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- Amines and other nitrogen compounds
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- Chemistry in daily life
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24 - Solid State Questions Answers
in a cubic closed packing of mixed oxide, it is found that lattice has O-2 ion & one half of the octahedral voids are occupied by trivalent cation A+3 & ONE EIGHTH of the FORMULA OF COMPOUND IS
1) AB2O4
2) A2B2O4
3) A2BO4
4)ABO4
the question is incomplete, after ONE EIGHTH, i guess, it will be one eighth of the tetrahedral void are occupied by B
so let no. of O = 1
then no. of octahedral voids = 1 and half of this = 1/2
similarly tetrahedral voids = 2 so 1/8 of it = 2*1/8 = 1/4
so formula = A.5B.25O1 or A2B1O4
if my assumption for question is correct then answer would be 3) here B is divalent positive ion.
what are d cordination no. of NaCl, ZnS, CsCl, CaF2,,Li2O upto four places, is there any trick to remember all these?
NaCl fcc structure 6 for Na and 6 for Cl
ZnS In ZnS , S forms the face centered cube and Zn occupies half of the tetrahedral holes..The unit cell can be devided into eight equal small cubes,each with 1 Zn and 4 S ions so the coordination no will be 4 for Zn and 4 for S
CsCl bcc structue 8 for Cs and 8 for Cl
CaF2 fcc structure of Ca with 100% filled tetrahedral voids by F, so coordination no. of Ca is 8 and that of F is 4
Li2O anti fluorite structure, fcc structure of O with 100% filled tetrahedral voids by Li so coordination of O is 8 and that of Li is 4
IF VOLUME OF 1 MOLE OF STRNTIUM CHLORIDE (HAS FLUORITE STR) IS X CM3 & VOL OF UNIT CELL IS Y CM3 D NA CONSTANT IS GIVEN BY
ANS ( X/Y) X4
since 1 unit cell of SrCl2 contains 4 molecules of the same so ratio will be 4X/Y
IN A FACE CENTRED CUBIC LATTICE A UNIT CELL IS SHARED EQUALLYBY HOW MANY UNIT CELLS
ANS 6
due to the reason that a unit cell contribute to 6 of the face attached memers
IN CRYSTAL OF WHICH ONE OF D FOLLOWING IONIC COMPD. WOULD U EXPECT MAXm DISTANCE B/W CENTRE OF ANION &CATION
1)CsI
2)CsF
3)LiF
4)LiI
accordong to my opinion it will be CsI
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
if x is length of body diagonal ,then distance b/w 2 nearest cation in rock salt str is
ans x/(61/2)
two nearest cations will be at corner and face centre so distance between these two will be y = a/21/2
here a is side length and according to the given condition x = a31/2
so a = x/ 31/2
so y = x / 61/2
in the crystal str of Fe3O4 Fe2+& Fe3+ occupy resp.
50% octatrahedral &12.5% tetrahedral viods
let atom of O = 1
so 50% Fe++ in octahedral void = 1*1/2 = 1/2
and 12.5 % Fe+++ in tetrahedral voids = 2*1/8 = 1/4
so total Fe = 1/2 + 1/4 = 3/4
so name of the compound = Fe3/4O = Fe3O4
d possible types of two dmensional lattice &3 D primitive unit cells are resp.
ans 5& 7
and in 3 D these are 7 crystal system, I have given an article for these 7 already
how many atoms are there in a cubic unitcell having one atom on each corner &two atoms on each body diagonal of cube?
& also if instead 2 there are 3 atoms at body diagonal?
corner atoms = 8*1/8 = 1
diagonal atoms = 4*2 = 8
total = 9
in second case calculate similarly