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