What are the others examples of NaCl type ionic salts ?

Most of the halides of alkali metals, oxides and sulphides of alkaline earth metals have this type of structures.

(1) Group 1^st Halides ie NaI, KCl, RbI, RbF ( except Cs halides) .

(2) Group 2^nd  oxide   MgO, CaO, BaO, SrO ( except BeO)

(3) Ammonium Halides ie NH₄Cl ,NH₄Br ,NH₄I etc

(4) Silver Halides ie AgF, AgCl, AgBr ,( except AgI)

(5) Other examples , TiO ,FeO, NiO etc

 

Note: Ferrous oxide also has sodium chloride, types structure in which O^-2 ions are arranged in ccp and Fe⁺² ions occupy octahedral ions. However, this oxide is always non – Stoichiometric and has the composition Fe_0.95 O  It can be explained on the assumption that some of the Fe⁺² ion are replaced by 2/3^rd as many Fe⁺³ ions in the octahedral voids.

Related topics:

SODIUM CHLORIDE (ROCK SALT TYPE) STRUCTURE:

ZINCE BLENDE (ZnS) TYPE STRUCTURE (SPHELERITE):

THE WURTIZE STRUCTURES (ZnS):

CAESIUM CHLORIDE (CsCl) STRUCTURE:

CALCIUM FLORIDE (FLORITE) STRUCTURE:

ANTIFLUORITE STRUCTURE (REVERE OF FLUORITE) Na2O:

SPINEL STRUCTURE :

 

 

 

 

 

What are structural information of Dimomd ?

Diamond structure is ZnS type structure  in which carbon atoms forms  a face centred cubic (FCC/CCP) lattice as well as  four out of eight (50%) or alternate tetrahedral voids are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four other. No discrete molecule can be discerned (identified) in diamond .the entire crystal is giant molecule a unit cell of which is shown as below.

Note : Only those atoms which form four covalent bond produce a repeated 3D structure using only covalent bonds.

Lattice of Diamond is ZnS type structure.

(1) C- form FCC/CCP (4-atoms)

(2) C- atoms present at the (50%) alternative tetrahedral voids (4-atoms)

(3) Total Number of one lattice unit is eight (8) hence molecular formula of diamond is (C8) (i.e. Z= 8)

(4) Number of C-C bond in lattice cell is = 4×4= 16

(5) Number of C-C bond per carbon atom is 16/8=2

(6) The distance between two Corbin atom is d_C-C = a√3/4 and the radius of carbon atom = dc-c/2 = r_c = a√3/2x4

(8) Packing efficiency (PE = π√3/10= 0.34 or 34%):

(9) Voids = 66 % 

Additional Information: 

Related Questions:

What are the normal spinel structures?

What are the inverse spinel structures?

How to calculate radius ratio of square Voids?

How to calculate radius ratio triangular voids?

How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

What are structural information of Dimomd ?

In diamond, carbon atoms occupy FCC/CCP lattice point as well as alternate tetrahedral voids. If edge length of unit cell is 3.56pm, the radius of carbon atom is ?

The edge length of unit cubic cell of diamond is 356.7 pm and this cubic unit cell contains 8 carbon atoms calculate (A) Distance between two carbon atoms, assuming them to spheres in contact (B) Radius of carbon atoms (C) Fraction of the total volume that is occupied by the carbon atoms.

SPINEL STRUCTURE :

(1) Structure of FeO, Fe₂O₃, and Fe₃O₄  :

(1) FeO: This oxide is non-stoichiometric and has a composition FexO (Generally ‘x’ varying from0.92 to 0.97). The oxide ions form a cubic close packing. The octahedral voids are occupied by Fe2+ but a small number of Fe²⁺ is replaced by Fe³⁺ ions. Three Fe²⁺will be replaced by two Fe³⁺ to maintain electrical neutrality but then what we obtain is an iron-deficient crystal.

(2) Fe₂O₃: If all Fe2+ are replaced by Fe3+, the ratio between Fe:O will be 2 : 3 (since 3 Fe²⁺ are replaced by 2Fe³⁺⁾ and not 1 : 1. As such, we obtain Fe₂O₃.

