# On Computation of Face Index of Certain Nanotubes

Authors: Ansheng Ye; Aisha Javed; Muhammad Kamran Jamil; Kanza Abdul Sattar; Adnan Aslam; Zahid Iqbal; Asfand Fahad
Journal: Discrete Dynamics in Nature and Society
 (2020)

---

## Abstract

Topological index is a number that can be used to characterize the graph of a molecule. Topological indices describe the physical, chemical, and biological properties of a chemical structure. In this paper, we derive the analytical closed formulas of face index of some planar molecular structures such asTUC4, TUC4C8S, TUHC6, TUC4C8R, and armchair TUVC6.


---

## Body

## 1. Introduction

In this time of rapid technological development, the pharmacological techniques have evolved rapidly during the recent years. Consequently, a large number of new drugs and chemical compounds have been obtained. A huge amount of work is required to study the biological, chemical, and pharmacological aspects of these new drugs and chemical compounds. This workload is becoming more and more cumbersome as it requires sufficient tools, reagents, human resources, and a lot of time to check the performance of these new chemical compounds. However, the developing countries cannot afford these equipment and reagents to check up these biochemical properties and are resultantly unable to compete with the developed world in the areas of medical science and industry. To some extent, the chemical graph theory solved this problem as it assists to measure the pharmaceutical, chemical, and physical properties of the chemical compounds. Fortunately, previous research has revealed that chemical properties of a molecule such as boiling point, melting point, and toxicity are closely related to their molecular structures (see [1, 2]). This relationship is one of the key reasons for the development of the mathematical chemistry. In the chemical graph theory, a molecular structure can be represented in the form of a graph G=VG,EG, where vertices V=VG and edges E=EG of a graph G show the atoms and the bonds of a molecular structure, respectively.A topological index (TI) is an invariant that is assigned to a molecular structure (graph) and is used to characterize the molecule. It may be thought as a convenient device which converts a chemical constitution into a unique number, which is independent of the way in which the corresponding graph has been drawn or labeled. TIs were employed in developing a suitable correlation between the chemical structure and chemical or biological activities and physical properties. Several researchers working in the area of chemical and mathematical sciences have introduced TIs, such as the Wiener index, Randić indices, Zagreb indices, PI index, eccentric index, atom-bond connectivity index, and forgotten index, which have been used to predict the characteristics of the nanomaterials, drugs, and other chemical compounds. There are several papers to calculate the topological indices of some special molecular graphs [3–12].The notions of a planar graph, its faces, and an infinite face are well known in the literature. LetG=VG,EG,FG be a finite simple connected planar graph, where VG, EG, and FG represent the vertex, edge, and face sets, respectively. A face f∈FG is incident to an edge e∈EG if e is one of those which surrounds the face. Similarly, a face f∈FG is incident to a vertex v in G if v is at the end of one of those incident edges; the incidency of v to the face f is represented by v∼f. The face degree f in G is given as df=∑v∼fdv. For the notions and notations not given here, we refer [13] to the readers.Recently, Jamil et al. [14] introduced a novel topological index named as the face index. The face index helped to predict the energy and the boiling points of selected benzenoid hydrocarbons with the correlation coefficient r>0.99. For a graph G, the face index (FI) can be defined as(1)FIG=∑f∈FGdf=∑v∼f∈FGdv.In this paper, we calculate the face index of some special molecular graphs which have been widely used in drugs.


## 2. Main Results

In this section, we investigate the exact formulas of the face index for the molecular structures of vastly studied nanotubes with wide range of applications:TUC4C8S,TUC4C8R,TUHC6,TUC4, and TUVC6. To find the face indices of the molecular graphs of these nanotubes, we partitioned the face set depending on the degrees of each face.


