Bismuth tri-iodide (BiI3) is an inorganic compound. It is the product of the reaction of bismuth and iodine, which once was of interest in qualitative inorganic analysis. Layered BiI3 crystal is considered to be a three-layered stacking structure, where bismuth atom planes are sandwiched between iodide atom planes, which form the sequence I-Bi-I planes. The periodic stacking of three layers forms rhombohedral BiI3 crystal with R-3 symmetry [15] [16] . The successive stacking of one I-Bi-I layer forms hexagonal structure with symmetry [17] [18] . A single crystal of BiI3 has been synthesized by Nason and Keller [19] . Figure 1 shows one unit of bismuth tri-iodide.

The graph of a single unit of bismuth tri-iodide contains six 4-cycles of which two are on the top, two are in

the middle and two at the bottom. The unit cells of bismuth tri-iodide can be arranged either linearly or in a sheet form. A linear arrangement with m unit cells is called an m-bismuth chain; mn unit cells arranged into m rows and n columns is called an m × n bismuth sheet.

We first obtain a lower bound for the metric dimension of an m-bismuth chain.

Lemma 1. Let G be an m-bismuth chain. Then b(G) ≥ 7m + 5.

Proof. The two 2-degree vertices of each 4-cycle are equidistant from every vertex of G. So one of them must belong to any resolving set. Similarly the two 1-degree vertices incident with a common vertex are equidistant from every vertex of G. So one of them must belong to any resolving set. Hence b(G) ≥ 7m + 5.

Our next result realizes the lower bound obtained in Lemma 1.

Theorem 1. Let G be an m-bismuth chain. Then b(G) ≤ 7m + 5.

Proof. The cycles in an m-bismuth chain are listed as, and, and. Let p_i, q_i, , , and, and be the vertices as indicated in Figure 2. We claim that is a resolving set. Let u, v Î V\W.

Case 1:, for some j,.

If u and v are adjacent, then resolves them. If u and v are non-adjacent in, then they are equidistant from x_j. In this case either x_j?1 or x_j+1 resolves u and v. This argument holds for. If j = 1 or 2m, then or resolves u and v respectively. The argument is similar if or, and.

Case 2: and.

In this case u and v are resolved by q_s,. The argument is similar if and or and. In both cases a leaf vertex resolves u and v.

Lemma 1 and Theorem 1 solve the metric dimension problem for an m-bismuth chain completely. Thus we have the following result.

Theorem 2. Let G be an m-bismuth chain. Then b(G) = 7m + 5.

In the proof of Theorem 2, we constructed a metric basis W consisting of a 2-degree vertex from each 4-cycle and a leaf vertex from each of the pair of leaf incident with a 4-cycle.

Let W^* be the set of all 2-degree vertices and all the leaves vertices. Clearly W^* is a resolving set being a superset of the constructed resolving set W.

It is easy to see that representations of any two vertices u, v Î V\W^* will differ in at least two distinct places. Hence W^* is a fault-tolerant metric basis.

Theorem 3. Let G be an m-bismuth chain. Then.

We state the results for m × n-bismuth sheet.

Theorem 4. Let G be an m × n-bismuth sheet. Then.

Theorem 5. Let G be an m × n-bismuth sheet. Then.

Lead chloride is a halide crystal which occurs naturally in the form of mineral cotunnite. It is used in the production of infra red transmitting glass and basic chloride of lead known as patteson’s white lead, perry, ornamental glass called aurene glass, stained glass. It is also used as an intermediate in refining bismuth (Bi) ore, it is used in the synthesis of organometallic, lead titanate and barium titanate. The structure of lead chloride is orthorhombic dipyramidal.

The graph of a single unit of lead chloride is obtained from that of bismuth tri-iodide by joining just one 2-degree vertex of each of the 4-cycles to a new vertex.

As in the case of bismuth tri-iodide, chains and sheets of lead chloride are defined.

Lemma 2. Let G be an m-lead chloride chain. Then.

Proof. Every pair of 1-degree vertices incident with a common vertex are equidistant from every vertex of the graph. Hence at least one vertex must belong to any resolving set. Hence.

We adopt a labeling scheme in which vertices are listed level wise (levels 1 to 7) from top to bottom. The notation will depict the ith vertex in level k.

Theorem 6. Let G be an m-lead chloride chain. Then.

Proof. Let p_i, q_i, , be the 1-degree vertices as shown in Figure 3. We claim that is a resolving set. Let u, v Î V\W.

Case 1: u, v belong to the same level.

In view of the symmetry of the graph, it is enough to consider vertices in levels 1 to 4. Any two vertices on level 1 are resolved by either or. Now we consider any two vertices on level 2. Without loss of generality let us assume that, and,. In this case either or resolves u and v. In fact if i = 3s ? 1, 1 ≤ s ≤ m, then and. Similar argument will hold if i = 3s or i = 3s +1. The argument is similar for any two vertices in the level 3 except for the pairs of vertices and. These pairs are equidistant, respectively, from and. The pair is resolved by and the pair is resolved by. The similar argument holds for the level 4.

Case 2: u, v belong to different levels.

Different level vertices are resolved by any p_i or q_j,.

Lemma 2 and Theorem 6 solve the metric dimension problem for an m-lead chloride chain completely. Thus we have the following result.

Theorem 7. Let G be an m-lead chloride chain. Then.

Theorem 8. Let G be an m-lead chloride chain. Then.

Quartz is a crystalline form of silicon dioxide with formula SiO₂. A quartz network can be built in various ways. The quartz network of dimension 1 is a 12-cycle. The quartz network of dimension n consists n² number of 12-cycles and these 12-cycles are arranged in the shape of a rhombus. Figure 4 depicts a quartz network of

dimension 3.

Theorem 2.11. Let G be the graph of quartz. Then b(G) = 2.

Proof. It is easy to see that {a, b} is a resolving set. This follows by using the underlying binarytree structure rooted at a. Vertices at different levels are resolved by the root and vertices at the same level are resolved by b.

Theorem 2.12. Let G be the graph of quartz. Then b′(G) = 4.