Question

Which of the following are thermodynamically stable?
A. C (diamond)
B. C (graphite)
C. ${P_4}$ (white)
D. ${P_4}$ (black)

Answer 1

The term ‘Thermo’ means heat and ‘dynamics’ means motion. Therefore, thermodynamics is concerned with heat motion. It deals with changes accompanying all types of physical and chemical processes.
Thermodynamics is based on three generalizations known as the First, Second, and Third law of thermodynamics. Since thermodynamic law deals with energy they apply to all the phenomena of nature.
The laws of thermodynamics are not concerned at all with the atomic or molecular structure of matter.
When a system is in its lowest energy state then it is said to be thermodynamically stable.

Complete step by step answer:
The first option is C (diamond). Diamond is an allotrope of carbon. Each carbon atom in a diamond is $s{p^3}$ hybridized and is bonded tetrahedrally to four other carbon atoms by simple covalent bonds. Thus, it has a three-dimensional network of strong covalent bonds in which $C - C$ bond length is 154 pm. The reaction from which diamond converts to graphite is spontaneous and the $\Delta G$ value is negative for this reaction. During the transition from diamond to graphite, it contains a large activation energy barrier due to which diamond is not likely to transition on its own and it is kinetically stable and thermodynamically unstable.

The second option is C (graphite). It is also an allotrope of carbon. Graphite, unlike diamond, has a two-dimensional sheet-like structure in which all the carbon atoms are $s{p^2}$ hybridized. Each carbon atom is bonded to the three other carbon atoms through covalent bonds forming hexagonal planar rings. The $C - C$ bond distance in rings is 142 pm indicating strong bonding. In diamond, it contains no delocalized electrons but in graphite, it contains one delocalized electron per carbon due to which it causes greater attraction hence giving stronger bonds and more stability to the structure. Thus, graphite is thermodynamically stable.

The third option is ${P_4}$ (white). It is one of the allotropic forms of phosphorus that exists as ${P_4}$ units. The four $s{p^3}$ hybridized phosphorus lie at the corners of the regular tetrahedron. Each phosphorus atom is linked to three other P- atoms by covalent bonds so that each P- atom completes its octet. The tetrahedral structure produces a higher angular strain which makes the white phosphorus more reactive and less stable.

The fourth option is ${P_4}$ (black). Black phosphorus has two forms $\alpha$ and $\beta$ forms. The structure of black phosphorus has an orthorhombic pleated honeycomb structure due to which it is less reactive and it's interlinked six-membered rings where each atom is bonded to three other atoms makes it thermodynamically more stable.

So, the correct answer is Option B,D .

Note: Various forms of the same element which have the same chemical properties but differ in their physical properties are called its allotropic forms and the phenomenon is called allotropy.
All the allotropes occur in the solid-state and differ in the arrangement of atoms.
All the varieties melt to give the same liquid.