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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)
1. 104 pm
2. 125 pm
3. 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)
The empirical formula of the compound is
1. MX
2. MX2
3. M2X
4. M5X14
Answer: b
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)
1) 39.27%
2) 68.02%
3) 74.05%
4) 78.54%
Answer: D
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,
SR2 + QR2 =SQ2
Or
L2+L2 =(4r)2
r = L/(2√2 )
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.