a) For the complex [Fe(CN)], write the hybridisation type, magnetic character and spin nature. (Atomic number: Fe = 26)
b) Draw one of the geometrical isomers of the complex [Pt(en)Cl] which is optically active.
b) Draw one of the geometrical isomers of the complex [Pt(en)Cl] which is optically active.
Answer: (a) d²sp³; paramagnetic (1 unpaired); low-spin. (b) The cis-isomer (both Cl adjacent) — chiral, hence optically active.
Solution
(a) Fe = 3d; strong-field CN pairs to t: hybridisation \textbf{dsp}; \textbf{paramagnetic} with 1 unpaired electron (μ ≈ 1.73 BM); \textbf{low-spin} (inner-orbital).
(b) \textit{cis}-[Pt(en)Cl] — the two Cl on adjacent octahedral vertices; this arrangement has no symmetry plane, so it is chiral (optically active). Draw the octahedron with the two Cl cis and the two en chelate rings completing the sphere.
(b) \textit{cis}-[Pt(en)Cl] — the two Cl on adjacent octahedral vertices; this arrangement has no symmetry plane, so it is chiral (optically active). Draw the octahedron with the two Cl cis and the two en chelate rings completing the sphere.
CBSE Class 12 Board · Subjective