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Question

Assertion :ΔfHo for white phosphorus has been taken to be zero, whereas black phosphorus is most stable allotrope of phosphorus and this is only exception. Reason: The reference standard state of white phosphorus is taken despite the fact that this form not being the most stable allotrope but simply the most reproducible form of phosphorus.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Assertion is incorrect but Reason is correct
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Solution

The correct option is

A

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements.

Standard enthalpy of formation is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements in their most stable forms (Standard states 1 bar and usually 298 K).

The standard enthalpy of formation of elements in their pure state at 1 bar and usual temperature of 298 K is assumed to be zero.

Out of known allotropes of phosphorus black phosphorus is the most stable one but the standard enthalpy of formation of white phosphorus is zero. The exception comes into the picture due to the difficulty in the formation of black phosphorus. White phosphorus is relatively easier to form so the standard enthalpy of formation for white phosphorus is zero instead of most stable black phosphorus.

Although black phosphorus is the most stable allotrope of phosphorus, we take white phosphorus as a standard state and ΔHof as zero and this is the only exception in the standard state of enthalpy

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