To explain bonding in complexes valence bond theory was proposed by Linus Pauling. The main postulates of VB theory are as follows:

(i) The central atom loses a requisite number of electrons to form the cation. The number of electrons lost is equal to the valence of the resulting cation.

(ii) The central cation makes available a number of vacant orbitals equal to its coordination number for the formation of dative bonds with the ligands.

(iii) The cation orbitals hybridize to form new set of equivalent hybrid orbitals with definite directional characteristics.

(iv) The non-bonding metal electrons occupy the inner d-orbitals and do not participate in the hybridization.

(v) In the presence of strong ligands such as CN^-, NO, CO, the d-electrons are rearranged vacating some d-orbitals (when the number of d-electrons are more than 3 only) which can participate in hybridization.

(vi) In the presence of weak ligands such as F^-, Cl^-, H₂O, etc., the d-electrons are not rearranged.

(vii) The d-orbitals involved in the hybridization may be either (n - 1)d orbitals or outer d-orbitals.

The complexes formed by the involvement of (n - 1)d orbitals in hybridization are called inner orbital complexes or low spin complexes. The complexes formed by the involvement of d-orbitals of outer orbit are called outer orbital complexes or high spin complexes.

(viii) Each ligand contains a lone pair of electrons. A dative bond is formed by the overlap of a vacant hybrid orbital of metal ion and a filled orbital of ligand.

(ix) The complex will be paramagnetic, if any unpaired electrons present, otherwise diamagnetic.

(x) The number of unpaired electrons in a complex gives out the geometry of the complexes or vice versa.

The complex ion \([Cu(NH_3)_4]^{2+}\) has

Square planar configuration with one unpaired electron

The correct answer is option 2. Square planar configuration with one unpaired electron.

The given complex is \([Cu(NH_3)_4]^{2+}\).

Here, the central metal atom is Copper (Cu), its electronic configuration is

\(_{29}Cu = _{18}Ar]4s^2 3d^9\)

Let the oxidation number of Cu be x

\(x + 0 = 2\)

or, \(x = 2\)

The electronic configuration of \(Cu^{2+}\) is

The ammonia ligands are weak field ligands, which means that they do not split the d orbitals into two energy levels. The four \(NH_3\) ligands undergo \(L\rightarrow M\) sigma bond to form 4 hybrid orbitals which lead to the dsp² hybridization resulting in a square planar geometry.

The square planar geometry of [Cu(NH₃)₄]²⁺ means that the four ammonia ligands occupy the four equatorial positions of the complex ion. This leaves the two axial positions empty. The two electrons in the d orbitals will occupy the two degenerate d orbitals that are oriented along the axial positions. This gives the complex ion a square planar configuration with one unpaired electron.