Explain each of the following observations: (a) At roomtemperature I2 is a solid, Br2 is a liquid, and Cl2 and F2 areboth gases. (b) F2 cannot be prepared by electrolytic oxidationof aqueous F - solutions. (c) The boiling point of HFis much higher than those of the other hydrogen halides.(d) The halogens decrease in oxidizing power in the orderF2 > Cl2 > Br2 > I2.
Explain each of the following observations: (a) At room
temperature I2 is a solid, Br2 is a liquid, and Cl2 and F2 are
both gases. (b) F2 cannot be prepared by electrolytic oxidation
of aqueous F - solutions. (c) The boiling point of HF
is much higher than those of the other hydrogen halides.
(d) The halogens decrease in oxidizing power in the order
F2 > Cl2 > Br2 > I2.
(A) I2 has atoms which have the highest atomic radii in their family, and thus they have higher covalent bond and lower mobility thus they solidify at room temperature. i.e they have higher melting and boiling points.
Br2 is heavy but relatively less than iodine, they have high boiling points but low melting points and thus are liquid in room temperature.
Cl2 & F2 are very small as compare to other elements/atoms of their family and thus they are gaseous in nature.
(B) First of all we should know that F- cannot stay freely in water, it exists as HF in water
Now when we electrolyte the solution there are two components ready to be electrolysed, water and HF. Their potentials and reactions are shown
O2 + 4H+ + 4e- -----> 2H2O (+1.23 V)
F2 + 2e- -----> 2F- (+2.87 V)
These are written in the reduction direction; as oxidation reactions the voltages are flipped, and it's much easier to oxidize water than to oxidize fluoride ion, thus we cannot prepare F2 by electrolyticing.
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