chem test 3

2pz orbitals

meet head on , which means stronger interaction, more stable, so they will have lower energy (sigma 2p) than those meeting edge on.

degenerate orbitals

orbitals that have the same energy. fill according to hund's rule.

antibonding MO

sigma star orbital. ot's orbital forces the electron to spend most of its time away from the area between the nuclei, placing an e- in that orbital makes the molecule less stable. more energy.

important MO principle

the core orbital of an atom make no contribution to the stability of the molecules that contain the atom. the MO that are important are those formed when valence shell orbitals are combined. concentrate on 2s and 2p valence orbitals.

Adding electrons to MO

they are added one at a time starting with the lowest energy molecular orbital. H2 molecule is more stable than isolated H atoms. the energy of h2 molecule is lower than the isolated H atoms

B2, C2, N2 MO

when 2s orbital on one atom interacts with 2pz

describe the MO formed by the overlap of the following atomic orbitals a. 2s+2s b. 2px+2px c. 2py+2py d. 2pz+2pz e. 2s+2pz

(a) (σ2s) and (σ∗2s) (b) (πx) and (π*x) (c) (πy) and (π*y) (d) (σp) and (σ*p) (e) (σs−p) and (σ*s-p)

write electron config for the following diatomic molecules and calculate bond order: a. H2 b. C2 c. N2 d. O2 e. F2

(a) H2 has 2 e- (σ1s)2 BO = 2/2 = 1 (b) C2 has 8 e- (σ2s)2 (σ∗2s)2 (π x)2 (π y)2 BO = (6-2)/2 =2 (c) N2 10 e- (σ2s)2 (σ∗2s)2 (π x)2 (π y)2 (σp)2 BO = (8-2)/2 = 3 (d) O2 12 e- (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)1 (π*y)1 BO = (8-4)/2 = 2 (e) F2 14 e- (σ2s)2 (σ∗ 2s)2 (σp)2 (π x)2 (π y)2 (π* x)2 (π* y)2 BO = (8-6) 2 = 1

list all of the IMF that each would have a. SO2 b. CH3OH c. ICl3 d. SF4

(a) SO2: dispersion and dipole-dipole (b) CH3OH: dispersion, dipole-dipole, and hydrogen bonding (c) ICl3: dispersion and dipole-dipole (d) SF4: dispersion and dipole-dipole

color portrayal

. The Cu(NH3)4(2+) complex ion has a blue color because it absorbs light in the yellow portion of the spectrum. The CrO4(2-) ion appears yellow because it absorbs blue light. T

. Compare the positions of the Co2+ with Co3+ and Fe2+ with Fe3+ ions in the spectrochemical series. Explain why the value of delta generally increases with the charge on the transition metal ion.

As the charge of the metal increases, the ligands are pulled closer to the metal by attraction of the more highly charged metal ion to the electrons of the ligands. The closer the ligands are, the more they interact with the eg orbitals of the metal, increasing their energy relative to the t2g orbitals.

CrO4(2-) ions are bright yellow. In what portion of the visible spectrum do the ions absorb light?

A compound that looks yellow, like CrO4(2-), absorbs the opposite color of yellow, which is violet

is O22- paramagnetic ?

A molecule is paramagnetic if it contains unpaired electrons. The peroxide ion, O22- has the electron configuration: (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)2 (π*y)2 All the electrons are paired. The peroxide ion is not paramagnetic.

When CrO4(2-) reacts with acid to form Cr2O7(2-) ions, the color shifts from bright yellow to orange. Does this mean that the light absorbed shifts toward a higher or a lower frequency?

A shift in color from yellow to orange indicates that the light being absorbed shifts from violet to blue. This is a shift to lower energy of light, which corresponds to a lower frequency.

carboxylic acids with the general formula ch3(ch2)nCO2H have a NONPOLAR Ch3CH2 tail and a POLAR CO2H head. What effect does increasing the value of n have on the solubility of carboxylic acids in polar solvents, such as water? what about solubility in a nonpolar solvent?

