A. XeF4
B. XeOF4
C. XeOF2
D. Na4XeO6
¨ Xenon can exhibit a range of oxidation states in its compounds, including +2, +4, +6, and +8. The oxidation state of xenon in a compound is determined by the number of valence electrons it shares with other atoms.
¨ In XeF4 (xenon tetrafluoride), the oxidation state of xenon is +4. Each of the four fluorine atoms contributes one electron to form a single bond with the xenon atom, giving a total of eight valence electrons and a formal oxidation state of +4.
¨ In XeOF4 (xenon oxytetrafluoride), the oxidation state of xenon is +6. The molecule contains one xenon atom, four fluorine atoms, and one oxygen atom. The xenon atom shares electrons with each of the four fluorine atoms to form single bonds, and with the oxygen atom to form a double bond. The four fluorine atoms contribute four electrons, and the oxygen atom contributes two electrons, giving a total of 14 valence electrons and a formal oxidation state of +6.
¨ In XeOF2 (xenon oxydifluoride), the oxidation state of xenon is +4. The molecule contains one xenon atom, two fluorine atoms, and one oxygen atom. The xenon atom shares electrons with each of the two fluorine atoms to form single bonds, and with the oxygen atom to form a double bond. The two fluorine atoms contribute two electrons, and the oxygen atom contributes two electrons, giving a total of 10 valence electrons and a formal oxidation state of +4.
¨ In Na4XeO6 (sodium hexoxoxenonate), the oxidation state of xenon is +8. The molecule contains one xenon atom, six oxygen atoms, and four sodium atoms. The xenon atom shares electrons with each of the six oxygen atoms to form double bonds, and each oxygen atom also forms a single bond with a sodium atom. The six oxygen atoms contribute 12 electrons, and the four sodium atoms contribute four electrons, giving a total of 32 valence electrons and a formal oxidation state of +8.
Submitted by :- Jamil Ahmed