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Question
Explain the formation of a chlorine molecule on the basis of electronic theory of valency.
Solution
The atomic number of chlorine is 17 and its electronic configuration is 2, 8, 7. It needs one more electron to complete its octet. Chlorine gets this electron by sharing with another atom. So each chlorine atom shares one electron to form a chlorine molecule.
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- \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - C - H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{.....}|\phantom{.....}|\\
\ce{H}\ce{H-C-H}\ce{H}\\
|\\
\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{.....}\backslash\phantom{..}|\\
\phantom{....}\ce{H}\phantom{......}\ce{C - H}\phantom{}\\
\phantom{.......}|\\
\phantom{.......}\ce{H}\\
\end{array}\] - \[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\\
\end{array}\]
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