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

Class 12 Chapter 1

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

Class 12 Chapter 1

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)`

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

(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

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%`

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

Class 12 Chapter 1

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`

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

Class 12 Chapter 1

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

Class 12 Chapter 1

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 .

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

Class 12 Chapter 1

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 %`