chemical bonding h3

7/30/2019 Chemical Bonding H3

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Homework 7

Spring 2008

1)

a) Energy of an electron in a 1-D box.

En =n

2h

2

8mL

2

E1 =h2

8meL2=

(6.6261034J s)2

8(9.109 1031

kg)(5 5.2921011

m)2= 8.60510

19J

b) Energy of each electron the box representing an H2 molecule.

E1(H2) =(6.626 10

34J s)

2

8(9.1091031kg)(5 + 0.75)

2(5.292 10

11m)

2= 5.22 10

19J

c) Bond formation energy of forming the H2 molecule.

E = 2E1(H

2) 2E

1(H) = 2(5.2210

19) 2(8.605 10

19) = 6.76 10

19J

d) Convert bond formation energy and compare to 430kJ/mol.

E= (6.76 1019J)(6.02310

23mol

1)/1000 = 407.3kJ/mol

The approximation is not perfect but of the right magnitude. This agreement suggests that

delocalization of the electron density in H2, which lowers the kinetic energy of the ground state, plays a

key role in covalent bond formation.

2)

a) Calculate the energy of an electron in H2+.

EB = E1(H2+

) E1(H)

let H me

E1(H) =He

4

3222 h2

=

(9.109 1031kg)(1.602 1019C)4

322(8.854 1012C2J1m1)2(1.0551034Js)2

= 2.1771021J

E1(H) = (2.1771021J)

6.0231023

103

= 1310.6kJ/mol

E1(H2+

) = EB + E1(H) = 210+ 1310.6 = 1520.6kJ/mol

b) Since bond formation energy for each electron are additive:

H22+

+ 2e EB H2+

+ e EB H2

EB(H2) = 2(210kJ/mol) = 420kJ/mol

c) Compared to experimental bond formation energy of 430kJ/mol, possible differences could be:

Bond lengths of the two molecules are different; H2+ has a longer bond than H2 Electron-electron repulsion has been neglected.


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3)

a) Secular DeterminantsFor Linear H3 For cyclic H3

E 0

E

0 E

= 0

E

E

E

= 0

b) Solve the Secular Determinants

( E)3 2

2( E) = 0 ( E)

3+ 2

3 3

2( E) = 0

( E)[( E)2 2

2]= 0 [(-E)- ]

2[( E)+ 2]= 0

E =, 2 E=,,+ 2

c)Remember that = 210kJ/mol

2 > 2 so the E+ for the cyclic geometry will be more stable as it is a lower energy.

We can also calculate the bond formation energy for both geometries.

EB (linear) = 2(+ 2) 2 EB(cyclic) = 2(+ 2) 2

= 2 2 = 4

= 590kJ/mol = 840kJ/mol

Cyclic H3+ is more stable.

d) Formation energy for cyclic geometry was performed in part c, compare to 849kJ/mol. The

Huckel approximation is a slightly less negative number, but is a very good approximation.

e) Note that H3 has 3 electrons so 2 electrons will go into the lowest energy state while 1 electron willhave to go into the higher energy state. H3

- has 4 electrons so 2 electrons will go into the lowest energy

state and 2 electrons will go into the next higher energy state.

For the linear structure we know that the possible energies are: , 2.

For H3

EB= 2(+ 2) + 3= 2 2= 594kJ/mol

For H3

-

EB= 2(+ 2) + 2 4= 2 2= 594kJ/mol

So linear H3 and H3- are stable when compared to each other, but not as stable as H 3

+.

For the cyclic structure we know that the possible energies are: ,,+ 2.

For H3

EB= 2(+ 2)+ () 3= 3= 630kJ/mol

For H3

-

EB= 2(+ 2)+ 2() 4= 2= 420kJ/mol

So cyclic H3 is more stable than linear form or either form of H3-, but still not as stable as H3

+.


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4)

a) Solve the secular determinant to find the energies.

E

E= 0

E =

b) =-80 kJ/mol so EB= 2(+ ) 2= 2= 160kJ/mol

c)

ground excited

(in phase) (out of phase)

Note that these orbitals also resemble the ground and first excited states of a particle in a box, as the

ground state has no node and the excited state has one node (as is also the case for the ground, , and

excited, *, states of H2).

d) Secular determinant is the same as that of linear H3+

e) =-80 kJ/mol so EB

(allyl) = 2(+ 2) 2= 2 2= 226kJ/mol

f)

ground excited


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g) Energy gained by delocalizing the three carbon atom pz electrons over the molecular orbital.

h)

Edeloc = Eallyl Eethene = 2 22= (2 2 2) 0.84= 67.2kJ/mol

chemical bonding h3

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