Match List I with List II

Choose the correct answer from the options given below:

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

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

A. \([Ni(CO)_4]\): I. \(sp^3\)

The valence shell electronic configuration of ground state Ni atom is \(3d^8 4s^2\). All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field \(CO\) ligands approach \(Ni\) atom. The empty \(4s\) and three \(4p\) orbitals undergo sp3 hybridization and form bonds with \(CO\) ligands to give \([Ni(CO)_4]\).

B. \([Ni(CN)_4]^{2-}\): II. \(dsp^2\)

In \([Ni(CN)_4]^{2-}\), there is \(Ni^{2+}\) ion for which the electronic configuration in the valence shell is \(3d^8 4s^0\). In presence of strong field \(CN^-\) ions, all the electrons are paired up. The empty \(3d\), \(4s\) and two \(4p\) orbitals undergo \(dsp^2\) hybridization to make bonds with \(CN^-\) ligands in square planar geometry. Thus \([Ni(CN)_4]^{2-}\) is diamagnetic. It is said to be a low spin inner orbital complex.

C. \([Fe(CN)_6]^{4-}\): IV. \(d^2sp^3\):

In \([Fe(CN)_6]^{4-}\), There is \(Fe^{2+}\) ion for which the electronic configuration in the valence shell is \(3d^6 4s^0\). In presence of strong field \(CN^-\) ions, all the electrons are paired up. The two empty \(3d\), one \(4s\) and three \(4p\) orbitals undergo \(d^2sp^3\) hybridization to make bonds with \(CN^-\) ligands in octahedral geometry. Thus \([Fe(CN)_6]^{4-}\) is diamagnetic. 

D. \([MnF_6]^{4-}\): III. \(sp^3d^2\):

In \([MnF_6]^{4-}\), there is \(Mn^{2+}\) ion for which the electronic configuration in the valence shell is \(3d^5 4s^0\). In presence of weak field \(F^-\) ions, electrons are not paired up. Thus, the  one \(4s\), three \(4p\) and two \(5d\) orbitals undergo \(sp^3d^2\) hybridization to make bonds with \(F^-\) ligands in octahedral geometry. Thus \([MnF_6]^{4-}\) is diamagnetic and outer orbital complex.