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

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– + 14^H+ + 6^e– →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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