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+2 votes
60.2k views
in Chemistry by (48.5k points)
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A crystal is made up of metal ions 'M1 ' ana 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1 ' occupies 50% of octahedral voids and the cation 'M2 ' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1 ' and 'M2 ' are, respectively :

(1) +2, +4

(2) +3, +1

(3) +1, +3

(4) +4, +2

2 Answers

+1 vote
by (15.0k points)
selected by
 
Best answer

Correct option is (1) +2, +4

In ccp lattice oxide ion = corners + face centered

\(= \frac 18 \times 8 + \frac 12 \times 6\)

\(=4\)

Number of (O) = (O)4

In ccp lattice,

Number of octahedral void = 4

Number of tetrahedral void = 8

(M1) → 50% of octahedral void

Number of (M1) = \(\frac{50}{100} \times 4\)

= 2

= (M1)2

(M2) → 12.5 % of tetrahedral void

Number of (M2) = \(\frac{12.5}{100} \times 8\)

= 1

= (M2)1

Hence formula must be (M1)2(M2)O4.

For whole atom to be neutral(as oxidation state of O = -2),

O.N. of M1 = +2

O.N. of M2 = +4

(M1)2(M2)O4

= 2 x (+2) + 1 x (+4) + 4 x (-2)

= 4 + 4 + (-8)

= 8 - 8

= 0

+3 votes
by (48.1k points)

Correct answer is (1) +2, +4

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