MCQ
Among the following, the compound that is both paramagnetic and coloured is
- A$K_2Cr_2O_7$
- B$(NH_4)_2[TiCl_6]$
- ✓$VOSO_4$
- D$K_3[Cu(CN)_4]$
$Cr ^{-6}=[ Ar ] 3 d ^0$
In $\left( NH _4\right)_2\left[ TiCl _6\right]$, the oxidation state of $Ti$ is $+4$.
$Ti ^{+4}=[ Ar ] 3 d ^0$
In $VOSO _4$, the oxidation state of $V$ is $+4$.
$V ^{+4}=[ Ar ] 3 d ^1$
In $K _3\left[ Cu ( CN )_4\right]$, the oxidation state of $Cu$ is $+1$.
$Cu ^{+1}=[ Ar ] 3 d ^{10}$
In $VOSO _4, V ^{+4}$ configuration is $d ^1$, so it has one unpaired d-electron so it is paramagnetic and colored. All other compounds don't have any unpaired $d$ electrons, so they are diamagnetic in nature.
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$'C'$ is water soluble
Correct structure of $A$ and $D$ are
$(x)\begin{array}{*{20}{c}}
{O\,\,\,}\\
{||\,\,\,}\\
{C{H_3} - S - O - H}\\
{||\,\,\,\,}\\
{O\,\,\,\,}
\end{array}$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,O}\\
{\,\,\,\,\,\,||}\\
{(y)\,\,\,C{H_3} - C - O - H}
\end{array}$
$(z)\,\, CH_3 -OH$
Sucrose $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ Glucose $\quad$ Fructose
$C _{6} H _{12} O _{6} \stackrel{\text { Enzyme B }}{\longrightarrow} 2 C _{2} H _{5} OH +2 CO _{2}$
Glucose
In the above reactions, the enzyme $A$ and enzyme $B$ respectively are :-