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Chapter 9: Q68P (page 375)

Rank the members of each set of compounds in order of increasing the ionic character of their bonds. Use polar arrows to indicate the bond polarity of each:

(a) PCI3 , PBr3, PI3

(b) BF3 , NF3 , CF4

(c) SeF4 , TeF4, BrF3

Short Answer

Expert verified

The intermolecular force between the molecules are:

  1. PCI3 , PBr3 , PI3 : The order of the ionic character is:
    PBr3< PCI3< PI3.
  2. BF3 , NF3 , CF4 : The order of the ionic character is:
    NF3 < CF4 < BF3
  3. SeF4 , TeF4, BrF3 :The order of the ionic character is:
    SeF4 <BrF3 <TeF4 .

Step by step solution

01

Intermolecular Forces

Intermolecular forces are the interaction which are formed by the attraction of the two having opposite charges(partial positive and partial negative charge). The opposite charge are formed by the presence of the electron-negative atom in the molecule. Due to the presence of an electron-negative atom, there is an induced partial positive charge is generated on the electron-positive charge (or less electron-negative charge) atom. There will be an attraction between the both oppositely charges to form a bond.

Dipole-dipole Interaction: These interactions are the attractive forces which present in the polar molecules. Polar molecules have permanent dipoleswhich are produced by the presence of the opposite charge of the atom.

Ion-dipole interaction: These interactions are present in the ion and polar molecules. These interactions are similar to the dipole-dipole interaction.

London dispersion interaction: It is the weak bond which is formed for a short period of time. The bond is formed between the polar and non-polar molecules due to the phenomena of induction.

02

Explanation

PCI3, PBr3 , PI3 : In this molecule, halogens are involved with the electropositive metal P-atom. As we know, halogens are the most electronegative atom which tends to form a polar molecule and form an ionic bond. The size of Br-atom is larger than Cl-atom whereas Cl-atom is larger than F-atom. So, the order of the ionic character is: PBr3 <PCI3< PI3 .

BF3 , NF3 , CF4 : In BF3 molecule, there is the formation of the polar molecule due to the presence of electronegative F-atom with electropositive B-atom in which the difference between the electronegativity is large then CF4 and NF3 . In NF3 , F-atom is the most electronegative atom forming a bond with electronegative N-atom forming a dispersion force in which the difference between the electronegativity of N-atom and F-atom. In CF4 , there is the formation of the polar molecule due to the presence of electronegative F-atom with electropositive C-atom in which the difference between the electronegativity.

The order of the ionic character is:

NF3 < CF4 < BF3

SeF4 , TeF4 , BrF3 :In SeF4 molecule, there are electronegative F-atom belongs to halogen group form bond with Se-atom forming dipole-dipole interaction in which the difference in the electronegativity is less. In TeF4 , electronegative F-atom bonding with metal electropositive Te-atom forming dipole-dipole interaction but the difference between electronegativity is large then SeF4 and BrF3 . In BrF3, electronegative F-atom bonded with less electronegative Br-atom forming dipole-dipole interaction. The order of the ionic character is: SeF4 <BeF3 <TeF4 .

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Most popular questions from this chapter

Using the periodic table only, arrange the elements in each set-in order of decreasing EN:

(a) N, P, Si;

(b) Ca, Ga, As.

The average C-H bond energy in CH4 is 415kJ/mol. Use table 9.2(p. 353) and the following to calculate the average C-H bond energy in ethane(C2H6;C-C bond), in ethene (C2H4;C=C bond), and ethyne (C2H2;Cโ‰กC):

C2H6(g)+H2(g)โ†’2CH4(g)โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„ฮ”Ho=-65.07kJ/molC2H4(g)+2H2(g)โ†’2CH4(g)โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„ฮ”Ho=-202kJ/molC2H2(g)+3H2(g)โ†’2CH4(g)โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„โ€„ฮ”Ho=-376.74kJ/mol

Use Figure 9.20, p. 364, to indicate the polarity of each bond with partial charges

(a) Br-Cl

(b) F-Cl

(c) H-O

(d) Se-H

(e) As-H

(f) S-N.

Heats of reaction calculated from bond energies and from heats of formation are often, but not always, close to each other.

a) Industrial ethanol (CH3CH2OH ) is produce by a catalytic reaction of ethylene (H2C=CH2 ) with water at high pressure and temperatures. Calculateฮ”Horx for this gas-phase hydration of ethylene to ethanol, using bond energies and then using heats of formation.

b) ethylene glycol is produced by the catalytic oxidation of ethylene to ethylene oxide, which then reacts with water to form ethylene glycol:

The ฮ”Horx for this hydrolysis step, based on heat of formation, is -97kJ/mol. Calculate ฮ”Horx for the hydrolysis using bond energies.

c) why are two values relatively close for the hydration in part (a) but not close for the hydrolysis in part(b).

Even though so much energy is required to form a metal cation with a 2+ charge, the alkaline earth metals form halides with general formula

MX2, rather than MX.

(a) Use the following data to calculate the of MgCl:

Mg(s) Mg(g) = 148 kJ

Cl2(g) 2Cl(g) -243 kJ

Mg(g) (g) + e- =738 kJ

Cl(g) + (g) = -349 kJ

of MgCl = 783.5 kJ/mol.

(b) Is MgCl favoured energetically relative to Mg and ? Explain.

(c) Use Hessโ€™s law to calculate โˆ†Hยฐ for the conversion of MgCl to and Mg ( of = -641.6 kJ/mol).

(d) Is MgCl favoured energetically relative to ? Explain.
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