Question

$KF$ combines with $HF$ to form $KH{F_2}$. This compound contains the species:
A.${K^ + },\,{F^ - }\,\,and\,{H^ + }$
B.${K^ + },\,{F^ - }\,\,and\,HF$
C.${K^ + }\,and\,{\left[ {H{F_2}} \right]^ - }$
D.${\left[ {KHF} \right]^ + }and\,{F_2}$

Answer 1

Hint: The reaction happens between hydrogen fluoride and potassium fluoride, which gives $KH{F_2}$ . Now the species will contain some ions which will be predicted if we mix the formed compound in water and see what it's electrolytic products are. Here as we have given $KH{F_2}$ here we know that it is made up of potassium, hydrogen and fluorine but the ionic species are different.

Complete step-by-step answer:
Let’s say that when $KH{F_2}$ mixed in water we see some ionic species according to its electrolytic reaction. Here, $KH{F_2}$ produces ${K^ + }\,and\,{\left[ {H{F_2}} \right]^ - }$ but it conducts electricity only in fused form. There are some electrolytes which conduct electricity when they are mixed with water but some conduct electricity in a fused state. The movement of charged species towards oppositely charged electrodes completes the circuit and electricity flows. At the end it was found that generally cathode gets thinner and anode gets bulkier.
Fused state is a state where no water is produced or involved. If we consider $KH{F_2}$ in this case when this compound changes in liquid state without having a water molecule then it is called as the fused state.

Therefore, Option (C) is correct.

Note: In some cases sodium chloride also conducts electricity in a fused state. Conducting electricity means that the ions will move towards oppositely charged electrodes. There are two electrode cathodes and anodes. Cathode is negatively charged and anode is positively charged. When electricity is passed through the solution the cation (positively charged ions) move toward cathode while anions (negatively charged ions) move towards anode.