
What is the relative number of atoms?
A compound of Xe and F is found to have 53.5% of Xe. What is the oxidation number of Xe in this compound?
(a) -4
(b) 0
(c) +4
(d) +6
Answer
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Hint: To solve this question we first need to know what the oxidation state is. The degree of oxidation of a chemical compound is determined by its oxidation state. Also, the relative number of atoms is also known as the empirical formula of the compound.
Complete answer:
Now, let us first understand what is the relative number of atoms. The formula of the chemical compound that depicts the simplest positive integer ratio of the atoms present gives the relative number of atoms in that compound.
Now, to determine the oxidation state of Xe, we first need to determine the molecular formula of the compound.
The atomic mass of Xenon is 131.2 u.
So, the relative number of atoms in the compound containing 53.5% of Xe is = $\dfrac{53.5}{131.2}=0.4$.
The percentage composition of fluorine is 46.5%.
The atomic mass of fluorine is 19 u.
So, the relative number of atoms in the compound containing 46.5% of Fe is = $\dfrac{46.5}{19}=2.4$.
Hence, the ratio of the relative number of atoms of Xe and F in the compound is
\[\dfrac{Xe}{F}=\dfrac{0.4}{2.4}\]
This can be simplified as
\[\dfrac{Xe}{F}=\dfrac{1}{6}\]
So, the compound's molecular formula is $Xe{{F}_{6}}$.
Now, since fluorine is a halide and has 7 electrons in its valence shell, its usual oxidation state is -1. And the oxidation state of a neutral molecule is 0.
So,
Xe + (-6) = 0
Xe = +6
Hence the oxidation state of xenon in xenon hexafluoride $Xe{{F}_{6}}$ is option (d) +6.
Note:
-It should be noted that there are some exceptions while assigning the oxidation states of atoms in a molecule.
- When a hydrogen atom is bonded to a less electronegative atom, its oxidation number is -1.
- When an oxygen atom is bonded to a more electronegative atom, it exists in a peroxide ion, its oxidation number changes.
- When a Group VII(A) atom is bonded to a more electronegative atom, its oxidation number changes.
Complete answer:
Now, let us first understand what is the relative number of atoms. The formula of the chemical compound that depicts the simplest positive integer ratio of the atoms present gives the relative number of atoms in that compound.
Now, to determine the oxidation state of Xe, we first need to determine the molecular formula of the compound.
The atomic mass of Xenon is 131.2 u.
So, the relative number of atoms in the compound containing 53.5% of Xe is = $\dfrac{53.5}{131.2}=0.4$.
The percentage composition of fluorine is 46.5%.
The atomic mass of fluorine is 19 u.
So, the relative number of atoms in the compound containing 46.5% of Fe is = $\dfrac{46.5}{19}=2.4$.
Hence, the ratio of the relative number of atoms of Xe and F in the compound is
\[\dfrac{Xe}{F}=\dfrac{0.4}{2.4}\]
This can be simplified as
\[\dfrac{Xe}{F}=\dfrac{1}{6}\]
So, the compound's molecular formula is $Xe{{F}_{6}}$.
Now, since fluorine is a halide and has 7 electrons in its valence shell, its usual oxidation state is -1. And the oxidation state of a neutral molecule is 0.
So,
Xe + (-6) = 0
Xe = +6
Hence the oxidation state of xenon in xenon hexafluoride $Xe{{F}_{6}}$ is option (d) +6.
Note:
-It should be noted that there are some exceptions while assigning the oxidation states of atoms in a molecule.
- When a hydrogen atom is bonded to a less electronegative atom, its oxidation number is -1.
- When an oxygen atom is bonded to a more electronegative atom, it exists in a peroxide ion, its oxidation number changes.
- When a Group VII(A) atom is bonded to a more electronegative atom, its oxidation number changes.
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