Answer
Verified
28.5k+ views
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${P_4}(s) + 5{O_2}(g)\xrightarrow{{}}{P_4}{O_{10}}(s)$.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main
Other Pages
Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main
Differentiate between mass and inertia class 11 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
An air capacitor of capacity C10 mu F is connected class 12 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Mulliken scale of electronegativity uses the concept class 11 chemistry JEE_Main