Chapter 2 Practice Problems

Shapes predicted for H2Se, BF4-, NH4+

1. Bent 2. tetrahedral 3. Tetrahedral

Use molecular orbital diagrams to determine the number of unpaired electrons in (a) O2 (c) BN, (d) NO2.

a)2c)1d)1

predict that silicon-oxygen compounds are likely to contain networks of tetrahedra with Si-O single bonds and not discrete molecules with Si=O double bonds.

ΔS favors the gaseous state. As can be readily calculated for CO2, two pπ-pπ E=O double bonds are energetically more favorable than forming two extra E-O single bonds and this comes from the good overlap of the 2pAOs involved (Herr Professor Schrödinger can tell you more). For SiO2 the 3p-3p overlap is less; it has been proposed that the 2s2p core e⁻s prevent good overlap; 3p are also more diffuse. The formation of two additional Si-O single bonds is therefore energetically preferred as may be calculated

(a) Construct the form of each molecular orbital in linear [HHeH]^2+ using basis atomic orbitals on each atom and considering successive nodal surfaces. (b) Arrange the MOs in increasing energy.

(a) and (b) The number of molecular orbitals is equal to the total number of atomic orbitals. There are three atoms each with a 1s orbital. Therefore, there are 3 possible molecular orbitals: bonding, antibonding and non-bonding. The bonding orbital is the lowest in energy because all of the orbitals have the same phase. The non-bonding orbital is formed by one of the hydrogen's 1s orbital is in phase with helium's 1s but at the same time the other hydrogen's 1s orbital is out of phase. The antibonding is the highest in energy because all of the atoms alternate phases.

What shapes predicted for SO3, SO3-2, IF5

1. trigonal planar 2. trigonal pyramidal 3. Square pyramidal

Four elements arbitrarily labelled A,B,C, and D have electronegativities 3.8, 3.3, 2.8, and 1.3, respectively. Place the compounds AB, AD, BD, and AC in order of increasing covalent character

AD < BD < AC < AB ( increasing covalent character)tip: smaller is the electronegativity difference , more covalent is the compound

describe character for as mainly F or mainly S for the frontier orbitals

Molecular orbitals formed have more corresponding character depending upon the energy relation . HOMO is closed to energy of F orbitals so it must have more chara8 of fluorine . On the other hand LUMO has close energy to sulfur so it has more character of sulfur.

the common forms of nitrogen and phosphorus are N2(g) and P4(s) respectivly. Account for the difference in terms of the single and multiple bond enthalpies.

Single bond enthalpies have very less values when compared to multiple bond enthalpies.....because more energy is required to break the multiple bonds than single bonds Consider methane, CH4. It contains four identical C-H bonds, and it seems reasonable that they should all have the same bond enthalpy. However, if you took methane to pieces one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds. Every time you break a hydrogen off the carbon, the environment of those left behind changes. And the strength of a bond is affected by what else is around it. In cases like this, the bond enthalpy quoted is an average value. In the methane case, you can work out how much energy is needed to break a mole of methane gas into gaseous carbon and hydrogen atoms. That comes to +1662 kJ and involves breaking 4 moles of C-H bonds. The average bond energy is therefore +1662/4 kJ, which is +415.5 kJ per mole of bonds. That means that many bond enthalpies are actually quoted asmean (or average) bond enthalpies, although it might not actually say so. Mean bond enthalpies are sometimes referred to as "bond enthalpy terms".

Determine the bond orders of (a) S2, (b) Cl2, and (c) NO from their molecular orbital configurations and compare the values with the bond orders determined from Lewis structures. (NO has orbitals like those of O2)

The S2 will be going to follow the same bond order as the previous group atom O2 Molecular orbital configuration of O2 O2 = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p2 Number of electrons in bonding orbitals = 4 + 2 = 6 Number of electrons in anti-bonding orbitals = 2 BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-2) = 2 The Cl2 will be going to follow the same bond order as the previous group atom F2 Molecular orbital configuration of F2 O2 = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p4 Number of electrons in bonding orbitals = 4 + 2 = 6 Number of electrons in anti-bonding orbitals = 4 BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-4) = 1 Hence bond order of Cl2 will be 1 Molecular orbital configuration of NO NO = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p1 BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-1) = 2.5 The lewis structure of NO has a coordinate bond in order to satisfy the electron configuration, ut has two fixed bonds and one coordinate bond from O to N

In which of the species ICl6 and SF4 is the bond angle closest to that predicted by the VSEPR model?

bond angle in ICl6 is approximate 90 degree which is closest to that predicted by the VSEPR model. according to VSEPR model angle in ICL6 must be 90 degree. But in SF4 angle is 102 degree but accoding to VSEPR angle must be 109 degree

2.10 Use the concepts from Chapter 1, particularly the effects ofpenetration and shielding on the radial wavefunction, to account forthe variation of single-bond covalent radii with position in theperiodic table.

decreasing covalent radius going across because Zeff. down is due to principle quantum number.

Based on the MO discussion of NH3 in the text, find the average NH bond order in NH3 by calculating the net number of bonds and dividing by the number of NH groups.

no. of NH bonds are 3 & number of NH groups =3hence bond order = 1

Solid phosphorus pentachloride is an ionic solid composed on PCl4+ cations and PCl6- anions, but the vapor is molecular. What are the shapes of the ions in the solid?

the shapes of the ions in the solid are PCl4+ cations has tetrahedral because there are 4 P-Cl bonds which are arranged at 109.8 degrees ( sp3 hybridized) . and PCl6- anions has octahedral because there are 6 P-Cl bonds which are arranged at 90 degrees to each other ( sp3d2 hybridized)

Use the Ketelaar triangle in Fig. 2.38 and the electronegativity values in Table 1.7 to predict what type of bonding is likely to dominate in (a) BCl3, (b) KCl, (c) BeO.

x coordinate=average electronegativity y coordinate=electronegativity difference