Question
C forms ccp, A is at 25%terahedral viod &B is at 50% octahedral viod . if the particles along one of the fce diagonal are removed then formula?
ans A8B8C13
normal formula has C atom = 1
A atom = 2*1/4 = 1/2
B atom = 1*1/2 = 1/2
so structure = C2AB = C4A2B2
now we have to remove atoms from one of the face diagonal so
remove 2 corner C, so removed C = 1/8*2 = 1/4 and 1 face centre = 1/2 sum = 3/4
nothing else
so new formula = C4-3/4A2B2 = A8B8C13
Read 1 Solution.
C forms ccp(=fcc)
hence Z=4 where C occupies points at corners as well as at faces
Now Octahedral voids will also be equal to Z i.e. O.V.=4
So Tetrahedral voids will be 2 times Z = 2*4=8 i.e. T.V.=8
According to the question,
A occupies 25% of T.V. = (25/100) * 8 = 2
& B occupies 50% of O.V. = (50/100) * 4 = 2
So, A:B:C = 2:2:4 is expexted to decide the formula as A2B2C4
But, the question mentions that atoms along one of the face diagonal has been removed. It means that since C atoms were present at the faces and at the corners, hence 3 of C atoms went missing from one face diagonal. (i.e. 1 at face-centre & 2 at corners) . So total contribution of C that went missing from one face was 1*1/2(Face) + 2*1/8(Corner) = 3/4
But originally there were 4 C contribution. So we subtract 3/4 C contribution i.e. 4 - (3/4) = 13/4
Therefore, the new ratio of contributions is A:B:C = 2:2:13/4
To convert it into a whole number ratio, multiply it by 4. So new ratio is A:B:C = 8:8:13
And so the answer is A8B8C13
Thanks & Regards,
Nishant Srivastava (IIT-JEE Faculty)
9860673153