Question

Which of the following sets represent the normal physical states of the elements concerned at ${25^\circ }C$ and one $bar$ pressure with $\Delta {H^\circ }(formation) = 0$
A.$C$ (diamond) ,${S_8}(s),Na(s)$
B.$C$ (diamond) , $C$ (graphite) , $B{r_2}(l)$
C.$C$ (graphite) , $B{r_2}(g),{P_4}$ (white)
D.$C$ (graphite) , $B{r_2}(l),{I_2}(s)$

Answer 1

All the elements that occur in nature are present in any of the physical states. They can be solid, liquid or gaseous in nature. One element can occur in various different forms in nature that can form at different conditions of temperature and pressure.

Complete step by step answer:
Standard Enthalpy of formation $(\Delta {H^\circ })$ : The change in enthalpy that occurs when one mole of a substance is formed from its elements under same conditions is called standard enthalpy of formation.
Allotropes- The property due to which an element exists in two or more forms which differ in their physical and chemical properties is known as allotropy and the various forms are called allotropes. These allotropes differ in the arrangement of atoms or number of atoms in the compound.
The physical states of elements that occur naturally do not have any value of enthalpy of formation. Since they occur naturally no energy is needed for their formation.
The allotropes of carbon are diamond, graphite, coal, charcoal and lampblack. The allotrope of carbon which is most stable at room temperature and at $1atm$ pressure is called the standard form of carbon. Graphite is the most stable form of carbon and has zero enthalpy of formation. Diamond is a less stable form of carbon as compared to graphite.
Iodine in its natural state is a solid at $298K$ and $1atm$ pressure. Also bromine is a liquid at the standard state. Bromine is the only non–metallic element that exists as liquid under ordinary conditions. It is a reddish-brown liquid.
In Option A, diamond is given as the allotrope of carbon. So it is not the correct option.
In option B, both diamond and graphite are mentioned. So it is also the wrong option.
In option D bromine is in a gaseous state. So this is also not correct .
In the case of phosphorus, white phosphorus is taken as the standard state of phosphorus.
Sulphur is a pale yellow solid at ordinary conditions having formula ${S_8}$ .
So the correct option is D.
In Option A, diamond is given as the allotrope of carbon. So it is the wrong option. In option B, both diamond and graphite are mentioned. So it is also the wrong option. In option bromine is in a gaseous state. So this is also the wrong option.

Note: Diamond and graphite are allotropes of carbon. The allotropes of phosphorus are red phosphorous, black phosphorus, white phosphorous. Remember all elements in their standard states have zero enthalpy of formation. There are only two elements that are liquid in its natural state-mercury and bromine.