- A$1.6$
- ✓$2.6$
- C$3.6$
- D$4.6$
$n=3$
$\alpha=\frac{i-1}{n-1}$
$0.6=\frac{i-1}{3-1}$
$i=2.2$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{C{H_3} - CH = CH - C{H_2} - CH - C{H_3}}
\end{array}$ $\longrightarrow $ $C{H_3} - CH = CH - C{H_2}C{O_2}H$
$(I)\,[Kr]\,5s^1$ $(II)\, [Rn]\,5f^{14}\,6d^1\,7s^2$
$(III)\,[Ar]\,3d^{10}\,4s^2\,4p^5$ $(IV)\,[Ar]\,3d^6\,4s^2$
Consider the following Statements
$(i)\, I$ shows variable oxidation State
$(ii)\, II$ is a $d-block$ element
$(iii)$ The compound formed between $I$ and $III$ is covalent
$(iv)\,IV$ shows single oxidation state
Which Statement is True $(T)$ or False $(F)$?