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JEE Main Mock Test Series 2020 & Previous Year Papers

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Question 1: The arrangement of X ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X– is 250 pm, the radius of A+ is    (IIT JEE 2013)
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(a) 104 pm
(b) 125 pm
(c) 183 pm
(d) 57 pm
Answer: a
Solution:
A+ is present in the octahedral void of X-.
For octahedral Void
r+/r- = 0.414
or
r+ = 0.414 r- = 0.414 × 250 pm ≈104 pm
Hence, the correct option is a.

Question 2: A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below.    (IIT JEE-2012)
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The empirical formula of the compound is
(a) MX
(b) MX2
(c) M2X
(d) M5X14
Answer: b
Solution:
X- ions are present at the corners and  face centers:
Number of x- ions per lattice = 1/8 (Number of ions present at corners of cube) + ½ (Number of ions present at face centers) = 1/8(8) + ½(6) = 4
M+ ions are present at the four edge centers and at body center. corners and on face centers:
Number of M+ ions per lattice = 1/4(Number of ions present at edge centers of cube) + (Number of ions present at body center) = 1/4(4) + 1 = 2
Now: X: M = 4:2 = 2:1
So, the formula should be MX2
Hence, the correct option is b.

Question 3: The packing efficiency of the two-dimensional square unit cell shown below is   (IIT JEE-2010)
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(a) 39.27%
(b) 68.02%
(c) 74.05%
(d) 78.54%
Answer: d
Solution:
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Total number of the atoms per unit cell = ¼(4) +1 = 2
Let, Radious of the atom = r
SR = QR =PQ = SP =L
&
QS = 4r
Now, in triangle SRP,
SR+ QR2 =SQ2
Or
L2+L2 =(4r)2
r = L/(2√2 )
Or
Packing Efficiency = [(2× Volume of one atom)/(Total volume of the unit cell)]×100
=100 × {2π[L/(2√2 )]2 }/L2
= (π/8 )×100 = 78.54%
Hence, the correct option is D.

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