As the length of the nonpolar tail increases, the polar carboxylic "head" of the molecule becomes less significant in determining the physical properties of the compound. Therefore, the carboxylic acids with longer "tails" are less soluble in polar solvents such as water and more soluble in the nonpolar solvent, carbon tetrachloride.

. The difference between the energies of the t2g and eg sets of atomic orbitals in an octahedral or tetrahedral crystal field depends on both the metal atom and the ligands that form the complex. Which of the following metal ions would give the largest difference? (a) Rh3+ (b) Cr3+ (c) Fe3+ (d) Co2+ (e) Mn2+

BASED ON SPECTROCHEMICAL SERIES Rh3+ . The overall order would be Rh3+ > Cr3+ > Fe3+ > Co2+ > Mn2+

should the h2+, h2-, and h22- ions be more or less stable thanthe neutral h2 atom?

Calculate the bond order for the molecules. The molecule with the highest bond order is more stable. (a) H2 + 1 electron (σ1s)1 BO = 1/2 (b) H22- 4 electrons (σ1s)2 (σ*1s)2 BO = 0 (c) H2- 3 electrons (σ1s)2 (σ*1s)1 BO = 1/2 H2 (BO = 1) is more stable than H2 + and H22-.

is O2- more stable than o2? which has a larger bond order?

Calculate the bond order for the molecules. The molecule with the highest bond order is more stable. O2 12 e- (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)1 (π*y)1 BO = 2 O2- 13 e- (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)2 (π*y)1 BO =1.5 (3/2) O2 with a bond order of 2 is more stable than O2- with a bond order of 1.5.

why is O2 more stable than O22-?

Calculate the bond order for the molecules. The molecule with the highest bond order is more stable. O2 12 e- (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)1 (π*y)1 BO = 2 O22- 14 e- (σ2s)2 (σ∗2s)2 (σp)2 (π x)2 (π y)2 (π*x)2 (π*y)2 BO = 1 O2 is more stable than O22-.

Explain why Cu(NH3)4(2+) complexes have a deep-blue color if they don't absorb blue light. What light do they absorb?

Cu(NH3)4 2+ absorbs orange light (the opposite color of blue). The remaining light is reflected and the combination of these colors is perceived as blue in color.

One of the Fe(H2O)6(2+) and Fe(CN)6(4-) complex ions is high-spin and the other is lowspin. Which is which?

Fe(CN)6(4-) is low spin and Fe(H2O)6(2+) is high spin. Both are Fe2+ ions and CN- creates a larger splitting than H2O according to the spectrochemical series.

why are hydrocarbons not soluble in water?

Hydrocarbons do not possess any hydrophilic groups.

why do induced dipole-induced dipole forces increase as the number of electrons in a molecule increases?

In larger atoms where electron configurations involve more electrons, the atoms become more polarizable. The extent of "induced dipoles" increases as the atoms become more polarizable, which acts to increase the interaction process.

Which of the following ligands would give the largest value of Delta-o? (a) CN- (b) NH3 (c) H2O (d) OH- (e) F-

Metal ions at one end of the continuum are called strong-field ions, because the splitting due to the crystal field is unusually strong. Ions at the other end are known as weak-field ions. BASED ON SPECTROCHEMICAL SERIES CN- . The overall order would be CN- > NH3 > H2O > OH- > F-

Ni2+ forms a complex with the dimethylglyoxime (DMG) ligand that absorbs light in the blue-green portion of the spectrum. What is the color of the Ni(DMG)2 complex?

