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# Match the type of unit cell given in Column I with the features given in Column II.  Column I Column IA.Primitive cubic unit cell1.Each of the three perpendicular edges compulsorily have the different edge length,i.e $\mathrm{a}\ne \mathrm{b}\ne \mathrm{c}$B.Body centred cubic unit cell2.Number of atoms per unit cell is one.C.Face centred cubic unit cell3.Each of the three perpendicular edges compulsorily have the same edge length,i.e.$\mathrm{a}=\mathrm{b}=\mathrm{c}.$D.End centred orthorhombic unit cell4.In addition to the contribution from the corner atoms the number of atoms present in a unit cell is one.  5.In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three.

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a
A - 1,2 ; B - 2,3 ; C - 3,4 ; D - 3,5
b
A - 2,3 ; B - 3,4 ; C - 3,5 ; D - 1,4
c
A -1,2,3 ; B - 2,3 ; C - 3,4,5 ; D - 3,5
d
A - 1,3 ; B - 1,3 ; C - 3,5 ; D - 1,4

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detailed solution

Correct option is B

A. →(2,3)  B. →(3,4)  C. →(3,5)  D. →(1,4)A. For primitive unit cell, a = b = c     Total number of atoms per unit cell = 1/8 x  8 = 1    Here, 1/8 is due to contribution of each atom present at corner.B. For body centred cubic unit cell, a = b = c    This lattice contain atoms at corner as well as body centre.    Contribution due to atoms at corn€r = 1/ 8 x 8 = 1 Contribution    Due to atoms at body centre = 1C. For face centred unit cell, a = b = c     Total constituent ions per unit cell present at corners                   =18×8=1   Total constituent ions per unit cell present at face centre                  =12×6=3D.For end centered orthorhombic unit cell,a≠b≠c   Total contribution of atoms present at corner                  =18×8=1   Total contribution of atoms present at end centre                  =12×2=1Hence, other than corner it contain total one atom per unit cell.

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