(3) Fe₃O₄: This is obtained by replacing exactly two thirds of Fe²⁺ by Fe³⁺ (in FeO structure).The remaining Fe²⁺ ions and 50% of Fe³⁺ ions occupy the octahedral voids. The remaining Fe³⁺ ions occupy tetrahedral voids. If in the structure of Fe₃O₄, the Fe²⁺ ions are replaced by divalent cations such as Mg^2+, Zn^2+, etc., the compounds obtained are called ferrites. In ferrites, divalent cations occupy tetrahedral voids and trivalent cations occupy octahedral voids. This structure is called spinel structure.

(2) NORMAL SPINEL (AB₂O₄ ) STRUCTURE:

Example of Spinel is a MgAl₂O₄.( mineral) In it oxide ions (O^-2) are arranged in ccp with Mg⁺² ions occupying tetrahedral voids and Al⁺³ ions in a set of octahedral voids.

Many ferrites (such as ZnFe₂O₄) also possess spinel structure. These are very important magnetic materials and are used in telephone and memory loops in computers.

(3) INVERSE SPINEL STRUCTURE (Fe₃O₄-Magnetite):

In Fe₃O₄, Fe⁺² and Fe⁺³ ions are present in the ratio 2:1. it may be considered as having composition FeO.Fe₂O₃. In Fe₃O₄ Oxide arranged in ccp. Fe⁺² ions occupy octahedral voids while Fe⁺³ ions are equally distributed between octahedral and tetrahedral voids

MgFe₂O₄ also has structure similar to magnetite. In this Mg⁺² ions are present in place of Fe⁺² ion in Fe₃O₄. Magnetite has inverse spinet structure.

ANTIFLUORITE STRUCTURE (REVERE OF FLUORITE) Na2O:

The compound having A₂B formula are compounds having anti fluorite structure :

Anti fluorite structure is having arrangement of cations and anions opposite to the fluorite structure Li₂O has an anti fluorite structure.

(1)  In the crystal structure of Li₂O, the O^-2 ions constitute a cubic close packed lattice (fcc structure) and the Li⁺ ions occupy all the tetrahedral voids

(2)  Each oxide ion, O^-2 ion is in contact with 8 Li⁺ ions and each Li⁺ ions having contact with 4 oxide ion. Therefore, Li₂O has 4:8 coordination
(3) Stoichiometric ratio  of Na₂O is 2:1
(4)  radius ratio , Packing efficiency, density and void %:

 Other Examples: – Na₂O, K₂O, K₂S, Na₂S, Rb₂O, Rb₂S

Note: Metals like Al, Ag, Au, Cu, Ni, and Pt have ccp structure and Be, Mg, Co and Zn have a hcp structure. And Noble gases (except He has a hcp structure) have ccp structure.

CALCIUM FLORIDE (FLORITE) STRUCTURE:

The salient features of fluorite structure are:

(1)  The Ca⁺² ions are arranged in ccp arrangement, i.e. these ions occupy all the corners and the centres of each face of the cube

(2)  The F^– ions occupy all the tetrahedral holes.

(3)  Since there are two tetrahedral holes for each Ca⁺² ion and F^- ions occupy all the tetrahedral holes, there will be two F^- ions for each Ca⁺² ions, thus the stoichiometry of the compound is 1:2.

(4) Each Ca⁺² ion is surrounded by 8F^- ions and each F^- ions is surrounded by 4Ca⁺² ions. The Coordination number of Ca⁺² ion is eight and that of F^- ion is four, this is called 8:4 Coordination.

 (5) Each unit cell has 4 calcium ions and 8 fluoride ions so formula of unit cell is Ca₄F₈ which is explained as  below

      No. of Ca⁺² ions = 8(at corners)´1/8 + 6 (at face centres)´1/2

      No. of F ions = 8 (within the body)´1 = 8

      Thus the number of CaF₂ units per unit cell is 4.

(6) Radius Ratio, packing efficiency ,void % and density:

Other examples: of structure are SrF₂, BaCl₂, BaF₂, PbF₂, CdF₂, HgF₂, CuF₂, SrCl₂, etc.

CAESIUM CHLORIDE (CsCl) STRUCTURE:

The caesium chloride crystal is composed of equal number of caesium (Cs⁺) and Chloride Cl^- ions. The radii of two ions (Cs⁺ = 169 pm and Cl^- = 181 pm) led to radius ratio of Cs+ to Cl- as 0.93 which suggest a body centred cubic structure having a cubic hole

The salient features of this structure are as follows:

(1)  The chloride ion form the simple cubic arrangement and the caesium ions occupy the cubic interstitial holes. In other words Cl^- ions are at the corners of a cube whereas Cs⁺ ion is at the centre of the cube or vice versa

(2)  Each Cs⁺ ion is surrounded by 8 Cl^- ions and each Cl^- ion in surrounded by 8 Cs⁺ ions. Thus the Co – ordination number of each ion is eight. 