Ni(DMG)2 would appear to be the opposite color of blue-green, which is red or red-orange

explain the trends in the following data that show the melting point and dipole moment for the hydrogen halides HCl , -114 BP, 1.08 HBr, -87 BP, 0.18 HI, -51 BP, 0.45

One of the intermolecular forces in the hydrogen halides is the dipole-dipole force. As the dipole moment decreases, so too does the strength of the dipole-dipole force. As the strength of the intermolecular forces decreases the easier it is for the molecules to break free of the rigid solid lattice. The easier it is to break the solid structure, the lower the melting point must be. However, this is NOT the trend that is observed. The other force involved in hydrogen halides is dispersion. Dispersion forces generally increase both with molecular weight and polarizability of a species. The iodine atom is much heavier and is more polarizable (with many more electrons farther away from the nucleus) than the chlorine atom, therefore the dispersion forces in HI will be greater than in HCl. This is what causes the HI to have a higher melting point than HCl.

How much energy is associated with the absorption of a typical photon, such as a photon with a wavelength of 480 nanometers?

Since we know the wavelength of the photon, we can calculate its frequency using v= c/wavelength use the new found v to calculate the energy of a single photo E = hv --> J multiply by avogadro's number to get kJ/mol

order in increasing solubility in water (polar) a.NaCl b.CH3CH2C2CH c.CH3CH2OH d.CH2COOH

Solubility NaCl > CH3COOH > CH3CH2OH > CH3CH2CH2CH3 NaCl is an ionic compound which is readily soluble in polar water. The acid group ion CH3COOH makes this quite soluble, but not as much as an ionic species. The -OH group on CH3CH2OH is not as polar as the acid, making this less soluble. There are no hydrophilic groups on CH3CH2CH2CH3, making it the least soluble.

In an octahedral field what do the orbitals in a t2g set have in common? What do the orbitals in the eg set have in common?

The (eg) 0dx2 - y2 and dz2 orbitals on the metal ion at the center of the cube lie between the ligands---The eg orbitals (the dx2-y2 and dz2 orbitals) have the lobes of the orbitals oriented directly on the coordinate axes and therefore interact strongly with the electrons from the ligands. the (t2g) dxy, dxz, and dyz orbitals point toward the ligands. The coordinate axes are where the ligands are located so these orbitals do not interact strongly with the electrons from the ligands.

predict the order in which the boiling points of the following compounds should increase. Explain your reasoning. a. NH3 b. Ph3 c. AsH3

The predicted order of boiling temperatures is PH3 < AsH3 < NH3. NH3 undergoes hydrogen bonding. The other compounds experience dipole-dipole interactions and dispersion forces. Both PH3 and AsH3 have dipole-dipole forces, but the dipoles are very similar, so these forces would be the same for both. Since dispersion forces are directly related to molar mass, the remaining compounds are from left to right in order of increasing molar mass

explain why the dif between the energies of the sigma2p bonding and antibondinh orbitals is larger in difference than the difference between Pi-x or Pi-y bonding and Pi-x and Pi-y antibonding?

The stronger the interaction between a pair of atomic orbitals, the larger the difference between the energies of the bonding and antibonding orbitals formed. The σp molecular orbitals result from the head-on overlap of 2pz atomic orbitals. This is a stronger interaction than the sideways overlap of the 2px and 2py atomic orbitals that results in the πx and πy molecular orbitals. Therefore the difference in energy between the σp and the σ*p will be greater than the difference in energy between the πx πy and π*x π*y molecular orbitals.

I2 dissolving in CCl4

When I2 molecules dissolve in CCl4 solvent, the I2 units become surrounded by layer upon expanding layer of solvent molecules. Each successive solvent layer becomes less and less different from the nature of the bulk solvent.

diamagnetic

atoms or molecules in which the electrons are paired. they are repelled by both poles of a magnet.

pi bonds

concentrate the electrons above and below the axis rather than along the axis.

He2 MO

each helium atom has a 1s2 confi, which means there would be equal pairs of e- in both sigma and sigma* orbitals, making the energy equal of that of isolated He atoms, nothing to hold the atoms together

bonding molecular orbital

electrons that are placed in this orbital spend most of their time in the region directly between the two nuclei. they are called sigma MO bc looks like an s orbital when viewed along the H-H bond. Placing an e- in the sigma bonding MO stabilizes the molecule

paramagnetic

have one or more unpaired electrons . attracted to a magnetic field