6. PROPUESTA : MODIFICACIÓN DEL MARCO TEÓRICO COBIT . 54
6.3. Mapeo de las Metas TI y los Procesos Relacionados con las TI
6.3.3. Entregar, dar Servicio y Soporte
1. Draw the shape (bondary surfaces) of the following orbitals ;
(a) 2py (b) 3dz2 (c) 3dx2-y2 (Show coordinate axes in you sketches.) 2. Discuss the similarities and differences between a 1s and 1 2s orbital.
3. For each of the following pair of hydrogen orbitals, indicated which is higher in energy ; (a) 1s, 2s ; (b) 2p, 3p (c) 3dxy, 3dyz, (d) 3s, 3d (e) 4f, 5s 4. Which orbital in each of the following pairs is lower in energy in amny-electron atom ?
(a) 2s, 2p (b) 3p, 3d (c) 3s, 4s (d) 4d, 5f 5. Explain the meaning of the symbol 4d6.
6. The ground state electron configurations listed here are incorrect. Explain what mixtakes have been made in each and write the correct electron configuration.
Al : 1s22s22p43s23p3 B : 1s22s22p5 C : 1s22s22p6
7. Draw orbital diagrams for atoms with the following electronic configuraiton : (a) 1s22s22p5 (b) 1s22s22p6 (c) 1s22s22p63s23p64s23d7
8. Two p orbitals from one atom and two p orbitals from another atom are combined to form molecuar orbitals. How many MOs will result from this combination ? Explain.
9. Show the shapes of bonding and antibonding MOs formed by combination of (a) two s orbital (b) two p orbitals (side to side)
10. How do the bonding and antibonding MOs formed from a given pari to AOs compare to each other with respect to (a) energy (b) presence of nodes (c) internuclear electron density ?
11. Arrange the following species in order of increasieng stability : Li2, Li2+, Li2 -Justify your choice with a molecular orbitla energy level diagram.
12. Use molecular orbitla thoery to explain why the Be2 molecule cues not exist.
13. Explain why the bond order of N2 is greater than N2+, but the bodn order of O2 is less than that of O2+. 14. Compare the relative stability fo the followign species and indicating their magnetic properties (diamagnetic
or paramagnetic) ; O2, O2+, O2- (superoxide), O22- (peroxide ion)
15. Explain the significance of bond order. Can bond order be used for quantiative compoarisons of the strengths of chemcial bonds ?
16. What is the energy gap in band theory ? compare its size in conductors, semiconductors and insulators.
17. Which of the following substances exhibit H-bonding ? Draw the H bonds between two molecules of the substnace where appropriate :
(A) CH3CH2OH (B) CH C OH
||
O
3 (C) CH3 C|| CH3
O
(D) CH3 C|| NH2 O
18. How can one nonpolar molecule inducea dipole in a nearby nonpolar molecule ?
19. What type(s) of intermolecualr forces exist between the following pairs ? (a) HBr and H2S (b) Cl2 and CBr4, (c) I2 and NO3- and (d) NH3 and C6H6 ?
(Hint : To identify intermolecular forces, it is useful to classify the species being considered as (1) nonpolar molecules, (2) polar molecules, and (3) ions. Keep in mind that dispersion forces exist between all species.)
SOLUTIONS
1. (a) 2py
(b) 3dz2
(c) 3dx2–y2
2. Similarities between 1s and 2s
(1) both are non-directional. (2) spherical in shape
(3) the density of charge cloud is maximum at the nucleus and decrease with increase in distance from the nucleus both in 1s and 2s.
Difference :
(1) The principal quantum no. for 1s orbital is 1 but in 2s is 2.
(2) the 2s have 1 node or nodal plane but 1s have no any node.
(3) the energy of 2s orbital is higher than 1s orbital.
3. (a) 2s (b) 3p (c) 3dxy & 3dz have equal energy.
(d) for hydrogen 3s and 3d orbital have identical energy.
(e) 5s has higher energy than 4f orbital for hydrogen.
4. According to Aufbau rule, energy of the orbital increases
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s ...
In multi-electron systems, the relative order of energies can also be predicted with the help of ‘n + 1’
rule or BOHR-BURY RULE
According to this
(i) in neutral atoms, a subshell with lower value of (n + 1), has lower energy
(ii) If two subshell have equal value of (n + 1), the subshell with lower value of n has lower energy (a) out of 2s and 2p, 2s is filled first ; its n + 1 is lower or 2s has lower energy than 2p.
(b) Energy of 3p is lower than 3d
3p 3d
n = 3 n = 3
= 1 = 2
n + 1 = 4 n + 1 = 5
(c) 3s is lower in energy than 4s
3s 4s
n = 3 n = 4
= 0 = 0
n + 1 = 3 n + 1 = 4
(d) 4d is lower in energy than 5f
4d 5f
n = 4 n = 5
= 2 = 3
n + 1 = 6 n + 1 = 8
5. 4d6
Principal energy level = 4 Number of electrons = 6 Subshell = d
‘4’ indicates the principal energy level. ‘4d’ indicates the d-subshell of the 4th valence shell. 6 indicated the number of electrons present in ‘4d’ subshell.
