1. We need the geometry associated with each hybridization and then count the number of $90^\circ$ angles between bond pairs.

2. Analyze each option:

   A. $dsp^3$ hybridization

   • Geometry: trigonal bipyramidal
   • Bond directions: 3 equatorial, 2 axial
   • Angles:
     • equatorial-equatorial = $120^\circ$
     • axial-axial = $180^\circ$
     • axial-equatorial = $90^\circ$
   • Number of $90^\circ$ bond pair angles: $2 \times 3 = 6$ (each of the 2 axial bonds is at $90^\circ$ to each of the 3 equatorial bonds)

   B. $sp^3d^2$ hybridization

   • Geometry: octahedral
   • There are 6 bond pairs.
   • Each bond is at $90^\circ$ to 4 others and $180^\circ$ to 1 opposite bond.
   • Total number of $90^\circ$ angles: $\frac{6 \times 4}{2} = 12$ (division by 2 avoids double counting)

   C. $dsp^2$ hybridization

   • Geometry: square planar
   • There are 4 bond pairs in one plane.
   • Adjacent bonds make $90^\circ$.
   • Number of $90^\circ$ angles: $4$

   D. $sp^3d$ hybridization

   • This also corresponds to trigonal bipyramidal geometry.
   • So number of $90^\circ$ angles is again: $6$

3. Compare all counts:

   • $dsp^3$: $6$
   • $sp^3d^2$: $12$
   • $dsp^2$: $4$
   • $sp^3d$: $6$

4. Therefore, the maximum number of $90^\circ$ angles is for octahedral geometry, i.e. $sp^3d^2$ hybridization.

$\boxed{\text{Option B}}$