(3)  For exact fitting of Cs⁺ ions in the cubic voids the ratio r Cs⁺/rCl^-  should be equal to 0.732. However, actually the ratio is slightly larger (0.93). Therefore packing of Cl^- ions slightly open up to accommodate Cs⁺ ions.

(4)  The unit cell of caesium chloride has one Cs⁺ ion and one Cl^- ion as calculated below

      No. of Cl^- ion = 8(at corners) ´1/8 = 1

      No. of Cs⁺ ion = 1(at body centre)´1=1

      Thus, number of CsCl units per unit cell is 1

(5)  Relation between radius of cation and anion and edge length of the cube,

Effect of temperature on crystal structure:

Increase of temperature decreases the coordination of number, e.g. upon heating to
760 K, the CsCl type crystal structure having coordination 8:8 changed to NaCl type crystal structures having coordination 6:6.

Effect of pressure on crystal structure:

Increase of pressure increases the Co – ordination number during crystallization e.g. by applying pressure, the NaCl type crystal structure having 6:6 coordination number changes to CsCl type crystal having coordination number 8:8

Other common examples  of this type of structure are CsBr, CsI, TlCl, TlBr, TlI and TlCN

      Higher coordination number in CsCl(8:8) suggest that the caesium chloride lattice is more stable than the sodium chloride lattice in which Co – ordination number is 6:6. Actually the caesium chloride lattice is found to be 1% more stable than the sodium chloride lattice. Then the question arises why NaCl and other similar compounds do not have CsCl type
lattice – This is due to their smaller radius ratio. Any attempt to pack 8 anions around the relatively small cation (Li⁺, Na⁺, K⁺, Rb⁺) will produce a state in which negative ions will touch each other, sooner they approach a positive ion. This causes unstability to the lattice.   

THE WURTIZE STRUCTURES (ZnS):

It is an alternate form in which ZnS occurs in nature. The main features of this structure are

                                    A unit cell representation of wurtzite structure

(1) Sulphide ions have HCP arrangement and zinc ions occupy tetrahedral voids.

(2)  Only half the alternate tetrahedral voids are occupied by Zn⁺² ions.

(3)  Coordinate no. of Zn⁺² ions as well as S^-2 ions is 4. Thus, this structure has 4 : 4 coordination.

(4)  No. of Zn⁺² ions per unit cell: 

                                 = 4(within the unit cell) ´1 + 6(at edge centres) ´1/3  = 6

(5) No. of S^-2 ions per unit cell =

                                 = 12(at corners) ´1/6 +2 (at face centres) ½ +3 (within the unit cell)1=6

                            Thus, there are 6 formula units per unit cell.

ZINCE BLENDE (ZnS) TYPE STRUCTURE (SPHELERITE):

The zinc sulphide crystals are composed of equal number of Zn⁺² and S^2- ions. The radii of the two ions (Zn⁺² = 74 pm and S^-2 = 184 pm) led to the radius r⁺/r^- as 0.40 which suggests a tetrahedral arrangement.

The salient features of this structure are as follows:

(1)The Sulphide ions are arranged in ccp arrangement, i.e. sulphide ions are present at the corners and the centres of each face of the cube  

(2)  Zinc ions occupy tetrahedral hole. Only half (50 %) of the tetrahedral holes are occupied by Zn⁺² so that the formula of the zinc sulphide is ZnS i.e. the stoichiometry of the compound is 1:1 (Only alternate tetrahedral holes are occupied by Zn⁺²)

(3)  Since the void is tetrahedral, each zinc ion is surrounded by four sulphide ions and each sulphide ion is surrounded tetrahedrally by four zinc ions. Thus zinc sulphide has [4:4] Coordination.