### 2.1. Face Index ofTUC4C8Sn,q,r and TUC4C8Rn,q,r Nanotubes

The 2-dimensional lattice ofTUC4C8Sn,q,r is constructed by the alternatingly positioned squares C4 and octagons C8 (see Figure 1(a)), where n,q, and r represent the number of rows, octagons in each row, and squares in each row. A TUC4C8Sn,q,r nanotube can be constructed by rolling the 2D lattice of carbon atoms and can be seen in Figure 1(b).Figure 1
(a) 2D lattice ofTUC4C8S7,4,4. (b) 3D nanotube TUC4C8S.
(a)(b)Firstly, we prove the following formula which provides the exact values of the face index forTUC4C8Sn,q,r.Theorem 1.
LetG=TUC4C8Sn,q,r, where n,q,r≥1, be the 2-dimensional lattice of TUC4C8S nanotube; then, the face index of G is given as(2)FIG=24nq+12nr+4q+12r.Proof.
LetG be the 2-dimensional lattice of TUC4C8Sn,q,r nanotube with n number of rows and let q and r be the number of octagons and number of squares in each row, respectively. In TUC4C8Sn,q,r, the total number of faces in one row is q+r. Let fj denote the faces having ∑w∼fjdw=j and fj denote the number of faces with degree j. From Figure 1(a), it can be noticed that 2D lattice of TUC4C8S7,4,4 contains four types of internal faces f12, f20, f22, and f24 and an external face, f∞. When TUC4C8Sn,q,r has n rows, then sum of vertex degrees of external face is 8q+12r. The number of internal faces in each row is given in Table 1.
The face index ofTUC4C8Sn,q,r is(3)FIG=∑w∼f∈FGdw=∑w∼f12dw+∑w∼f22dw+∑w∼f24dw+∑w∼f∞dw=f1212+f2222+f2424+8q+12r=nr12+2q22+qn−224+8q+12r=12nr+44q+24nq−48q+8q+12r=24nq+12nr+4q+12r.
This completes the proof.Table 1
Numbers off12, f20, f22, and f24 with given number of rows.
nf12f20f22f241Rq——22r—2q—33r—2qQ44r—2q2q...............NNr—2q(n−2)qTheorem 2.
Letn,q,r≥1 and H=TUC4C8Rn,q,r be the graph of 2-dimensional lattice of TUC4C8R nanotube. Then, the face index of the graph H is given as(4)FIH=24nq+6nr+12q+18r.Proof.
Consider aTUC4C8Rn,q,r nanotube with n number of rows, q number of octagons, and r number of squares in each row as shown in Figure 2(a). The 2-dimensional lattice (H) of TUC4C8Rn,q,r is shown in Figure 2(b). In H, the total number of faces in one row is q+r. Let fj and fj denote the face with degree j and the number of faces with degree j, respectively. From the structure of H, one can notice that there are three types of internal faces f11, f12, and f24 and an external face f∞. The external face has degree 12q+2r. Table 2 illustrates the number of internal faces in TUC4C8Rn,q,r based on the degree of each face.
The face index of the graphH=TUC4C8Rn,q,r is(5)FIH=∑w∼f∈FHdw=∑w∼f11dw+∑w∼f12dw+∑w∼f24dw+∑w∼f∞dw=|f11|11+|f12|12+|f24|24+12q+2r=r11+n−1r212+nq24+12q+2r=11r+6nr−6r+24nq+12q+2r=24nq+6nr+12q+7r.
This completes the proof.Figure 2
(a) 3D nanotubeTUC4C8R. (b) 2D lattice of TUC4C8R.
(a)(b)Table 2
Numbers off11, f12, and f24 with given number of rows.
nf11f12f241R—Q2Rr/22q3RR3q4R3r/24q552r5q............NRn−1r/2Nq


### 2.2. Face Index ofTUC4p,q Nanotube

Forp≥2 and q≥3, the 2-dimensional lattice of TUC4p,q nanotube is obtained by the Cartesian product of the path Pp and the cycle Cq. For p=8 and q=4, the example of TUC4p,q is shown in Figure 3(b).Figure 3
(a) NanotubeTUC4p,6. (b) Nanotube TUC48,4.
(a)(b)Theorem 3.
LetK be the graph of TUC4p,q nanotube structure, where p≥2 and q≥3. Then, the face index of K is equal to(6)FIK=18q;ifp=2,34q;ifp=3,2q8p−7;ifp>3.Proof.
We will prove the result forp>3. Let K denote the graph of TUC4Cp,q nanotube structure. From Figure 3, we can notice that the graph K contains three types of internal faces, namely, f3q,f14, and f16, and an external face of degree 3q. By applying the definition and using the values from Table 3, the face index of K can be computed as(7)FIK=∑w∼f∈FKdw=∑w∼f3qdw+∑w∼f14dw+∑w∼f16dw=f3q3q+f1414+f1616=23q+142q+16p−3q=2q8p−7.
This completes the proof.Table 3
Numbers off11, f12, and f24 with given number of rows.
TUC4k,qf3qf12f14f16k = 22Q——k = 32—2q—k = 42—2qQk = 5—2q2q...............k = p2—2q(p−3)q


