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