To determine in which molecule/ion all bonds are not equal, we need to analyze their structures and bonding.

Concept: Bond equality in a molecule depends on its symmetry and the nature of the bonds (single, double, triple). Resonance structures can also lead to equal bond lengths that are intermediate between single and double bonds.

Why (C) is correct:

Let's analyze each option:

A) ${\mathrm{XeF}}_4$: Xenon tetrafluoride has a square planar geometry. All four Xe-F bonds are equivalent due to the high symmetry of the molecule. The hybridization of Xe is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

B) ${\mathrm{BF}}_4^{-}$: Tetrafluoroborate ion has a tetrahedral geometry. All four B-F bonds are equivalent due to the high symmetry of the molecule. The hybridization of B is ${\mathrm{sp}}^3$.

C) ${\mathrm{C}}_2{\mathrm{H}}_4$ (Ethene): Ethene has a planar structure with a carbon-carbon double bond and four carbon-hydrogen single bonds. The C=C bond length (approx. 1.34 Å) is different from the C-H bond length (approx. 1.09 Å). Therefore, not all bonds are equal.

D) ${\mathrm{SiF}}_4$: Silicon tetrafluoride has a tetrahedral geometry. All four Si-F bonds are equivalent due to the high symmetry of the molecule. The hybridization of Si is ${\mathrm{sp}}^3$.

Thus, in ${\mathrm{C}}_2{\mathrm{H}}_4$, not all bonds are equal.

Correct Answer: (C)