The Xe-F Bonds in XeF2, XeF4, and XeF6 are Equivalent. Justification is as follows:

Introduction:
In order to understand why the Xe-F bonds in XeF2, XeF4, and XeF6 are equivalent, we need to examine the molecular structures and bonding in these compounds.

Molecular Structures:
1. XeF2: In XeF2, xenon (Xe) is surrounded by two fluorine (F) atoms. The molecule has a linear structure, with the two F atoms arranged on opposite sides of the Xe atom.

2. XeF4: In XeF4, xenon (Xe) is surrounded by four fluorine (F) atoms. The molecule has a square planar structure, with the four F atoms arranged symmetrically around the Xe atom.

3. XeF6: In XeF6, xenon (Xe) is surrounded by six fluorine (F) atoms. The molecule has an octahedral structure, with the six F atoms arranged symmetrically around the Xe atom.

Bonding in XeF2:
In XeF2, each fluorine atom forms a single covalent bond with the xenon atom. This bond involves the sharing of one pair of electrons between Xe and F. The bond length and bond strength of these Xe-F bonds are equal.

Bonding in XeF4:
In XeF4, each fluorine atom forms a single covalent bond with the xenon atom. Similar to XeF2, each Xe-F bond in XeF4 involves the sharing of one pair of electrons between Xe and F. The bond length and bond strength of these Xe-F bonds are also equal.

Bonding in XeF6:
In XeF6, each fluorine atom forms a single covalent bond with the xenon atom. Once again, each Xe-F bond in XeF6 involves the sharing of one pair of electrons between Xe and F. The bond length and bond strength of these Xe-F bonds are equal.

Conclusion:
Based on the molecular structures and bonding in XeF2, XeF4, and XeF6, we can conclude that the Xe-F bonds in these compounds are equivalent. In all three compounds, each fluorine atom forms a single covalent bond with the xenon atom, involving the sharing of one pair of electrons. The bond lengths and bond strengths of these Xe-F bonds are equal in all three compounds.