Skip to content
Question 49 of 49

Q.Answer any two questions out of four questions:
(A)

(i) Though the aqueous solution of K2Cr2O7 is used as primary standard solution, aqueous solution of KMnO4 is used as secondary standard solution. — Explain.
(ii) Write the electronic configuration of 62Sm and also its possible oxidation states.
(iii) K2[PtCl6] does exist but not K2[NiCl6]. Explain.
(B)
(i) Give the reason behind lanthanoid contraction.
(ii) Identify the following ions in aqueous solution as coloured or colourless/light coloured: Mn2+, MnO4^-, Nd3+ (Atomic No. of Nd = 60)
(C)
(i) H2O2 is added to acidified aqueous solution of K2Cr2O7. Explain the corresponding chemical reaction with observable colour change.
(ii) Compound 'A' is formed when lanthanoid (Ln) reacts with element carbon (C) in inert atmosphere. Compound 'A' reacts in dilute acid medium to form 'X'. Mention the formula of A and X.
(D)
(i) Find out the value of 'n' in [Ni(CO)n] using EAN rule, if oxidation number of Ni is zero. (Given: Z(Ni) = 28)
(ii) Give an example of alloy of lanthanoids and mention its use.
(iii) Arrange the following oxides in increasing order of their acidic nature: MnO, MnO2, Mn2O3, Mn2O7
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026Subjective· 6mImportance★★★★★
100% · 49/49 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

A weighs-directly-without-standardising oxidant (K2Cr2O7) is a primary standard while KMnO4 is secondary; Sm is [Xe]4f^6 6s^2 (+2/+3); Pt(IV) but not Ni(IV) is stabilised by Cl-; lanthanoid contraction stems from poor 4f shielding; CrO5 is blue; Ni(CO)4 satisfies EAN; and Mn-oxide acidity rises with oxidation state.

(A)(i) Primary vs secondary standard

K2Cr2O7 can be obtained in a high state of purity, is non-hygroscopic, stable and does not decompose on storage, so it can be weighed directly to make a solution of exactly known concentration -> a primary standard. KMnO4 usually contains impurities (traces of MnO2), is not perfectly stable (it slowly decomposes in light and reacts with traces of organic matter/water), so its exact concentration is not known from weighing and its solution must first be standardised against a primary standard -> a secondary standard.

(A)(ii) 62Sm: [Xe] 4f^6 6s^2; common oxidation states +3 (characteristic) and +2.

(A)(iii) In K2[PtCl6] platinum is in +4. Being a 5d element, Pt readily attains and stabilises the +4 state, so [PtCl6]^2- exists. K2[NiCl6] would require Ni(+4); Ni(IV) is a very strong oxidising state that the weak-field, easily-oxidised Cl- ligand cannot stabilise (Cl- would be oxidised to Cl2), so K2[NiCl6] does not exist.

(B)(i) Lanthanoid contraction is the steady decrease in atomic and ionic (M3+) radii along the lanthanoid series. The 4f electrons shield one another very poorly from the increasing nuclear charge; as atomic number rises, the effective nuclear charge felt by the outer electrons increases and pulls them inward, causing the contraction.

(B)(ii) Mn2+ (3d^5): pale/light pink (coloured but faint); MnO4- (Mn is formally d^0): intensely coloured (purple) due to charge-transfer (not d-d); Nd3+ (4f^3): coloured (f-f transitions).

(C)(i) H2O2 added to acidified K2Cr2O7 gives a deep blue chromium peroxide CrO5 (chromium(VI) oxide peroxide):

K2Cr2O7 + 4H2O2 + H2SO4 -> 2CrO5 + K2SO4 + 5H2O

Observation: orange dichromate solution turns transient blue (the blue CrO5 is best seen/stabilised in an ether layer).

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.