Consider the following complex compounds.
(i) [Pt(NH3)2(SCN)2]
(ii) [Co(NH3)3(NO2)3]
(iii) [Pt(en)Cl2]
(iv) [Cr(en)2Br2]^+
(v) [Rh(en)3]^3+
(vi) [CoCl2Br2]^2-
Then calculate sum o

Question

Consider the following complex compounds.
(i) [Pt(NH3)2(SCN)2]
(ii) [Co(NH3)3(NO2)3]
(iii) [Pt(en)Cl2]
(iv) [Cr(en)2Br2]^+
(v) [Rh(en)3]^3+
(vi) [CoCl2Br2]^2-
Then calculate sum of total number of geometrical isomers in all above complex compounds.

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 6**<BR>SOLUTION: NUMBER OF GEOMETRICAL ISOMERS FOR EACH ARE: (I) 2, (II) 2, (III) 0, (IV) 2, (V) 0, (VI) 0. SUM OF GEOMETRICAL ISOMERS = 6.. Step 1: Co^3+ is a 3d^6 transition metal ion. In [CoF6]^3-, F^- is a weak-field ligand, so the complex is high-spin and utilizes outer d-orbitals for hybridization (sp^3d^2, outer orbital complex). Step 2: In [Co(NH3)6]^3+, NH3 is a strong-field ligand that forces pairing, so the complex is low-spin and utilizes inner d-orbitals for hybridization (d^2sp^3, inner orbital complex). Step 3: Therefore, the statement that [CoF6]^3- is an outer orbital complex and [Co(NH3)6]^3+ is an inner orbital complex is correct, matching option (d).

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