### 2.3. Zig-ZagTUHC6n,q Nanotube

Consider the graphK of TUHC6n,q zig-zag polyhex nanotube structure, where n denotes the number of rows and the number of hexagons in each row is represented by q. Figure 4 illustrates the nanotube TUHC6n,q and its 2-dimensional structure.Figure 4
(a) 3D nanotubeTUHC6n,q. (b) 2D lattice of TUHC65,5.
(a)(b)Theorem 4.
Forn,q≥1, let K represent the 2-dimensional graph of TUHC4n,q structure. The face index of K is(8)FIK=18nq+8q.Proof.
LetTUHC6n,q be a polyhex nanotube with n number of rows and q number of hexagons in each row and K be the 2-dimensional graph of TUHC6n,q structure. The molecular graph of TUHC6n,q is shown in Figure 4. Let fj denote the face having degree j, i.e., ∑w∼fjdw=j, and let fj denote the number of fj. The molecular graph of TUHC6n,q contains two types of internal faces f17 and f18 and an external face f∞. When TUHC6n,q has n rows, then the face degree of f∞ is 10q. The number of internal faces with the given number of rows is listed in Table 4.
The face index of the graphK=TUHC6n,q is(9)FIK=∑w∼f∈FKdw=∑w∼f17dw+∑w∼f18dw+∑w∼f∞dw=f1717+f1818+10q=2q17+n−2q18+10q=34q+18nq−36q+10q=18nq+8q,which is the required result.Table 4
The cardinalities of the faces with given degree for given number of rows.
nf16f17f181q——2—2q—3—2qQ4—2q2q............N—2q(n−2)q


### 2.4.TUVC6n,q Nanotube

The 2-dimensional graph ofTUVC6n,q is shown in Figure 5(b), and the TUVC6n,q nanotube can be constructed by rolling this lattice of carbon atoms (Figure 5(a)), where n represents the number of rows and q is the hexagons in each row.Figure 5
(a) Armchair 3D nanotubeTUVC6n,q. (b) 2D lattice of TUVC64,4.
(a)(b)Theorem 5.
LetL=TUVC4n,q, where n,q≥1, be the graph of 2-dimensional lattice of TUVC6n,q armchair polyhex nanotube. The face index of L is(10)FIL=36nq−8q.Proof.
LetL represent the 2-dimensional molecular graph of TUVC6n,q with n number of rows and q number of hexagons in each row. From Figure 5(b), we can easily notice that L contains 2 types of internal faces f16 and f18 and an external face, and the degree of external face is 14q. The cardinalities of internal faces with given degree and given number of rows are explained in Table 5.
The face index ofL=TUVC6n,q is(11)FIL=∑w∼f∈FLdw=∑w∼f16dw+∑w∼f18dw+∑w∼f∞dw=f1616+f1818+14q=2q16+2n−3q18+14q=32q+36nq−54q+14q=36nq−8q,and the proof is complete.Table 5
The cardinalities off14, f16, and f18 with given number of rows.
nf14f16f181q——2—2qQ3—2q3q4—2q5q............N—2q(2n−3)q


## 2.1. Face Index ofTUC4C8Sn,q,r and TUC4C8Rn,q,r Nanotubes

The 2-dimensional lattice ofTUC4C8Sn,q,r is constructed by the alternatingly positioned squares C4 and octagons C8 (see Figure 1(a)), where n,q, and r represent the number of rows, octagons in each row, and squares in each row. A TUC4C8Sn,q,r nanotube can be constructed by rolling the 2D lattice of carbon atoms and can be seen in Figure 1(b).Figure 1
(a) 2D lattice ofTUC4C8S7,4,4. (b) 3D nanotube TUC4C8S.
(a)(b)Firstly, we prove the following formula which provides the exact values of the face index forTUC4C8Sn,q,r.Theorem 1.
LetG=TUC4C8Sn,q,r, where n,q,r≥1, be the 2-dimensional lattice of TUC4C8S nanotube; then, the face index of G is given as(2)FIG=24nq+12nr+4q+12r.Proof.
LetG be the 2-dimensional lattice of TUC4C8Sn,q,r nanotube with n number of rows and let q and r be the number of octagons and number of squares in each row, respectively. In TUC4C8Sn,q,r, the total number of faces in one row is q+r. Let fj denote the faces having ∑w∼fjdw=j and fj denote the number of faces with degree j. From Figure 1(a), it can be noticed that 2D lattice of TUC4C8S7,4,4 contains four types of internal faces f12, f20, f22, and f24 and an external face, f∞. When TUC4C8Sn,q,r has n rows, then sum of vertex degrees of external face is 8q+12r. The number of internal faces in each row is given in Table 1.
The face index ofTUC4C8Sn,q,r is(3)FIG=∑w∼f∈FGdw=∑w∼f12dw+∑w∼f22dw+∑w∼f24dw+∑w∼f∞dw=f1212+f2222+f2424+8q+12r=nr12+2q22+qn−224+8q+12r=12nr+44q+24nq−48q+8q+12r=24nq+12nr+4q+12r.
This completes the proof.Table 1
Numbers off12, f20, f22, and f24 with given number of rows.
nf12f20f22f241Rq——22r—2q—33r—2qQ44r—2q2q...............NNr—2q(n−2)qTheorem 2.
Letn,q,r≥1 and H=TUC4C8Rn,q,r be the graph of 2-dimensional lattice of TUC4C8R nanotube. Then, the face index of the graph H is given as(4)FIH=24nq+6nr+12q+18r.Proof.
Consider aTUC4C8Rn,q,r nanotube with n number of rows, q number of octagons, and r number of squares in each row as shown in Figure 2(a). The 2-dimensional lattice (H) of TUC4C8Rn,q,r is shown in Figure 2(b). In H, the total number of faces in one row is q+r. Let fj and fj denote the face with degree j and the number of faces with degree j, respectively. From the structure of H, one can notice that there are three types of internal faces f11, f12, and f24 and an external face f∞. The external face has degree 12q+2r. Table 2 illustrates the number of internal faces in TUC4C8Rn,q,r based on the degree of each face.
The face index of the graphH=TUC4C8Rn,q,r is(5)FIH=∑w∼f∈FHdw=∑w∼f11dw+∑w∼f12dw+∑w∼f24dw+∑w∼f∞dw=|f11|11+|f12|12+|f24|24+12q+2r=r11+n−1r212+nq24+12q+2r=11r+6nr−6r+24nq+12q+2r=24nq+6nr+12q+7r.
This completes the proof.Figure 2
(a) 3D nanotubeTUC4C8R. (b) 2D lattice of TUC4C8R.
(a)(b)Table 2
Numbers off11, f12, and f24 with given number of rows.
nf11f12f241R—Q2Rr/22q3RR3q4R3r/24q552r5q............NRn−1r/2Nq


