The number of compounds/species in which as an orbital is used in hybridization is[Ag(NH3)2]+, [NiCl4]2–, Fe(CO)5, [Cu(NH3)4]2+, [Ni(CN)4]2–, [Cr(H2O)6]3+, [Co(H2O)6]2+

To answer this question, we need to understand the concept of hybridization in chemistry. Hybridization is the concept of mixing atomic orbitals into new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory.

The d orbital is used in hybridization in transition metals and their compounds.

Let's analyze each compound:

1. [Ag(NH3)2]+: Silver (Ag) is in the +1 oxidation state. It does not use d orbitals in hybridization.

2. [NiCl4]2–: Nickel (Ni) is in the +2 oxidation state. It uses sp3 hybridization, which involves the 3d, 4s, and 4p orbitals.

3. Fe(CO)5: Iron (Fe) is in the 0 oxidation state. It uses dsp3 hybridization, which involves the 3d, 4s, and 4p orbitals.

4. [Cu(NH3)4]2+: Copper (Cu) is in the +2 oxidation state. It does not use d orbitals in hybridization.

5. [Ni(CN)4]2–: Nickel (Ni) is in the +2 oxidation state. It uses dsp2 hybridization, which involves the 3d, 4s, and 4p orbitals.

6. [Cr(H2O)6]3+: Chromium (Cr) is in the +3 oxidation state. It uses d2sp3 hybridization, which involves the 3d, 4s, and 4p orbitals.

7. [Co(H2O)6]2+: Cobalt (Co) is in the +2 oxidation state. It uses d2sp3 hybridization, which involves the 3d, 4s, and 4p orbitals.

So, the number of compounds/species in which d as an orbital is used in hybridization is 5: [NiCl4]2–, Fe(CO)5, [Ni(CN)4]2–, [Cr(H2O)6]3+, [Co(H2O)6]2+.