CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A crystal is made up of metal iron 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation ‘'M1' occupies 50% of octahedral voids and the cation 'M2'occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation number of 'M1' and 'M2' are, respectively:


A

+2,+4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

+3,+2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

+4,+2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

+1,+3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

+2,+4


Answer:(A)

Explanation for correct option:

In ccp lattice oxide ion =corners + face centered

=18×8+12×6=4

Number of (O) =4(O)4

In ccp lattice ,

Number of octahedral void=4

Number of tetrahedral void=8

(M1)50% of octahedral void

Numberof(M1)=50100×4=2(M1)2

(M2)12.5%of tetrahedral void

Numberof(M2)=12.5100×8=1=(M2)1

Hence formula must be (M1)2(M2)O4

For whole atom to be neutral(as oxidation state of O=-2), O.N.ofM1=+2O.N.ofM2=+4

As, (M1)2(M2)O4=2×(+2)+1×(+4)+4×(-2)=4+4+(-8)=8-8=0

Therefore, correct option is A


flag
Suggest Corrections
thumbs-up
21
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Close Packing
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon