1. Assign oxidation number to the underlined elements in each of the following species:
(a) CaO)
(d) KMnO4
(b) NaHSO4
(e) NaH PO4
Fe304
(f) NaBH4

Explanation:

When it comes to assigning oxidation number to elements, the following rules should be observed;

1. Oxidation number of a neutral compound is zero and is equal to the sum of the oxidation numbers of the element.

2. Oxidation number of hydrogen is +1 except in hydrides where it is -1. Oxidation number of oxygen is -2 except in peroxides where it is -1.

3. For some elements, their oxidation number is equal to their group number.

(a) CaO

Let the oxidation number of Ca be x

Using rules 1 and 2;

Oxygen = -2

x + (-2) = 0

x - 2 = 0  ; x = 2

(d) KMnO4

K belongs to group 1 hence has oxidation number of +1

O = -2

Let the oxidation number of Mn be x

+ 1   +  x   + 4(-2) = 0

1 + x -8 = 0

x = +7

(b) NaHSO4

Na belongs to group 1 hence has oxidation number of +1

O = -2

Hydrogen = +1

Let the oxidation number of S be x

+ 1   +  1  + x   + 4(-2) = 0

2 + x -8 = 0

x = +6

(e) NaH PO4

Na belongs to group 1 hence has oxidation number of +1

O = -2

Hydrogen = +1

Let the oxidation number of P be x

+ 1   +  1  + x   + 4(-2) = 0

2 + x -8 = 0

x = +6

Fe304

Let the oxidation number of Fe be x

Using rules 1 and 2;

Oxygen = -2

3x + 4(-2) = 0

3x - 8 = 0  ; x = 8/3

For Fe3O4, two Fe atoms have an oxidation state of +3 and one of +2

(f) NaBH4

Na belongs to group 1 hence has oxidation number of +1

Hydrogen = +1

Let the oxidation number of B be x

+ 1   + x   + 4(1) = 0

1 + x + 4 = 0

x = -5

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole