Question

Write oxidation state and coordination number of [(3)42]

+ ion​

Answer 1

Explanation:

Oxidation state = +3. Coordination number = 6.

Co(III):3d

6

:t

2g

6

e

g

o

(ii) Oxidation state = +3. Coordination number = 6.

Cr(III):3d

3

:t

2g

3

(iii) Oxidation state +2. Coordination number = 4

Co(II):d

7

:e

4

t

2

3

(iv) Oxidation state = + 2. Coordination number = 6

Mn(II):d

5

t

2g

3

e

g

2