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.

Principle of atom conservation (PAOC):

We know that, in a chemical reaction, the atoms of each element remain conserved. If atoms are conserved, moles of atoms will also be conserved. This is known as the principle of atom conservation.

Total number of moles of an atoms of an element on reactant side = Total number of moles of atoms of an on product side.

Illustration of POAC: Take a chemical reaction as example.

(A) According to the principle of atom conservation (POAC) for K atoms:

    Total moles of K atoms in reactant = total mole of K atoms in products

or

 Moles of K atoms in KClO₃ = moles of K atoms in KCl.

Now, since 1 molecule of KClO₃ contains 1 atom of K, 1mole of KClO₃ contains 1 mole of K similarly, 1mole of KCl contains 1mole of K.

     Thus, mole of K atoms in KClO₃ = 1 × moles of KClO₃

        And mole of K atoms in KCl = 1 × moles of KCl

Hence            1 × moles of KClO₃ = 1 × moles of KCl

 (B) Similarly applying the principle of atom conservation for O atoms:

         Mole of O in KClO₃ = 3 ×moles of KClO₃

 And moles of O in O₂ = 2 × moles of O₂ 

      3 × moles of KClO₃ = 2 × moles of O₂

Illustrative example (1): Write the POAC equations for all the atoms in the following reaction.

Illustrative solution:

Applying POAC on (P) atoms:

                                                          P à H₃PO₄

Numbers of moles of (P) atoms in P₄ = Numbers of moles of (P) atoms in H₃PO₄

                                                         4x nP₄= 1x nH₃PO₄

Applying POAC on (H) atoms:

                                       1x nHNO3 = 3 x nH₃PO₄ + 2 x nH₂O

Applying POAC on (N) atoms:

                                       1 x nHNO₃ = 1 x nNO₂

Applying POAC on (O) atoms:

                                       3 x nHNO₃ = 4 x nH₃PO₄ +2 x nNO₂ + 1nH₂O

Related Questions:

Illustrative example (2) 27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.

Illustrative example (3)0.32 gm mole of LiAlH4 in ether solution was placed in a flask and 74 gm (1 mole ) of t- butyl alcohol was added . The product is LiAlHC12H27O3. Calculate mass of product if all lithium atoms are converted.

Illustrative example (4) All the carbon atom present in KH3(C2O4)2.2H2O weighting 254 gm is converted to CO2. How many gram of CO2 were obtained.

Illustrative example (5) For BaCl2.XH2O , 2.1 gm of Compound gives 2 gm of anhydrous BaSO4 (M.mass =233)upon treatment with H2SO4 then calculate value of X.

Illustrative example (6) In a determination of P an aqueous solution of NaH2PO4 is treated with a mixture of Ammonium and magnesium ions to precipitate magnesium Ammonium phosphate (Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate. Which is weighed. A solution of NaH2PO4 yield 3.33 gm of Mg2P2O7. What is weighed of NaH2PO4 was present originally.

DEGREE OF HARDNESS OF HARD WATER:

(1)The amount of hardness causing substances (soluble salts of calcium or magnesium) in a certain volume of water measures the extent of hardness or degree of hardness.

(2) Hardness of water is always calculated in terms of calcium carbonate although this is never responsible for causing hardness of water because of its insoluble character.

(3) The reason for choosing CaCO₃ as the standard for calculating hardness of water is the ease in calculation as its molecular weight is exactly 100.Thus the amount of various hardness causing substances in terms of CaCO₃ can be calculated on the basis of the following relations.

Thus the various types of harnesses in a water sample may be calculated as below.

Temporary hardness =Hardness due to Ca(HCO₃)₂ + Hardness due to Mg(HCO₃)₂

Permanent hardness = Hardness due to CaCl₂ + due to CaSO₄ + due toMgCl₂ + due toMgSO₄

Degree of hardness is usually expressed as parts per million (ppm) and thus may be defined as the number of parts by weight of CaCO₃ (equivalent to calcium and magnesium salts) present in a million (10⁶) parts by weight of water.

From the above definition, we can say that;

ILLUSTRATIVE EXAMPLE (1): Determine the degree of hardness of a sample of water containing 30 ppm of MgSO₄.

SOLUTION: 1 MgSO₄ =1CaCO₃

                      120 ppm = 100 ppm

                      Hence 30 ppm = 25 ppm   Ans