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Chemical Bonding - Preparatory Chemistry - Structures | CHE 100, Assignments of Chemistry

Material Type: Assignment; Class: Preparatory Chemistry; Subject: Chemistry (CHE); University: Monroe Community College; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/09/2009

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Na + F Na [ F ] NaF
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Mg Mg
N [N]
Mg Mg Mg N
N [N]
Mg Mg
2+
3-
2+
3-
2+
32
Cl [Cl]
Ba + Ba BaCl
Cl [Cl]
-
2+
-
2
KK
O+ [O] KO
KK
+
2-
+
2
Br
[Br ]
Al + Br Al [Br ] AlBr
[Br ]
Br
-
3+ -
-
3
CHE100
Chemical Bonding
1.
(a)
(b)
(c)
(d)
(e)
pf3
pf4

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Na + F Na [ F ] NaF

.^.

..^

.^.

Mg Mg

N [ N ]

Mg Mg Mg N

N [ N ]

Mg Mg

Cl [ Cl ]

Ba + Ba BaCl

Cl [ Cl ]

K K

O + [ O ] K O

K K

Br

[Br ]

Al + Br Al [Br ] AlBr

[Br ]

Br

CHE

Chemical Bonding

(a)

(b)

(c)

(d)

(e)

  1. (a) OBr (^2)

O 1 x 6e = 6e Br 2 x 7e = 14e total = 20e

:Br O Br

: : :^ :

O

Br Br

(b) ClO 4 - Cl 1 x 7e = 7e O 4 x 6e = 24e charge 1e total = 32e

O

O Cl O

O

:^

[ [

Cl

O

O O O

(c) SO (^3) S 1 x 6e = 6e O 3 x 6e = 18e total = 24e

24e - 20e = 2 bonds 2e

32e - 24e = 4 bonds 2e

40e - 32e = 4 bonds 2e

32e - 24e = 4 bonds 2e

40e - 32e = 4 bonds 2e

O S O

O

O C O

O

O N O

O

S

C

N

O

O

O

O

O

O O

O

O

(d) NO 3 - N 1 x 5e = 5e O 3 x 6e = 18e charge = 1e total = 24e

[ [

(e) PO 4 3-

..^ ..

.^.

O

O P O

O

[ [

P 1 x 5e = 5e O 4 x 6e = 24e charge = 3e total = 32e 3-

P

O

O

O O

(f) H O 3 + H 3 x 1e = 3e O 1 x 6e = 6e charge = -1e total = 8e 14e - 8e = 3 bonds 2e

H O H

H

[ [

O

H H H

(g) SF (^2) S 1 x 6e = 6e F 2 x 7e = 14e total = 20e 24e - 20e = 2 bonds 2e

:F S F:

S

F F

(h) CO 3 2- C 1 x 4e = 4e O 3 x 6e = 18e charge = 2e total = 24e 32e - 24e = 4 bonds 2e

[ [

molecular: bent

e-pair: tetrahedral

molecular: tetrahedral

molecular: trigonal palnar

molecular: trigonal palnar

molecular: tetrahedral

molecular: bent

e-pair: tetrahedral

e-pair: tetrahedral

molecular: trigonal pyramidal

molecular: trigonal planar

  1. (a) H CCH 3 3 H 6 x 1e = 6e C 2 x 4e = 8e total = 14e

28e - 14e 2

= 7 bonds

24e - 12e 2

= 6 bonds

20e - 10e 2

= 5 bonds 28e - 18e 2

44e - 32e 2

52e - 26e 2

46e - 32e 2

32e - 24e 2

= 5 bonds

= 6 bonds

= 13 bonds

= 7 bonds

= 4 bonds

H H

H C C H

H H

(b) H CCH 2 2

H 4 x 1e = 4e C 2 x 4e = 8e total = 12e

H 2 x 1e = 2e C 2 x 4e = 8e total = 10e

H 10 x 1e = 10e C 4 x 4e = 16e total = 26e

B 1 x 3e = 3e F 3 x 7e = 21e total = 24e

H 2 x 1e = 2e C 1 x 4e = 4e O 2 x 6e = 12e total = 18e

H 2 x 1e = 2e S 1 x 6e = 6e O 4 x 6e = 24e total = 32e

H 3 x 1e = 3e P 1 x 5e = 5e O 4 x 6e = 24e total = 32e

H H

C C

H H

(c) HCCH

H C C H

(d) HCOOH

O

H C O H

(e) H SO 2 4

..^

O

H O S O H

O

(f) H PO 3 4

..^

O

H O P O H

O

H

(g) H CCH CH CH 3 2 2 3

H H H H

H C C C C H

H H H H

(h) BF (^3)

..^

..^

F

B F

F