1. Find the steric number of the central atom

For halogen-centered interhalogen ions, we count:

• number of bonded atoms around the central atom,
• number of lone pairs on the central atom.

If the total number of electron pairs around the central atom is 6, the hybridization is $sp^3d^2$.

2. Check each option

Option A: $[ICl_2]^-$

Let iodine be central.

• Iodine has 7 valence electrons.
• Negative charge adds 1 electron.
• Total on central atom basis: $8$ electrons.
• Two $I-Cl$ bonds use 2 electron pairs for bonding.
• Remaining electrons on iodine give 3 lone pairs.

So around iodine:

• Bond pairs = 2
• Lone pairs = 3
• Total = $5$ electron pairs

Hence hybridization is $sp^3d$.

So, A is not correct.

Option B: $[ICl_4]^-$

Again iodine is central.

• Iodine has 7 valence electrons.
• Negative charge adds 1 electron.
• Total = $8$ electrons.
• Four $I-Cl$ bonds are formed.
• Remaining electrons on iodine = 2 electrons = 1 lone pair.

So around iodine:

• Bond pairs = 4
• Lone pairs = 2
• Total = $6$ electron pairs

Thus the electron pair geometry is octahedral and the hybridization is: $sp^3d^2$

So, B is correct.

Option C: $[IF_6]^-$

Iodine is central.

• Iodine has 7 valence electrons.
• Negative charge adds 1 electron.
• Total = $8$ electrons.
• Six $I-F$ bonds are formed.
• Remaining electrons on iodine = 1 lone pair? Let us check carefully using steric number formula.

Steric number can be found as: $\frac{V + M - C + A}{2}$ where

• $V=7$ for iodine,
• $M=6$ monovalent atoms,
• $C=1$ for negative charge.

So, $\text{SN} = \frac{7+6+1}{2} = 7$ Thus there are 7 electron pairs around iodine:

• 6 bond pairs
• 1 lone pair

Hence hybridization is $sp^3d^3$, not $sp^3d^2$.

So, C is not correct.

Option D: $[BrF_2]^-$

Bromine is central.

• Bromine has 7 valence electrons.
• Negative charge adds 1 electron.
• Total = $8$ electrons.
• Two $Br-F$ bonds are formed.
• Remaining electrons on bromine correspond to 3 lone pairs.

So around bromine:

• Bond pairs = 2
• Lone pairs = 3
• Total = $5$ electron pairs

Hence hybridization is $sp^3d$.

So, D is not correct.

3. Final conclusion

Only $[ICl_4]^-$ has 6 electron pairs around the central atom, so it has: $sp^3d^2$

Therefore, the correct option is B.