In a solid AB having the NaCl structure A atoms occupy the corners of the cubic unit cell . If all the faces centred atoms along one of the axis are removed then the resultant stoichiometry of the solid is

Question

$(a)\;AB_2\qquad(b)\;A_2B\qquad(c)\;A_4B_3\qquad(d)\;A_3B_4$

Answer 1

Na in NaCl has 8 corners and 6 face atoms. If we remove face centered atom of one axis , two face atoms are removed.

Thus A is at 8 corners and B is at 4 faces.

$\therefore A = \large\frac{8}{8} = 1$

$\Rightarrow B = \large\frac{4}{2} = 2$

Thus formula is $AB_2$

Hence answer is (a)

Answer 2

According to NaCl structure so 'A' occupies 8 corners of the cubic unit cell which contribute 1/8 each to this cube therefore 8*1/8 = 1 also 6 face centered atoms contribute 1/2 to this cube therefore 6*1/2=3 Therefore total contribution of atom'A' is 4 Now B occupies octahedral voids(NaCl structure) which contribute 1/4 therefore 1+12*1/4 = 4 So the formula is A4B4 or AB Now you are removing all the face centered atoms along one axis(any axis origin at centre of cube) so the number of atoms removed from cube are 2 so contribution from 'A' remains 8*1/8(corner atoms) + 4*1/2(facecentered atoms) = 3 so the resultant becomes A3B4

Comments

But one octahedral  void is at center also.
So no of octahedral void after removing  atoms from a axis will be 12×1/4=3 not 4!!!