The hybridization of Ag in the linear complex [Ag(NH)] is sp.

Explanation:
In order to determine the hybridization of Ag in the complex [Ag(NH)], we need to consider the electronic configuration of Ag and the ligand NH.

1. Electronic Configuration of Ag:
Silver (Ag) belongs to the 5th period of the periodic table and has an atomic number of 47. The electronic configuration of Ag is [Kr] 4d10 5s1.

2. Ligand NH:
The ligand NH, also known as ammonia, has the formula NH3. The central atom, nitrogen (N), has an atomic number of 7. The electronic configuration of N is 1s2 2s2 2p3.

3. Bond Formation:
In the complex [Ag(NH)], the nitrogen atom of NH3 acts as a ligand and donates a lone pair of electrons to form a coordinate bond with the Ag atom. The Ag atom uses its 5s1 and 4d10 electrons to form bonds.

4. Hybridization:
The hybridization of an atom is determined by the number of sigma bonds and lone pairs around it. In the case of Ag, it forms two sigma bonds with two nitrogen atoms in the NH3 ligands. Therefore, the hybridization of Ag is sp, as it involves one s orbital and one p orbital in its hybrid orbitals.

5. Linear Geometry:
The complex [Ag(NH)] has a linear geometry because the Ag atom is bonded to two nitrogen atoms in a straight line. This linear arrangement is consistent with the sp hybridization of the Ag atom.

Therefore, the hybridization of Ag in the linear complex [Ag(NH)] is sp.