Select the compound in which the chlorine shows oxidation state \[+7\].
A \[HCl{{O}_{4}}\]
B \[HCl{{O}_{3}}\]
C \[HCl{{O}_{2}}\]
D \[HClO\]
Answer
626.7k+ views
Hint: The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative, or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.
Complete step by step answer:
Here, we have four compounds of chlorine. So, we have to find the oxidation state of chlorine in each compound to find the answer that is in which compound chlorine shows oxidation state is \[+7\]. Let us calculate the oxidation state of the chlorine in each compound;
We have known that the oxidation state of hydrogen and oxygen is fixed, i.e. \[+1\] and \[-2\] respectively. Let \[x\] be the oxidation state of chlorine in each compounds;
In \[HCl{{O}_{4}}\], we have one chlorine atom, one hydrogen atom and four oxygen atom. So, we can write as
\[\Rightarrow 1+x+4(-2)=0\]
\[\Rightarrow x=+7\]
In,\[HCl{{O}_{3}}\], we have one chlorine atom, one hydrogen atom and three oxygen atoms. So, we can write as
\[\Rightarrow 1+x+3(-2)=0\]
\[\Rightarrow x=+5\]
In, \[HCl{{O}_{2}}\] we have one chlorine atom, one hydrogen atom, and two oxygen atoms. So, we can write as
\[\Rightarrow 1+x+2(-2)=0\]
\[\Rightarrow x=+3\]
In, \[HClO\] we have one chlorine atom, one hydrogen atom and one three oxygen atoms. So, we can write as
\[\Rightarrow 1+x+1(-2)=0\]
\[\Rightarrow x=+1\]
So, chlorine shows \[+7\]\[HCl{{O}_{4}}\]in only. Hence, the correct answer is option A.
Note: Like hydrogen and oxygen there are some more elements whose oxidation states are fixed in chemical compounds. Group 1 and group 2 metals in compounds have oxidation states is = +1 and +2, respectively. Also, the oxidation state of an element in free form is equal to zero.
Complete step by step answer:
Here, we have four compounds of chlorine. So, we have to find the oxidation state of chlorine in each compound to find the answer that is in which compound chlorine shows oxidation state is \[+7\]. Let us calculate the oxidation state of the chlorine in each compound;
We have known that the oxidation state of hydrogen and oxygen is fixed, i.e. \[+1\] and \[-2\] respectively. Let \[x\] be the oxidation state of chlorine in each compounds;
In \[HCl{{O}_{4}}\], we have one chlorine atom, one hydrogen atom and four oxygen atom. So, we can write as
\[\Rightarrow 1+x+4(-2)=0\]
\[\Rightarrow x=+7\]
In,\[HCl{{O}_{3}}\], we have one chlorine atom, one hydrogen atom and three oxygen atoms. So, we can write as
\[\Rightarrow 1+x+3(-2)=0\]
\[\Rightarrow x=+5\]
In, \[HCl{{O}_{2}}\] we have one chlorine atom, one hydrogen atom, and two oxygen atoms. So, we can write as
\[\Rightarrow 1+x+2(-2)=0\]
\[\Rightarrow x=+3\]
In, \[HClO\] we have one chlorine atom, one hydrogen atom and one three oxygen atoms. So, we can write as
\[\Rightarrow 1+x+1(-2)=0\]
\[\Rightarrow x=+1\]
So, chlorine shows \[+7\]\[HCl{{O}_{4}}\]in only. Hence, the correct answer is option A.
Note: Like hydrogen and oxygen there are some more elements whose oxidation states are fixed in chemical compounds. Group 1 and group 2 metals in compounds have oxidation states is = +1 and +2, respectively. Also, the oxidation state of an element in free form is equal to zero.
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