- APresence of strong field ligand in CuSO4⋅5H2O.
- BDue to absence of water (ligand), d - d transitions are not possible in CuSO4.
- CAnhydrous undergoes d - d transitions due to crystal field splitting.
- DColour is lost due to unpaired electrons.
Explanation:
In CuSO4⋅5H2O, water acts as ligand and causes crystal filed splitting. This makes d - d transitions possible. On the other hand, in CuSO4, due to absence of ligand crystal filed splitting is not possible.
Hence no colour is observed.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$HC \equiv CH\mathop {\xrightarrow{{{\text{1% HgS}}{{\text{O}}_{\text{4}}}}}}\limits_{20\% {H_2}S{O_4}} A\xrightarrow{{C{H_3}MgX}}B\xrightarrow{{[O]}}$

Sum of the number of methylene groups $\left(- CH _2-\right.$ ) and oxygen atoms in $R$ is. . . .

$\begin{array}{*{20}{c}}
{C{H_3} - C \equiv C - C{H_2} - \mathop C\limits_{\mathop {||}\limits_O } - Cl}
\end{array} \ \xrightarrow[{BaS{O_4}}]{{{H_2} - Pd}}$