1
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

The number of octahedral void(s) per atom present in a cubic close-packed structure is
A
1
B
3
C
2
D
4

Explanation

Number of octahedral voids is same as number of atoms.
2
MCQ (Single Correct Answer)

AIPMT 2012 Mains

Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are cooupied by a monovalent metal B. The formula of the oxide is
A
ABO2
B
A2BO2
C
A2B3O4
D
AB2O2

Explanation

Number of atoms in cubic close packing = 4 = O2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A2+ ions = 8 $$ \times $$ $${1 \over 4}$$ = 2

Number of octahedral voids = Number of B+ ions = N = 4

Ratio of ions will be,

O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB2O2
3
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
A
288 pm
B
408 pm
C
144 pm
D
204 pm

Explanation

Given $$a$$ = 408 pm

For the face centred cubic structure

r = $${a \over {2\sqrt 2 }}$$ = $${{408} \over {2\sqrt 2 }}$$ = 144 pm

$$ \therefore $$ Diameter = 2r = 2$$ \times $$144 = 288 pm
4
MCQ (Single Correct Answer)

AIPMT 2011 Mains

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y$$-$$) will be
A
275.1 pm
B
322.5 pm
C
241.5 pm
D
165.7 pm

Explanation

For NaCl crystal, $${{{r_ + }} \over {{r_ - }}} = 0.414$$

Given, r+ = 100 pm

$$ \therefore $$ $${{100} \over {{r_ - }}} = 0.414$$

$$ \Rightarrow $$ $${r_ - } = {{100} \over {0.414}}$$ = 241.5 pm

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