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Oxidation numbers of A, B and C are +2, +5 and –2 respectively. Possible formula of compound is :
A
$A_2(BC_2)_2$
B
$A_3(BC_4)_2$
C
$A_2(BC_3)_2$
D
$A_3(B_2C)_2$
Detailed Solution
In a neutral compound, the sum of oxidation numbers is zero.
For $A_3(BC_4)_2$: $3(+2) + 2[(+5) + 4(-2)] = 6 + 2(-3) = 0$. Correct.
$A_2(BC_2)_2$: $4 + 2(5 - 4) = 6$; $A_2(BC_3)_2$: $4 + 2(5 - 6) = 2$; $A_3(B_2C)_2$: $6 + 2(10 - 2) = 22$. None of these is zero.
(This is like $Ca_3(PO_4)_2$.)
So the possible formula is $A_3(BC_4)_2$.
For $A_3(BC_4)_2$: $3(+2) + 2[(+5) + 4(-2)] = 6 + 2(-3) = 0$. Correct.
$A_2(BC_2)_2$: $4 + 2(5 - 4) = 6$; $A_2(BC_3)_2$: $4 + 2(5 - 6) = 2$; $A_3(B_2C)_2$: $6 + 2(10 - 2) = 22$. None of these is zero.
(This is like $Ca_3(PO_4)_2$.)
So the possible formula is $A_3(BC_4)_2$.