- A$n = 6, m = -4$ to $+4,$ no. of electrons $= 18$
- B$n = 5, m = -2$ to $+4,$ no. of electrons $= 9$
- C$n = 5, m = -3$ to $+3,$ no. of electrons $= 9$
- ✓$n = 5, m = -4$ to $+4,$ no. of electrons $= 18$
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$(I)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
\,\,{C{H_3}} \\
{|\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - COOH} \\
{|\,\,\,\,\,\,} \\
\,\,{C{H_3}\,}
\end{array}$
$(IV)$ $(CH_3-CH_2)_3C-COOH$
$A$. $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$
$B$. $\mathrm{N} \rightarrow \mathrm{N}^{-}$
$C$. $\mathrm{O} \rightarrow \mathrm{O}^{2-}$
$D$. $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$
$E$. $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$
Choose the most appropriate answer from the options given below :