MCQ
Chlorine water on standing loses its colour and forms
- A$HCl$ only
- B$HCl$ and $HClO_2$
- ✓$HCl$ and $HOCl$
- D$HOCl$ and $HOCl_2$
✓
Answer
Correct option: C.
$HCl$ and $HOCl$
c
$\begin{align}
\,\,+I \\
\Rightarrow \,\,Cl{{O}^{-}} \\
B.O.\,=\,1.0 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$ $\begin{align}
\,+III\,\,\,\,\, \\
\,\,ClO_{2}^{-} \\
B.O.\,=\,1.5 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$
$\begin{align}
\,\,+I \\
\Rightarrow \,\,Cl{{O}^{-}} \\
B.O.\,=\,1.0 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$ $\begin{align}
\,+III\,\,\,\,\, \\
\,\,ClO_{2}^{-} \\
B.O.\,=\,1.5 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$
$\begin{align}
+V\,\,\,\,\, \\
\,\,ClO_{3}^{-} \\
B.O.\,=\,1.67 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$ $\begin{align}
\,+VII\,\,\,\,\, \\
\,\,ClO_{4}^{-} \\
B.O.\,=\,1.75 \\
Hyb.\,\,of\,Cl\,:\,s{{p}^{3}} \\
\end{align}$
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