The square-planar complex [Pd(Br)(NO₂)(SCN)(py)] has two ambidentate ligands (NO₂⁻ through N or O, SCN⁻ through S or N). Let a be the total number of isomers (geometrical and linkage), b the number of

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The square-planar complex [Pd(Br)(NO₂)(SCN)(py)] has two ambidentate ligands (NO₂⁻ through N or O, SCN⁻ through S or N). Let a be the total number of isomers (geometrical and linkage), b the number of unpaired electrons on Pd, and c the number of inner (n−1)d orbitals used in hybridisation. The value of (a ÷ 2) + b + c is

TestHub · Accepted Answer

Mabcd square planar has 3 geometrical isomers, and the two ambidentate ligands give 2 × 2 = 4 linkage forms, so a = 3 × 4 = 12. Pd²⁺ is d⁸ square planar, so b = 0 and c = 1 (dsp²). (12 ÷ 2) + 0 + 1 = 7.

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