1. Interpret the question

We need to identify the species in which all the bonds are not equal, i.e. the bonds are of more than one type/environment.

This depends on the molecular geometry and whether all bonded atoms occupy equivalent positions.

2. Check each option

A. $XeF_4$

• Central atom: Xe
• Total electron pairs around Xe = $4$ bond pairs $+ 2$ lone pairs
• Electron pair geometry: octahedral
• Molecular shape: square planar
• The two lone pairs occupy opposite positions, leaving four F atoms in equivalent positions.

So, all four $Xe-F$ bonds are equivalent.

A is not correct.

B. $BF_4^-$

• Central atom: B
• Four bond pairs, no lone pair
• Shape: tetrahedral
• All four F atoms are symmetrically equivalent.

So, all four $B-F$ bonds are equal.

B is not correct.

C. $SF_4$

• Central atom: S
• Total electron pairs = $4$ bond pairs $+ 1$ lone pair
• Electron pair geometry: trigonal bipyramidal
• Molecular shape: seesaw

In a trigonal bipyramidal arrangement:

• There are axial positions
• There are equatorial positions

The lone pair occupies an equatorial position to minimize repulsion. Thus in $SF_4$:

• $2$ fluorines are axial
• $2$ fluorines are equatorial

Axial and equatorial $S-F$ bonds are not equivalent. So all bonds are not equal.

C is correct.

D. $SiF_4$

• Central atom: Si
• Four bond pairs, no lone pair
• Shape: tetrahedral
• All four $Si-F$ bonds are equivalent.

D is not correct.

3. Final conclusion

Only $SF_4$ has bonds in different environments (axial and equatorial), so all the bonds are not equal.

Therefore, the correct option is:

$\boxed{C}$

4. Comparison with stored correct answer

Stored correct answer: $C$

My derived answer: $C$

They match.