Which of the following has the weakest \[C-X\] bond?
A.$C{{H}_{3}}F$
B.$C{{H}_{3}}Cl$
C.$C{{H}_{3}}Br$
D.$C{{H}_{3}}I$

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Hint: Before talking about the answer, you should know about the \[C-X\] bond. Here, \[C-X\] bond is a bond between carbon and halogen, where $C$ stands for carbon and $X$ stands for halogen.

Complete step by step answer:
Halogens are group 17 elements which have high electronegative atoms, some of them are Fluorine, Bromine, Chlorine and Iodine. It requires only one electron to complete its octet. The Carbon-Halogen bond is highly polar because carbon atoms are electropositive in nature.
As we go down the group, the size of the elements will start increasing. As the size is increased, the force of attraction between atoms will start decreasing, that means, as we go down the group, the bond strength will start decreasing.
As we know that fluorine is present in the starting of the group, that means the bond strength of fluorine is high and as we go down the group, iodine is present and bond strength of iodine is weak.
Therefore, the correct option is (D), that is, $C{{H}_{3}}I$ has the weakest \[C-X\] bond.


Additional information: The order of bond strength of carbon – halogen bond is
\[C{{H}_{3}}-F>C{{H}_{3}}-Cl>C{{H}_{3}}-Br>C{{H}_{3}}-I\]
The order of bond length of carbon – halogen bond is
$C{{H}_{3}}-I>C{{H}_{3}}-Br>C{{H}_{3}}-Cl>C{{H}_{3}}-F$
Fluorine is the most electronegative atom that means it will be smaller in size as compared to chlorine, bromine and iodine because the positive electron attracts more strongly to a negative electron which makes the size smaller.


Note:
The carbon – fluorine $(C-F)$ bond is the strongest bond because carbon is a $2p$ orbital and fluorine is also $2p$ orbital. So, the overlapping takes place perfectly which makes the bond stronger whereas in carbon – iodine $(C-Br)$ bond, the iodine has $5p$ orbital which means the overlapping is not so perfect and hence bond is weaker.