Chemistry color{red} ✎ PROBLEM SOLVING TECHNIQUES

### PST-1 CALCULATION OF PARTICLES PER UNIT CELL

Q 3168145005

A compounds formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. Drive the formula of the compounds .
Class 12 Chapter 1
Solution:

As A atoms are presents at the 8 corners of the cubes , therefore number of atoms of A in the unit cell  = 1/8 xx 8 = 1

As B atoms are present at the face centres of the 6 corners of the cube , therefore number of atoms of B in the unit cell  = 1/2 xx 6 = 3

therefore Ratio of atoms A : B = 1 : 3

Hence , the formula of the compounds is AB_3
Q 3108145008

An ionic compounds made up of atoms A and B has a face - centred cubic arrangement in which atoms A are at the corners and atoms B are at the face- centres. If one of the atoms is missing from the corner , what is the simplest formula of the compound ?.
Class 12 Chapter 1
Solution:

No. of atoms of A at the corners = 7  ( because one A is missing )

therefore Contribution atoms of A towards unit cell = 7 xx1/8 = 7/8

No. of atoms B at the face -centres = 6

therefore Contribution of atom B towards unit cell = 6xx1/2 = 3

Ratio of A : B = 7/8 : 3

 = 7 : 24

therefore Formula is A_7B_(24)

### PST -2 CALCULATION ON NUMBER OF VOIDS AND FORMULA OF COMPOUNDS

Q 3118345200

In the mineral , spinel , having the formula MgAl_2O_4, oxide ions are arranged in the cubic close packing , Mg^(2+) ions occupy the tetrahedral voids while Al^(3+) ions occupy the octahedral voids.

(i) What percentage of tetrahedral voids is occupied by Mg^(2+) ions ?
(ii) What percentage of octahedral voids is occupied by Al^(3+) ions ?
Class 12 Chapter 1
Solution:

According to the formula MgAl_2O_4, if there are 4 oxide ions, there will be 1 Mg^(2+) ion and 2 Al^(3+) ions . But if the 4 O^(2-) ions are in ccp arrangement , there will be 4 octahedral voids and 8 tetrahedral voids .

Thus 1 Mg^(2+) ion is present in one of the 8 tetrahedral voids

therefore % of tetrahedral voids occupied by Mg^(2+) = 1/8 xx100 =12.5 %

Similarly , 2 Al^(3+) ions are present in two octahedral voids out of 4 available.

therefore %  of octahedral voids occupied by Al^(3+) = 2/4 xx 100 = 50%

### PST-3 RELATIONSHIP BETWEEN THE NEAREST NEIGHBOUR DISTANCE AND EDGE OF UNIT CELL OF CUBIC CRYSTAL

Q 3148345203

In face - centred cubic (fcc) crystal lattice, edge length is 400p m. Find the diameter of the greatest sphere which can be fitted in to the interstitial void without distortion of the lattice .
Class 12 Chapter 1
Solution:

For fcc , radius of atoms (R) = a/(2 sqrt2) = (400)/(2 sqrt2) = 141.4 p m

As octahedral voids is bigger in size than the tetrahedral voids the greatest sphere will fit in to octahedral voids .
Radius of octahedral voids (r) = 0.414xx141.4 p m = 58.54 p m

therefore Diameter of the greatest sphere fitting into the voids = 2xx58.54 p m = 117.08 p m

### PST-4 RELATION BETWEEN DENSITY AND EDGE OF CUBIC CRYSTALS.

Q 3108345208

Sodium has bcc structure with nearest neighour distance 365.9 p m Calculate its density (Atomic mass of sodium = 23)
Class 12 Chapter 1
Solution:

For the bcc structure , nearest , neighour distance (d) is related to the edge (a) as d = sqrt3/2

or a = 2/sqrt3 d = 2/(1.732) xx 365.9 = 422.5 p m

For bcc structure Z = 2

For sodium M = 23

therefore rho = (Z xx M)/(a^3 xx N_0)

 = (2xx23 g mol^(-1))/((422.5xx 10^(-10) cm)^3 xx (6.02xx10^23) mol^(-1)) = 1.51 g//cm^3
Q 3148445303

The density of copper metal is 8.95 g cm^(-3) . If the radius of copper atom is 127.8 p m, is the copper unit cell a simple cubic a body - centred cubic or a face - centred cubic ? ( given At, mass of Cu = 63.54 g mol^(-1) and N_A = 6.02xx10^(23) mol^(-1) )
Class 12 Chapter 1
Solution:

If copper atom were simple cubic ,

a = 2 , r = 2xx127.8 p m = 255.6 p m

Z = 1

rho = (Z xx M)/(a^3 xx N_0)

 = (1xx63.54 g mol^(-1))/((255.6 xx 10^(-10) cm)^3 xx(6.02xx10^(23) mol^(-1))) = 6.34 g cm^(-3)

Actual rho = 8.95 g cm^(-3)

Hence , copper atom is not simple cubic .

If copper atom were body - centred

a = (4r)/sqrt3 = (4xx63.54 g mol^(-1))/(1.732) p m = 295.15 p m

Z = 2

rho = (Z xx M)/( a^3 xx N_0)

 = (2xx63.54 g mol^(-1))/((295.15xx10^(-10) cm)^3 xx (6.02xx10^(23) mol^(-1))) = 8.21 g cm^(-3)

Hence , copper atom is not body - centred .

If copper atom were face- centred

a = 2sqrt2 r = 2xx1.414xx127.8 p m = 361.4 p m

rho = (Z xx M)/(a^3 xx N_0) = (4xx63.54 g mol^(-1))/((361.4 xx 10^(-10) cm)^3 xx(6.02xx10^(23) mol^(-1))) = 8.94 g cm^(-3)

Hence copper is face - centred cubic .

### PST-5 CALCULATION OF PERCENTAGE OF METAL IONS WITH VARIABLE VALENCY FROM FORMULA OF NON-STOICHIOMETRIC COMPOUND

Q 3118545400

The composition of a sample of wustite is Fe_(0.93)O_(1.00). What percentage of the iron is present in the form of Fe (III) ?
Class 12 Chapter 1
Solution:

The composition is Fe_(0.93)O_(1.00)  instead of FeO because some Fe^(2+) ions have been replaced by Fe^(3+) ions. Let us first calculate the number of Fe^(2+) and Fe^(3+) ions present . The formula Fe_(0.93)O_(1.00) implies that 93 Fe atoms are combined with 100O atoms . Out of 93Fe atoms , suppose Fe atoms present as Fe^(3+) = x . Then Fe^(2+) = 93-x As the compound is neutral , total charge on Fe^(2+) and Fe^(3+) ions = total charge on O^(2-) ions . Thus

3xx x +2 (93-x) = 2xx100

or 3x+186 -2x = 200

or x = 14

Fe^(3+) = 14

Hence Fe^(2+) = 93-14 = 79

Thus the given formula , in fact , implies Fe_(0.79)^(2+) \ \ Fe_(0.14)^(3+) \ \ O_(1.0)^(2-)

Total molar mass  = 0.93xx56+1xx16 = 68.08 g

Fe^(3+) present = 0.14xx56 = 7.84 g

therefore % age of Fe present as Fe(III) = (7.84)/(68.08) xx 100 = 11.5 %