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Hint:Let's start by discussing whether the association is right or wrong.
Phosphorus is a chemical element with an atomic number 15 and a symbol P. Elemental phosphorus mostly exists in two major forms, white phosphorus and red phosphorus. But phosphorus is a highly reactive element, phosphorus never exists as a free element on Earth.
Complete step by step solution:
> The assertion in the given problem states “White phosphorus is more reactive than red phosphorus”. This assertion is true.
> The reason for this behavior lies in the structure of both white phosphorus and red phosphorus. They have the following structures
> As you can see that in red phosphorus 4 atoms are linked together and are also linked to other such consecutive groups of atoms, whereas, in case of white phosphorus the 4 atoms are only with each other and act as individual units. This crosslinking in red phosphorus makes it much more stable than white phosphorus.
> Additionally, in white phosphorus there is a high strain on the bonds because of its discreet packaging, whereas the strain is quite less in the case of red phosphorus.
Hence, the assertion is true.
> Now the statement for reason is “It readily catches fire in air to give dense white fumes of ${P_4}{O_{10}}$”
- The reaction of white phosphorus with air is shown below
${P_4}(s) + {O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$
Hence, the reason given is a true statement, but as we saw above has no relation to the assertion.
Hence, option B is the correct choice.
Note:In such types of problems, you can skip writing a balanced equation for reactions as we did in the reaction ${P_4}(s) + {O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$, as the purpose of showing this reaction is only to show the final product. But in other problems, especially in lab work it will be wise to write a balanced equation for every reaction. Example${P_4}(s) + 5{O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$.
Phosphorus is a chemical element with an atomic number 15 and a symbol P. Elemental phosphorus mostly exists in two major forms, white phosphorus and red phosphorus. But phosphorus is a highly reactive element, phosphorus never exists as a free element on Earth.
Complete step by step solution:
> The assertion in the given problem states “White phosphorus is more reactive than red phosphorus”. This assertion is true.
> The reason for this behavior lies in the structure of both white phosphorus and red phosphorus. They have the following structures
> As you can see that in red phosphorus 4 atoms are linked together and are also linked to other such consecutive groups of atoms, whereas, in case of white phosphorus the 4 atoms are only with each other and act as individual units. This crosslinking in red phosphorus makes it much more stable than white phosphorus.
> Additionally, in white phosphorus there is a high strain on the bonds because of its discreet packaging, whereas the strain is quite less in the case of red phosphorus.
Hence, the assertion is true.
> Now the statement for reason is “It readily catches fire in air to give dense white fumes of ${P_4}{O_{10}}$”
- The reaction of white phosphorus with air is shown below
${P_4}(s) + {O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$
Hence, the reason given is a true statement, but as we saw above has no relation to the assertion.
Hence, option B is the correct choice.
Note:In such types of problems, you can skip writing a balanced equation for reactions as we did in the reaction ${P_4}(s) + {O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$, as the purpose of showing this reaction is only to show the final product. But in other problems, especially in lab work it will be wise to write a balanced equation for every reaction. Example${P_4}(s) + 5{O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$.
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