Ncert Exercises & Solutions

Match the compounds (given in Column I) with the oxidation state of Cobalt (given in Column II) and assign the correct code.

	Column I (Compound)		Column II (Oxidation state of Co)
A.	$\left[\mathrm{Co}(\mathrm{NCS})\left(\mathrm{NH}_3\right)_5\right]\left(\mathrm{SO}_3\right)$	1.	+4
B.	$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{SO}_4$	2.	+2
C.	$\mathrm{Na}_4\left[\mathrm{Co}\left(\mathrm{S}_2 \mathrm{O}_3\right)_3\right]$	3.	0
D.	$\left[\mathrm{Co}_2(\mathrm{CO})_8\right]$	4.	+3

Codes:

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.

	Column I (Compound)		Column II (Oxidation state of Co)
A.	\(\left[\mathrm{Co}(\mathrm{NCS})\left(\mathrm{NH}_3\right)_5\right]\left(\mathrm{SO}_3\right)\)	1.	+4
B.	\(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{SO}_4\)	2.	+2
C.	\(\mathrm{Na}_4\left[\mathrm{Co}\left(\mathrm{S}_2 \mathrm{O}_3\right)_3\right]\)	3.	0
D.	\(\left[\mathrm{Co}_2(\mathrm{CO})_8\right]\)	4.	+3