Chiral complex from the following is
For identifying chiral complexes:
• Check for the absence of planes of symmetry or center of symmetry.
• Cis configurations with bidentate ligands (like en) often result in chiral complexes.
• Trans configurations are usually symmetric and achiral.
trans – [Co(NH3)4Cl2] +
cis – [PtCl2(en)2] 2+
cis – [PtCl2(NH3)2]
trans – [PtCl2(en)2] 2+
Chirality in coordination complexes occurs when the complex lacks a plane of symmetry.
- For the given complexes:
cis–[PtCl\(_2\)(en)\(_2\)]\(^{2+}\): The cis arrangement of ethylene diamine (en) ligands around the Pt center creates a chiral structure.
trans–[PtCl\(_2\)(en)\(_2\)]\(^{2+}\): The trans arrangement is symmetric, making the complex achiral.
cis–[PtCl\(_2\)(NH\(_3\))\(_2\)]: The complex has a plane of symmetry and is not chiral.
trans–[Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\): The trans arrangement of ligands makes the complex symmetric and achiral.
Final Answer: (1) cis–[PtCl\(_2\)(en)\(_2\)]\(^{2+}\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Consider the following sequence of reactions:
The major product $P$ is:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,