
Which incorrect statement?
A. All halogens form oxyacids
B. All halogen show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states
C. Hydrofluoric acid forms $KH{F_2}$ and ${K_2}{F_2}$ and attacks glass
D. Oxidising power is in order ${F_2} > C{l_2} > B{r_2} > {I_2}$
Answer
495.9k+ views
Hint: Halogens are the elements of group 17 of the periodic table. These are non-metals. Halogens are known to be highly reactive and their reactivity decreases down the group. The halogens existing as liquid are bromine, gases are fluorine and chlorine, and solid are iodine and astatine.
Complete step by step answer:
Let us discuss each point one by one.
All halogen form oxyacids.
Oxoacids or oxyacids can be defined as the compounds having an oxygen atom attached to a halogen atom. The oxoacids formed by halogens are $HOF,HCl{O_3},HBr{O_3},andHI{O_3}$. So, we can say that all the halogens form oxoacids.
All halogen show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states
If we talk about fluorine, then it exhibits \[ - 1, + 1\] oxidation in the compounds hydrogen fluoride and hypofluorous acid.
From here, we can say that chlorine, bromine and iodine exhibits \[ - 1, + 1, + 3, + 5, + 7\] oxidation states in the formation of oxoacids, hydrogen halides and interhalogen compounds etc.
So, all the halogen show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states except fluorine.
Hydrofluoric acid forms $KH{F_2}$ and ${K_2}{F_2}$ attacks glass
$KH{F_2}$ is used in the electrolysis to prepare fluorine and these are prepared by the KF and HF salts. The reaction can be represented as:
$2KH{F_2}(s) \to {H_2}(g) + {F_2}(g) + 2KF(s)$
So, the statement is correct.
Oxidising power is in order ${F_2} > C{l_2} > B{r_2} > {I_2}$
The trend given above for oxidising power is correct for halogen as according to the physical properties of halogens when we move down the group the oxidising power or strength decreases.
In the end, we can conclude that all the halogens show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states is incorrect.
So, the correct answer is Option B.
Note: When we discuss the physical and chemical properties of halogens like oxidation states, the formation of chemical compounds like halides etc. Astatine named element is not considered as a radioactive element.
Complete step by step answer:
Let us discuss each point one by one.
All halogen form oxyacids.
Oxoacids or oxyacids can be defined as the compounds having an oxygen atom attached to a halogen atom. The oxoacids formed by halogens are $HOF,HCl{O_3},HBr{O_3},andHI{O_3}$. So, we can say that all the halogens form oxoacids.
All halogen show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states
If we talk about fluorine, then it exhibits \[ - 1, + 1\] oxidation in the compounds hydrogen fluoride and hypofluorous acid.
From here, we can say that chlorine, bromine and iodine exhibits \[ - 1, + 1, + 3, + 5, + 7\] oxidation states in the formation of oxoacids, hydrogen halides and interhalogen compounds etc.
So, all the halogen show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states except fluorine.
Hydrofluoric acid forms $KH{F_2}$ and ${K_2}{F_2}$ attacks glass
$KH{F_2}$ is used in the electrolysis to prepare fluorine and these are prepared by the KF and HF salts. The reaction can be represented as:
$2KH{F_2}(s) \to {H_2}(g) + {F_2}(g) + 2KF(s)$
So, the statement is correct.
Oxidising power is in order ${F_2} > C{l_2} > B{r_2} > {I_2}$
The trend given above for oxidising power is correct for halogen as according to the physical properties of halogens when we move down the group the oxidising power or strength decreases.
In the end, we can conclude that all the halogens show \[ - 1, + 1, + 3, + 5, + 7\] oxidation states is incorrect.
So, the correct answer is Option B.
Note: When we discuss the physical and chemical properties of halogens like oxidation states, the formation of chemical compounds like halides etc. Astatine named element is not considered as a radioactive element.
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