Match ListI with List II

Choose the correct options from the options given below:

A-III, B-IV, C-II, D-I

The correct answer is (2) A-III, B-IV, C-II, D-I.

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.