- A$N\,-\,N$
- B$F\,-\,F$
- ✓$C\,-\,C$
- D$O\,-\,O$
In the given molecules fluorine, nitrogen and oxygen atoms have lone pair electons in Them these lone pairs creates lonepair-lone pair repulsion between $F-F,N-N$ and $O-O$ molecules. Due to which it will increase its bond length and decrease the bond energy. so, The correct option is $C-C$ as there are no lone pairs present on carson atom
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${H_3}C - C \equiv CH\,\xrightarrow[{HgS{O_4}}]{{{H_2}O,\,{H_2}S{O_4}}}\,\mathop {Intermediate}\limits_{(A)} \,$$ \to \,\mathop {\Pr oduct}\limits_{(B)} $
$H_2C = C = CH - CH = O$
$sp^3$ - $sp^2$ - $sp$
Given that
Salt || $K_{sp}$
$AgCl$ || $2 \times {10^{ - 10}}$
$AgBr$ || $5 \times {10^{ - 13}}$
$Ag$ || $9 \times {10^{ - 17}}$


$(I)$ $\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{O - S - {C_6}{H_4} - C{H_3}} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\,$ $(II)$ $\begin{array}{*{20}{c}} {O\,\,\,\,\,\,\,} \\ {||\,\,\,\,\,\,\,} \\ {O - S - {C_6}{H_5}} \\ {||\,\,\,\,\,\,\,} \\ {O\,\,\,\,\,\,\,} \end{array}\,$$(III)$ $N_3$ $(IV)$ $Br$