Let’s analyze each compound one by one and its corresponding properties.
Step 1: Analyzing Compound (A) [Ni(NH\(_3\))\(_6\)]\(^{2+}\).
The given complex is octahedral, and since it is formed by a \(Ni^{2+}\) ion (d\(^8\) configuration), it will have 2 unpaired electrons. Therefore, it is paramagnetic. The hybridization is \(sp^3d^2\) for an octahedral geometry. Thus, statement (A) is correct.
Step 2: Analyzing Compound (B) [Ni(CO)\(_4\)].
This complex has a tetrahedral geometry with \(Ni^{0}\) (d\(^8\) configuration). The \(CO\) ligand is strong and leads to no unpaired electrons, making it paramagnetic. Therefore, the hybridization is \(sp^3\) for a tetrahedral complex, and statement (B) is correct.
Step 3: Analyzing Compound (C) [Ni(CN)\(_4\)]\(^{2-}\).
For this complex, \(Ni^{2+}\) (d\(^8\)) is in a square planar geometry with \(dsp^2\) hybridization. The \(CN^-\) ligand is strong and leads to pairing of electrons, resulting in a diamagnetic complex. Thus, statement (C) is correct.
Step 4: Analyzing Compound (D) [NiCl\(_4\)]\(^{2-}\).
For this complex, \(Ni^{2+}\) (d\(^8\)) is in a tetrahedral geometry with \(sp^3\) hybridization. The chloride ion is weak, so the complex will be paramagnetic, with unpaired electrons. Therefore, statement (D) is correct.
Step 5: Conclusion.
The correct set of coordination compounds against the mentioned properties is option (C), as statements A, C, and D are correct.
Final Answer: Option (C) A, C and D.