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Interpretation : The given pairs of atoms have to be arranged in order of increasing bond polarity and the reasoning has to be explained.
Concept Introduction :
The tendency of an atom to attract shared electrons is called electronegativity. Chemists have assigned each
Answer to Problem SIII2RE
Bond polarities of given pairs of atoms are:
H-I: 0.56
H-Cl:1.06
H-F:1.88
Based on the bond polarities, the given pairs can be arranged as:
H-I<H-Cl< H-F
Reason- The electronegativity value increases from bottom to top of the periodic table. So fluorine is more electronegative than chlorine and iodine is least electronegative. So the bond polarity of H-F is highest and H-I is lowest.
Explanation of Solution
Given information:
Three pairs of atoms are given:
H-I, H-Cl, H-F
The electronegativity scale is used to compare individual atoms. Atoms that are more electronegative attract bonded electrons more strongly than atoms with less electronegative values. This results in formation of a polar bond. As the difference in electronegativity increases, the bond polarity increases.
H-I- From the electronegativity scale, electronegativity value of hydrogen is 2.10 and iodine is 2.66. Difference in electronegativity value = 2.66-2.10=0.56.
H-Cl-From the electronegativity scale, electronegativity value of hydrogen is 2.10 and chlorine is 3.16. Difference in electronegativity value=3.16-2.10=1.06.
H-F- From the electronegativity scale, electronegativity value of hydrogen is 2.10 and fluorine is 3.98. Difference in electronegativity value=3.98-2.10=1.88.
Reason- From the electronegativity values of metals, one can see that the scale increase from bottom to top of the periodic table. Since all these nonmetals are bonding with hydrogen, the difference in their electronegativities increases from bottom to top. Hence the bond polarity of H-F is highest and that of H-I is lowest. This can be represented as:
H-I< H-Cl< H-F
By comparing the electronegativity values of atoms, one can compare the polarity of bonds.
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