(4)  For exact fitting of Zn⁺² in the tetrahedral holes, formed by close packing of S^-2 ions, the ratio Zn⁺²/S^-2 should be 0.225. Actually this ratio is slightly large (0.40)

(5)  There are four Zn⁺² ions and four S^-2 ions per unit cell as calculated below:

      No. of S^-2 ions = 8(at corners)´1/8 + 6(at face centres)´1/2 = 4

      No. of Zn⁺² ions = 4 (within the body)´1 = 4

(6) Density, Packing efficiency (PE) and Void % ;

Examples: Thus, the number of ZnS units per unit cell is equal to 4. Some more examples of ionic solids having Zinc blende structures are CuCl, CuBr, CuI, AgI, BeS (beryllium sulphide).

ILLUSTRATIVE EXAMPLE:       If silver iodide crystallizes in a zinc blende structure with I^- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag⁺ ions.

SOLUTION:      In AgI, if there are nI^- ions, there will be nAg⁺ ions. As I^- ions form the lattice, number of tetrahedral voids = 2n. As there are nAg⁺ ions to occupy these voids, therefore fraction of tetrahedral voids occupied by Ag⁺ ions = n/2n = ½ = 50%.

SODIUM CHLORIDE (ROCK SALT TYPE) STRUCTURE:

The sodium chloride structure is composed of Na⁺ and Cl^- ions. The number of sodium ions is equal to that of Cl^- ions. The radii of Na⁺ and Cl^- ions 95 pm and 181 pm giving the radius ratio of 0.524

The radius ratio of 0.524 for NaCl suggest an octahedral void. Thus the salient features of this structure are as follows:

(1) Chloride ions (In a typical unit cell) are arranged in cubic close packing (ccp). In this arrangement, Cl^- ions are present at the corners and at the centre of each face of the cube. This arrangement is also regarded as face centred cubic arrangement (fcc).

(2)  The sodium ions are present in all the octahedral (Voids) holes.

(3) Since, the number of octahedral holes in ccp structure is equal to the number of anions, every octahedral hole is occupied by Na⁺ ions. So that the formula of sodium chloride is NaCl i.e. Stoichiometry of NaCl is 1:1.

(4 ) Since there are six octahedral holes around each chloride ions, each Cl^- ion is surrounded by 6 Na⁺ ions. Similarly each Na⁺ ion is surrounded by 6 Cl^- ions. Therefore, the coordination number of Cl^- as well as of Na⁺ ions is six. This is called 6:6 coordination.

(A) Nearest neighbor of Na⁺ and Cl^- ions is 6 (Six) at distance a/2.

(B) Next nearest Na⁺ and Cl^- ions is 12 at distance a/root 2

(5)  It should be noted that Na⁺ ions to exactly fit the octahedral holes, the radius ratio of sodium and chloride ions should be equal to 0.414. However, the actual radius ratio 0.524 exceeds this value. Therefore to accommodate large Na⁺ ions, the Cl^- ions move apart slightly i.e. they do not touch each other and form an expanded face centred lattice.

(6)  The unit cell of sodium chloride has 4 sodium and 4 chloride ions as calculated below

      No of sodium ions =  12 (at edge centres) ´1/4 + 1 (at body centre)´1= 4

      No of chloride ions = 8(at corner)´1/8+6 (at face centres) ´1/2 = 4

      Thus, the number of NaCl units per unit cell is 4.

(7)  The edge length of the unit cell of NaCl type of crystal is 2(r+R) where r = radii of Na⁺ ion and R is radii of Cl^-

Thus, the distance between Na⁺ and Cl^- ions = a/2

(8) Density and packing efficiency of NaCl are as:

Examples of NaCl type ionic salts:

Most of the halides of alkali metals, oxides and sulphides of alkaline earth metals have this type of structures.

(1) Group 1^st Halides ie NaI, KCl, RbI, RbF ( except Cs halides) .

(2) Group 2^nd  oxide   MgO, CaO, BaO, SrO ( except BeO)

(3) Ammonium Halides ie NH₄Cl ,NH₄Br ,NH₄I etc

(4) Silver Halides ie AgF, AgCl, AgBr ,( except AgI)

(5) Other examples , TiO ,FeO, NiO etc

Note: Ferrous oxide also has sodium chloride, types structure in which O^-2 ions are arranged in ccp and Fe⁺² ions occupy octahedral ions. However, this oxide is always non – Stoichiometric and has the composition Fe_0.95 O  It can be explained on the assumption that some of the Fe⁺² ion are replaced by 2/3^rd as many Fe⁺³ ions in the octahedral voids.