## 2.2. Face Index ofTUC4p,q Nanotube

Forp≥2 and q≥3, the 2-dimensional lattice of TUC4p,q nanotube is obtained by the Cartesian product of the path Pp and the cycle Cq. For p=8 and q=4, the example of TUC4p,q is shown in Figure 3(b).Figure 3
(a) NanotubeTUC4p,6. (b) Nanotube TUC48,4.
(a)(b)Theorem 3.
LetK be the graph of TUC4p,q nanotube structure, where p≥2 and q≥3. Then, the face index of K is equal to(6)FIK=18q;ifp=2,34q;ifp=3,2q8p−7;ifp>3.Proof.
We will prove the result forp>3. Let K denote the graph of TUC4Cp,q nanotube structure. From Figure 3, we can notice that the graph K contains three types of internal faces, namely, f3q,f14, and f16, and an external face of degree 3q. By applying the definition and using the values from Table 3, the face index of K can be computed as(7)FIK=∑w∼f∈FKdw=∑w∼f3qdw+∑w∼f14dw+∑w∼f16dw=f3q3q+f1414+f1616=23q+142q+16p−3q=2q8p−7.
This completes the proof.Table 3
Numbers off11, f12, and f24 with given number of rows.
TUC4k,qf3qf12f14f16k = 22Q——k = 32—2q—k = 42—2qQk = 5—2q2q...............k = p2—2q(p−3)q


## 2.3. Zig-ZagTUHC6n,q Nanotube

Consider the graphK of TUHC6n,q zig-zag polyhex nanotube structure, where n denotes the number of rows and the number of hexagons in each row is represented by q. Figure 4 illustrates the nanotube TUHC6n,q and its 2-dimensional structure.Figure 4
(a) 3D nanotubeTUHC6n,q. (b) 2D lattice of TUHC65,5.
(a)(b)Theorem 4.
Forn,q≥1, let K represent the 2-dimensional graph of TUHC4n,q structure. The face index of K is(8)FIK=18nq+8q.Proof.
LetTUHC6n,q be a polyhex nanotube with n number of rows and q number of hexagons in each row and K be the 2-dimensional graph of TUHC6n,q structure. The molecular graph of TUHC6n,q is shown in Figure 4. Let fj denote the face having degree j, i.e., ∑w∼fjdw=j, and let fj denote the number of fj. The molecular graph of TUHC6n,q contains two types of internal faces f17 and f18 and an external face f∞. When TUHC6n,q has n rows, then the face degree of f∞ is 10q. The number of internal faces with the given number of rows is listed in Table 4.
The face index of the graphK=TUHC6n,q is(9)FIK=∑w∼f∈FKdw=∑w∼f17dw+∑w∼f18dw+∑w∼f∞dw=f1717+f1818+10q=2q17+n−2q18+10q=34q+18nq−36q+10q=18nq+8q,which is the required result.Table 4
The cardinalities of the faces with given degree for given number of rows.
nf16f17f181q——2—2q—3—2qQ4—2q2q............N—2q(n−2)q


## 2.4.TUVC6n,q Nanotube

The 2-dimensional graph ofTUVC6n,q is shown in Figure 5(b), and the TUVC6n,q nanotube can be constructed by rolling this lattice of carbon atoms (Figure 5(a)), where n represents the number of rows and q is the hexagons in each row.Figure 5
(a) Armchair 3D nanotubeTUVC6n,q. (b) 2D lattice of TUVC64,4.
(a)(b)Theorem 5.
LetL=TUVC4n,q, where n,q≥1, be the graph of 2-dimensional lattice of TUVC6n,q armchair polyhex nanotube. The face index of L is(10)FIL=36nq−8q.Proof.
LetL represent the 2-dimensional molecular graph of TUVC6n,q with n number of rows and q number of hexagons in each row. From Figure 5(b), we can easily notice that L contains 2 types of internal faces f16 and f18 and an external face, and the degree of external face is 14q. The cardinalities of internal faces with given degree and given number of rows are explained in Table 5.
The face index ofL=TUVC6n,q is(11)FIL=∑w∼f∈FLdw=∑w∼f16dw+∑w∼f18dw+∑w∼f∞dw=f1616+f1818+14q=2q16+2n−3q18+14q=32q+36nq−54q+14q=36nq−8q,and the proof is complete.Table 5
The cardinalities off14, f16, and f18 with given number of rows.
nf14f16f181q——2—2qQ3—2q3q4—2q5q............N—2q(2n−3)q


## 3. Conclusion

In [14], using multiple linear regression, it has been shown that the novel face index can predict the π electron energy and boiling point of benzenoid hydrocarbon with a correlation coefficient greater than 0.99. Therefore, this index can be useful in QSPR/QSAR studies. In this paper, we have computed the novel face index of some nanotubes.



---

*Source: 3468426-2020-07-16.xml*