6. Al = 1s2 2s2 2p6 3s2 3p1
Here 3s2 3p3 is filled first before completing 2p subshell, which as to attain 6 electrons before electron start filling 3s or 3p.
B = 1s2 2s2 2p1 Atomic no. of B = 5 F = 1s2 2s2 2px2 2py2 2p1 [ Atomic no. of F = 9]
7. (a) F = 9 = 2, 7
(b) P = 15 = 2, 8, 5
(c) Co = 2, 8., 15, 2
8. For example if we take oxygen molecule both the oxygen atoms contains two p–orbitals.
If we take one p–orbital from one oxygen atom and another p–orbital from another oxygen atom then
If we take another set of p-orbitals each from each oxygen atom then
So we get 4 molecular orbitals from the combination of two p-orbitals from two atoms.
9. (a) From two s–orbitals
The formation of molecular orbitals by linear combination of two s orbital is as shown below
(b) From two p–orbital (side–side)
When two 2p orbitals which have mutually parallel axis interact to give rise molecular orbitals as shown below
side wise approach 10. In combine state
(A + B)2 = A2 + B2 + 2AB
Uncombined state
(A – B)2 = A2 + B2 – 2AB
Nuclear – nuclear repulsion will be more energy in more stability is less.
(a) Energy - Energy of the antibonding orbital is raised above the energy of the parent atomic orbitals that have combined and the energy of the bonding orbital has lowered than the parent orbitals. The total energy of two molecular orbitals however remains the same as that of the two original atomic orbitals.
(b) presence of node & (c) internuclear electron density - bonding molecular orbital most of the electron density is located between the nuclei of the bonded atoms and hence the repulsion between the nuclei is very low while in an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei, as a matter of fact there is a nodal plane (i.e., plane in which the electron density is zero.)
11. The electronic configuration of Li is 1s22s1. The M. O. energy level diagram of Li2 is as follows For Li2
Bond order = 2
1 (Nb – Na)
= 2
1(4 – 2) = 2 2= 1
For Li2+ one electron should be removed from the bonding molecular orbital so bond order will be
= 2
1 (Nb – Na)
= 2
1 (3 – 2) = 2 1 = 0.5
for Li2– one electron should be added to the antibonding molecular orbital so bond order will be
= 2
1(4 – 3) = 2 1 = 0.5
The greater the bond order of a diatomic molecule the more stable it will be
So in between Li2, Li2+, Li2– Li2 is more stable than both Li2+, Li2– stability order is Li2, Li2+, Li2– In Li2+, the positive charge is more that is the nucleus will attract the electrons so size will be less than Li2– & stability is more.
12. The electronic configuration of Be is 1s22s2 so the M. O. diagram is as follows
Here the Bond order = 2
1 (Nb – Na)
= 2
1(4 – 4) = 0
If the bond order is zero i.e., the molecule doesn’t exist.
13. The electronic configuration of N2 molecule is (one electron is removed from bonding molecular orbital) So the bond order of N2+ is less than N2.
The electronic configuration of O2 molecule is
(1s)2 (1s*)2 (2s)2 (2s*)2 (2px)2 (py2py)2 (pz2pz)2 (*y2py)1 (*z2pz1)
(i.e., One electron is removed from antibonding orbital i.e., p*) so bond order of O2+ is greater than O2
14. The electronic configuration of O2 is as follows O2
(1s)2 (1s*)2 (2s)2 (*2s)2 (2pz)2 (2px2 = py2) (2px1 = *2py1)
As the bond order increases stability increase so the stability order is O2+ > O2 > O2– > O22–
Magnetic Properties
for O2 there are two unpaired electron so it is paramagnetic in nature. For O2+ it also contain one unpaired electron so it is also paramagnetic in nature. For O2– it also has one unpaired electron so it is also paramagnetic in nature. but O22– is diamagnetic due to absence of unpaired electron.
15. Bond order = 2
1 (Nb – Na)
A positive bond order means a stable molecule while a negative (or) zero bond order means an unstable molecule. Bond length decreases bond order increases.
Yes bond order can be used for quantitative comparison of the strength of the chemical bonds.
16. The highest occupied energy bond is valence bond while the lowest unoccupied energy bond is conduction band. The energy gap between the top of the valence bond and the bottom of the conduction band is called the energy gap (Eg). In case of insulators the energy gap is very large and in case of semiconductor the energy gap is very small and in case of conductors the energy gap is zero.
Formation of energy bands in (a) Sodium crystal (b) Magnesium crystal
17. (A) (intermolecule hydrogen bonding)
(B) No H–bonding is present in the molecule
The energy of a hydrogen bond is of the order 40 kJ/mole.
(C) (intermolecule hydrogen bonding)
(D) (intermolecule hydrogen bonding)
18. If in an atom the electrons are moving at some distance from the nucleus. At any instant it is likely that the atom has a dipole moment created by the specific position of electrons. It is instantaneous dipole which lasts for just a fleeting moment and in the next instant the electron are in different positions and the atom has a new instantaneous dipole and so on.
19. (a) HBr & H2S dipole-dipole interaction (keesen force)
(b) Cl2 & CBr4 Instaneous dipole-instaneous induced dipole (London force) (c) I2 & NO3- dipole induce dipole
(d) NH & CH dipole induce dipole