2+(aq) ion.(Z=27)
Ans. Electronic configuration of the M2+ ion (Z=27) would be M2+(aq): (Ar) 3d7
It would contain three unpaired electrons. The 'spin only' magnetic moment is given by the relation :
μ = n (n + 2) BM = 3 (3 + 2) BM = 3.87 BM

 

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