Ncert Exercises & Solutions

Match the compounds given in Column I with the oxidation state of

cobalt has given in column II and assigns the correct code.

Column I (Compound)	Column II (Oxidation state of Co)
A. $[Co (NCS) {({NH}_{3})}_{5}] ({SO}_{3})$	1. +4
B. $[Co {({NH}_{3})}_{4} {Cl}_{2}] {SO}_{4}$	2. +2
C. ${Na}_{4} [Co {(S_{2} O_{3})}_{3}]$	3. 0
D. $[C o_{2} {(C O)}_{8}]$	4. +3

Codes

A B C D

1. 2 3 4 1

2. 3 1 5 2

3. 5 4 3 2

4. 4 1 2 3

Hint: Oxidation state of CO ligand is zero

Calculate the oxidation state of CO in different complexes as follows:

A. $[Co (NCS) {({NH}_{3})}_{5}] ({SO}_{3})$

Let oxidation state of Co is x.

$+ 2 = - 1 + 5 \times 0 + x$
$x = + 2 + 1 = + 3$

B. $[Co {({NH}_{3})}_{4} {Cl}_{2}] {SO}_{4}$

Let oxidation state of Co is x

$\Rightarrow x + 4 \times 0 + 2 \times (- 1) = + 2$
$\Rightarrow x - 2 = + 2$
$x = 4$

c. ${Na}_{4} [Co {(S_{2} O_{3})}_{3}]$

Let oxidation state of Co is x

$X + 3 \times (- 2) = - 4$
$x - 6 = - 4$
$x = - 4 + 6 = + 2$

D. $[C o_{2} {(C O)}_{8}]$

Let oxidation state of Co is x

$x - 8 \times 0 = 0$
$x = 0$

Hence, option 4th is the correct answer.