
Bad conductor of electricity is:
A. ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$
B. ${\text{HCl}}$
C. ${\text{HBr}}$
D. ${\text{HI}}$
Answer
570.3k+ views
Hint: To conduct electricity free ions are required. The conductivity of the ionic compound depends upon the ionization of that compound in aqueous solution.
Complete answer:
Conductivity is the ability to conduct electricity which depends upon the presence of free electrons as well as free ions.
Free ions produced when ionic compounds dissociation in water.
The ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ compound has high stability. The fluorine is the most electronegative element so, the hydrogen-fluorine bond is a very stable and thus the ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ does not ionize in aqueous solution.
${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ does not conduct electricity so, ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is a bad conductor of electricity and thus option (A) is correct.
The chlorine atom has less electronegative than fluorine so the hydrogen-chlorine bond is weak comparatively. HCl ionizes in the aqueous solution forming hydronium and chloride ions and so HCl conducts electricity. So, option (B) is incorrect.
The bromine atom has less electronegative than fluorine and chlorine so the hydrogen-bromine bond is weak comparatively. HBr ionizes in the aqueous solution forming hydronium and bromide ions and so, HBr conducts electricity. So, option (C) is incorrect.
The iodine atom has less electronegative than fluorine, chlorine and bromine so the hydrogen-iodine bond is weak comparatively. HI ionizes in the aqueous solution forming hydronium and iodide ions and so, HI conducts electricity. So, option (D) is incorrect.
Therefore, option (A) ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is correct.
Note: For the hydrogen halides, the conductivity can be decided on the basis of acidic strength. The HF is a weak acid it dissociates very less in water and thus a bad conductor whereas the hydrogen chloride, hydrogen bromide and hydrogen iodide are strong acids, they dissociate completely in water thus produce ions and conduct electricity.
Complete answer:
Conductivity is the ability to conduct electricity which depends upon the presence of free electrons as well as free ions.
Free ions produced when ionic compounds dissociation in water.
The ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ compound has high stability. The fluorine is the most electronegative element so, the hydrogen-fluorine bond is a very stable and thus the ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ does not ionize in aqueous solution.
${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ does not conduct electricity so, ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is a bad conductor of electricity and thus option (A) is correct.
The chlorine atom has less electronegative than fluorine so the hydrogen-chlorine bond is weak comparatively. HCl ionizes in the aqueous solution forming hydronium and chloride ions and so HCl conducts electricity. So, option (B) is incorrect.
The bromine atom has less electronegative than fluorine and chlorine so the hydrogen-bromine bond is weak comparatively. HBr ionizes in the aqueous solution forming hydronium and bromide ions and so, HBr conducts electricity. So, option (C) is incorrect.
The iodine atom has less electronegative than fluorine, chlorine and bromine so the hydrogen-iodine bond is weak comparatively. HI ionizes in the aqueous solution forming hydronium and iodide ions and so, HI conducts electricity. So, option (D) is incorrect.
Therefore, option (A) ${{\text{H}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is correct.
Note: For the hydrogen halides, the conductivity can be decided on the basis of acidic strength. The HF is a weak acid it dissociates very less in water and thus a bad conductor whereas the hydrogen chloride, hydrogen bromide and hydrogen iodide are strong acids, they dissociate completely in water thus produce ions and conduct electricity.
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