
In dark, which of the following reacts with hydrogen?
A. ${{F}_{2}}$
B. $C{{l}_{2}}$
C. ${{I}_{2}}$
D. $B{{r}_{2}}$
Answer
163.8k+ views
Hint: As we move down a group in halogen the reactivity of halogen elements decreases. Fluorine is most reactive among the halogens. The most reactive ones react even in the dark.
Complete Step by Step Answer:
The reactivity of halogens decreases as we move from fluorine to iodine. Fluorine has the smallest size thus it possesses strong repulsion between the electrons of two fluorine atoms and has the weakest bond. Thus fluorine is highly reactive and can easily react in dark with hydrogen gas.
Fluorine reacts with hydrogen gas in dark condition to form an acid of fluorine that is hydrogen fluoride. The reaction is given as follows:
${{F}_{2}}+{{H}_{2}}\to 2HF$
Thus in this reaction one mole of fluorine gas reacts with one mole of hydrogen gas to produce two moles of hydrogen fluoride as a product. Thus we can write that in dark fluorine or gas reacts with hydrogen gas.
Thus the correct option is A.
Note: Fluorine is the most electronegative element among halogens. It has the lowest boiling and melting point among halogens. It has the symbol of $F$.
Complete Step by Step Answer:
The reactivity of halogens decreases as we move from fluorine to iodine. Fluorine has the smallest size thus it possesses strong repulsion between the electrons of two fluorine atoms and has the weakest bond. Thus fluorine is highly reactive and can easily react in dark with hydrogen gas.
Fluorine reacts with hydrogen gas in dark condition to form an acid of fluorine that is hydrogen fluoride. The reaction is given as follows:
${{F}_{2}}+{{H}_{2}}\to 2HF$
Thus in this reaction one mole of fluorine gas reacts with one mole of hydrogen gas to produce two moles of hydrogen fluoride as a product. Thus we can write that in dark fluorine or gas reacts with hydrogen gas.
Thus the correct option is A.
Note: Fluorine is the most electronegative element among halogens. It has the lowest boiling and melting point among halogens. It has the symbol of $F$.
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