Question

9. (8 pts.) Four atoms were arbitrarily labelled D,E,F and G. Their Pauling electronegativity values are as follows D=3.5, E-3.3, F-2.4 and G-1.3. These elements form the following species: DD, DE, DF and DG a. Arrange the four species in order of decreasing covalent character: least covalent most covalent b. Of the four species, which is/are: ionic polar covalent nonpolar covalent

Answer 1

Answer: Hey dear student find the solution of your query, if you have any doubt feel free to ask. Thanks!! As we know: > The large electronegativity difference means the electron is gained or lost more than it is shared. > The small electronegativity difference means the attraction to atom is nearly the same. > Hence the lower the electronegativity difference means the greater the covalent character. The higher the electronegativity difference means the poor the covalent character. In the given question the Pauling electronegativity value of the elements are given as follows D=3.5 E=3.3 F= 2.4 G=1.3 The species form the given elements are: DD = Electronegativity difference = 3.5-3.5=0.0 DE = Electronegativity difference = 3.5-3.3=0.2 DF = Electronegativity difference=3.5-2.4=1.1 DG = Electronegativity difference = 3.5-1.3 = 2.2 a. The decreasing order of the given species on the bases of their covalent character: DD (most covalent) > DE > DF>DG (least covalent) Ionic compound: DG (Higher electronegativity difference) Polar covalent: DE (Small electronegativity difference but not equal to zero) Polar covalent: DD (electronegativity difference zero)