Match ListI with List II
| List I | List II |
| A. \([Co(NH_3)_6]^{3+}\) | I. \(sp^3\) |
| B. \([CoF_6]^{3-}\) | II. \(dsp^2\) |
| C. \([NiCl_4]^{2-}\) | III. \(d^2sp^3\) |
| D. \([Ni(CN)_4]^{2-}\) | IV. \(sp^3d^2\) |
Choose the correct options from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. A-III, B-IV, C-I, D-II
| List I | List II |
| A. \([Co(NH_3)_6]^{3+}\) | III. \(d^2sp^3\) |
| B. \([CoF_6]^{3-}\) | IV. \(sp^3d^2\) |
| C. \([NiCl_4]^{2-}\) | I. \(sp^3\) |
| D. \([Ni(CN)_4]^{2-}\) | II. \(dsp^2\) |
A. \([Co(NH_3)_6]^{3+}\):
Here the central metal is \(Co\), whose electronic configuration will be
\(_{27}Co = _{18}Ar] 4s^2 3d^7\)
Let the oxidation state of \(Co\) will be x
so, \(x +6(0)=3\)
or, \(x = 3\)
So, \(Co\) is in +III oxidation state so its configuration becomes

In the presence of \(NH_3\) a strong ligand, the \(3d\) electrons pair up leaving two \(d-orbitals\) empty. Hence, the hybridization is \(d^2sp^3\) forming an inner orbital octahedral complex.

B. \([CoF_6]^{3-}\)
Here the central metal is \(Co\), whose electronic configuration will be
\(_{27}Co = _{18}Ar] 4s^2 3d^7\)
Let the oxidation state of \(Co\) will be x
so, \(x + 6(-1)= 3\)
or, \(x = 3\)
So, \(Co\) is in +III oxidation state so its configuration becomes

In the presence of \(F\) a weak ligand, the \(3d\) electrons do not pair up. Hence, the hybridization is \(sp^3d^2\) forming an outer orbital octahedral complex.

C.\([NiCl_4]^{2-}\)
Here the central metal is \(Ni\), whose electronic configuration will be
\(_{28}Ni = _{18}Ar] 4s^2 3d^8\)
Let the oxidation state of \(Ni\) will be x
so, \(x + 4(-1)= -2\)
or, \(x = 2\)
So, \(Ni\) is in +II oxidation state so its configuration becomes

In the presence of \(Cl\) a weak ligand, the \(3d\) electrons do not pair up. Hence, the hybridization is \(sp^3\) forming an outer orbital octahedral complex.

D. \([Ni(CN)_4]^{2-}\)
Here the central metal is \(Ni\), whose electronic configuration will be
\(_{28}Ni = _{18}Ar] 4s^2 3d^8\)
Let the oxidation state of \(Ni\) will be x
so, \(x + 4(-1)= -2\)
or, \(x = 2\)
So, \(Ni\) is in +II oxidation state so its configuration becomes

In the presence of \(CN\) a strong ligand, the \(3d\) electrons pair up leaving two \(d-orbitals\) empty. Hence, the hybridization is \(dsp^2\) forming an inner orbital octahedral complex.
