Find the hybridization and number of coordinate bonds formed in...

The given complex ion is tetracyanonickelate(II), which is $[Ni(CN)_4]^{2-}$.

The central metal ion is Ni in +2 oxidation state. The atomic number of Ni is 28, so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.

Since $CN^-$ is a strong field ligand, it causes the pairing of the $3d$ electrons. The 8 electrons pair up to occupy four $3d$ orbitals, leaving one $3d$ orbital empty.

The empty orbitals available for hybridization are one $3d$, one $4s$, and two $4p$ orbitals. Thus, the hybridization is $dsp^2$, resulting in a square planar geometry.

Since there are four monodentate $CN^-$ ligands, each donates a pair of electrons to the central metal ion, forming a total of 4 coordinate bonds.

Therefore, the hybridization is $dsp^2$ and the number of coordinate bonds is 4.

Answer: $dsp^2$ and 4 coordinate bonds.