Nickel (Z = 28) combines with a uninegative monodentate ligand to form a diamagnetic complex [NiL$$_4$$]$$^{2-}$$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

Nickel has an atomic number of 28, so its electron configuration is [Ar] 4s² 3d⁸. In the complex [NiL₄]²⁻, the ligand L is uninegative (charge -1) and monodentate. The overall charge of the complex is -2. Let the oxidation state of nickel be x. Then, we write the equation: x + 4 × (-1) = -2. Solving this, x - 4 = -2, so x = +2. Therefore, nickel is in the +2 oxidation state.

The electron configuration of Ni²⁺ is [Ar] 3d⁸, meaning it has 8 electrons in the 3d orbitals. The complex is diamagnetic, which means it has zero unpaired electrons. Now, we must determine the geometry of the complex to understand the orbital splitting and electron pairing.

Since there are four ligands, the complex could be tetrahedral or square planar. In a tetrahedral field, the d-orbitals split into a lower-energy e set (two orbitals) and a higher-energy t₂ set (three orbitals). For a d⁸ configuration in a tetrahedral complex, the small splitting typically results in high-spin behavior. The e set can hold 4 electrons (all paired), and the remaining 4 electrons go into the t₂ set. With three orbitals in the t₂ set, Hund's rule applies: three electrons occupy each orbital singly, and the fourth electron pairs with one, resulting in two unpaired electrons. However, the complex is diamagnetic (zero unpaired electrons), so tetrahedral geometry is not possible.

Therefore, the complex must be square planar. In a square planar field, the d-orbitals split into four energy levels: dₓᵧ (highest energy), dₓ₂₋ᵧ₂ (next highest), d_z₂ (lower), and dₓᵧ/d_yz (degenerate, lowest). For a d⁸ configuration, the electrons fill the orbitals as follows: the lowest degenerate pair dₓᵧ/d_yz holds 4 electrons (two pairs), then d_z₂ holds 2 electrons (one pair), and dₓᵧ holds 2 electrons (one pair), leaving dₓ₂₋ᵧ₂ empty. All 8 electrons are paired, resulting in zero unpaired electrons, which matches the diamagnetic property.

For square planar geometry, the hybridization is dsp². This involves one d orbital (dₓ₂₋ᵧ₂), one s orbital, and two p orbitals, forming four hybrid orbitals directed towards the corners of a square.

Thus, the hybridization is dsp² and the number of unpaired electrons is zero.

Hence, the correct answer is Option A.