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Test: Packing Efficiency (NCERT)


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Test: Packing Efficiency (NCERT) for NEET 2022 is part of Topic-wise MCQ Tests for NEET preparation. The Test: Packing Efficiency (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Packing Efficiency (NCERT) MCQs are made for NEET 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Packing Efficiency (NCERT) below.
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Test: Packing Efficiency (NCERT) - Question 1

The edge length of a fcc  is 508 pm. If the radius of cation is 110 pm, the radius of anion is

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 1

For fcc,
2(r+ + r-) = a
2(110 + r-) = 508
r- = 508/2 - 110 = 144 pm

Test: Packing Efficiency (NCERT) - Question 2

A metal X crystallises in a face-centred cubic arrangement with the edge length 862 pm. What is the shortest separation of any two nuclei of the atom ?

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 2

For fcc arrangement , distance of nearest neighbour (d) is

= 609.6 pm

Test: Packing Efficiency (NCERT) - Question 3

The edge length of sodium chloride unit cell is 564 pm. If the size of Cl- ion is 181 pm. The size of Na+ ion will be

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 3

2(r+ + r-) = a
2(2Na+ + rCl-) = 564
2Na = 564/2 - 181 = 101 pm

Test: Packing Efficiency (NCERT) - Question 4

If the distance between Na+ and Cl- ions in NaCl crystals is 265 pm, then edge length of the unit cell will be?

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 4

In NaCl crystal , Edge length = 2 x distance between Na+ and Cl
=2 x 265 = 530 pm

Test: Packing Efficiency (NCERT) - Question 5

The radius of Na+ is 95pm and that of Cl- is 181 pm. The edge length of unit cell in NaCl would be (pm).

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 5

a = (r+ + r-) = 2(95 + 181) = 552 pm

Test: Packing Efficiency (NCERT) - Question 6

Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?157 pm

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 6

For fcc,
r = √2 / 4 x a = √2 / 4 x 361

= 127 pm

Test: Packing Efficiency (NCERT) - Question 7

Total volume of atoms present in a fcc unit cell of a metal with radius r is

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 7

a = 2√2 r
Volume of the cell = a3 = (2√2 r)3 =  16√2r3
No. of shperes in fcc = 8 x 1/8 + 6 x 1/2 = 4
Volume of the four spheres = 4 x 4/3πr3
= 16πr3

Test: Packing Efficiency (NCERT) - Question 8

The relation between atomic radius and edge length 'a' of a body centred cubic unit cell:

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 8

Distance between nearest neighbours, d = AD/2
In right angled △ABC,AC= AB2 + BC2 
AC2 = a2 + a2  or AC = √2a
Now in right angled △ADC,
AD2 = AC2 + DC2
∴ d = √3a / 2
Radius , r = d/2 = √3/4 a

Test: Packing Efficiency (NCERT) - Question 9

Edge length of unit cell of Chromium metal is 287 pm with the arrangement. The atomic radius is the order of:

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 9

In bcc lattice, r = √3/4
r = 
= 124.27 pm

Test: Packing Efficiency (NCERT) - Question 10

The fraction of total volume occupied by the atoms present in a simple cube is

Detailed Solution for Test: Packing Efficiency (NCERT) - Question 10

For simple cube,
Radius(r) = a / 2 [a = edge length]
Volume of the atom = 
Packing fraction = 
Volume of the sphere (atoms) in an unit cell.
For simple cubic, Z = 1 atom
Packing fraction = 

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