Ch 5. Bonding And Intermolecular Forces Flashcards
Which one of the following anions cannot behave as a Lewis base/ligand?
A) F-
B) ОН-
C) NO3-
D) BH4-
D)
A Lewis base/ligand is a molecule or ion that donates a pair of nonbonding electrons in order to even be a candidate Lewis base/ligand, a molecule must have a pair of nonbonding electrons in the first place. The ion in choice D does not have any nonbonding electrons.
Carbon aroms with nonbonding electrons are excellent Lewis bases/ligands. Therefore, which of the following molecules is not a potential Lewis base/ligand?
A) CO2
B) CO
C) CN-
D) CH3-
A
Which of the folowing is most likedly an ionie compound?
A) NO
B) HI
C) CIF
D) KBr
A diatomic compound is ionic if the electronegativities of the atoms are very different. Of the atoms listed in the choices, those in choice D have the greatest electronegativity difference (K is an alkali metal, and Br is a halogen); K will give up its lone valence electron to Br, forming an ionic bond.
Determine the hybridization of the central atom in each of the following molecules or ions from the previous example:
A) H2O
B) SO2
C) NH4(+)
D) PCl3
E) CO3(2-)
A) Hybridization of O is sp3.
B) Hybridization of S is sp2.
C) Hybridization of N is sp3.
D) Hybridization of P is sp3
E) Hybridization of C is sp2
Identify the mixture of compounds that cannot experience hydrogen bonding with each other
A) NH3/H2O
B) H2O/HF
C) HF/CO2
D) H2S/HCI
D
Hydrogen bonding occurs when an H covalently bonded to an F, O, or N electrostatically interacts with another F, O, or N (which doesn’t need to have an H). Therefore, choices A, B, and C can all experience hydrogen bonding. Choice D, however, cannot, and this is the answer.
An understanding of intermolecular forces is of critical importance because they govern so many physical properties of a substance. The property least likely to be influenced by intermolecular force strength is:
A) color
B) melting point
C) solubility
D) vapor pressure
Any physical property that involves separating molecules from one another will very much depend upon the strength of intermolecular forces. Molecules are spread out during melting (choice B), dissolving (choice C), and evaporation (choice D). Choice A is therefore the best choice here.
Of the following, which one will have the lowest meting point?
A) MgO
B) CH4
C) Cr
D) HF
B
Almost all ionic compounds are solids at room temperature. Therefore, choice A is eliminated.
Similarly, all metals except for mercury (Hg) are solids at room temperature, so eliminate choice C. Both answers B and D will be molecular solids. Hydrogen fluoride is able to hydrogen bond and will therefore have stronger intermolecular interactions than the nonpolar methane. Since choice B has the weakest intermolecular forces (London dispersion), it will be easiest to melt.
Which of these molecules has the strongest dipole moment?
A) PBr3O
B) PF5
C) CCI4
D) SF6
A
A bond has a dipole moment when the two atoms involved in the bond differ in electio. negativity. However, an entire molecule can only have a dipole if it contains bond dipoles and is asymmetrical. Choice A is tetrahedral and not all four substituents are the same.
Therefore, it is asymmetrical and has a small negative dipole in the direction of the most electronegative substituent, oxygen. The remaining choices are trigonal bipyramidal, let. rahedral, and octahedral respectively. All have identical substituents, are symmetrical, and have no net dipole moment.
A pure sample of which of the following ions/molecules will participate in intermolecular hydrogen bonding?
I. CH3CO2H
II. СО2
III. H2S
A) l only
B) II only
C) l and II
D) I and III
A
In order to participate in intermolecular hydrogen bonding, a molecule must be able to act as both a hydrogen bond donor and acceptor. In order to act as a hydrogen bond donor, a molecule must possess a hydrogen (H) atom covalently bound to a nitrogen (N), oxygen
(0), or fluorine (F) atom. In order to act as a hydrogen bond acceptor, a molecule must have an oxygen, nitrogen, or Auorine atom with an unshared pair of electrons. CH CO,H meets both of these requirements, and is therefore a valid choice. CO, does not possess any hydrogen atoms and is therefore an invalid option. While H,S may seem like an enticing choice, sulfur is not sufficiently electronegative to produce hydrogen bonding when cova-lently bound to hydrogen atoms.
