Among the following options, select the option in which each complex in
Set-I shows geometrical isomerism and the two complexes in
Set-II are ionization isomers of each other.
[en = H2NCH2CH2NH2]
Set-I Analysis:
Set-II Analysis:
\([Co(NH_3)_5Cl]SO_4\) and \([Co(NH_3)_5(SO_4)]Cl\): These two complexes are ionization isomers because they differ in the ion released in solution.
Conclusion:
Option (C) satisfies both conditions:
To solve the problem, we analyze both sets to find complexes showing geometrical isomerism in Set-I and ionization isomers in Set-II.
1. Geometrical Isomerism in Set-I:
- [Co(NH3)3(NO2)3] does not show geometrical isomerism as it contains identical ligands.
- [Co(en)2Cl2] contains two bidentate ethylenediamine (en) ligands and two chloride ligands; it exhibits cis and trans geometrical isomers.
Hence, these two complexes are suitable for geometrical isomerism.
2. Ionization Isomers in Set-II:
- [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl differ by interchange of ligands inside and outside coordination sphere, so they are ionization isomers.
Final Answer:
Set-I: [Co(NH3)3(NO2)3] and [Co(en)2Cl2]
Set-II: [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl
This matches option C.
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is