An ionic compound is created with anions A & B present at corners of cube in 3:1 ratio. All face-centers & body centre contain cation Z. Atomic weight of A, B and Z are respectively 20, 40 & 15 amu. Edge length of unit cell is 1A0 , density =d(amu(A0)3) and empirical formula of ionic compound is AaBbZc, here a, b and c are smallest integer value. Then value of (a+b+c+d) is E the value E15 is (considering unit cell is electrically neutral).

Rank of anion −A=68=34Rank of anion −B=28=14Rank of cation −z=1+62=4Hence, ionic compound empirical formula is A3B1C16 d=(68×20)+(28×40)+(4×15)(1)3amuA3d=85 So, a+b+c+d=3+1+16+85=105

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An ionic compound is created with anions A & B present at corners of cube in 3:1 ratio. All face-centers & body centre contain cation Z. Atomic weight of A, B and Z are respectively 20, 40 & 15 amu. Edge length of unit cell is $1 \overset{0}{A}$ , density $= d (\frac{a m u}{{(\overset{0}{A})}^{3}})$ and empirical formula of ionic compound is $A_{a} B_{b} Z_{c}$ , here a, b and c are smallest integer value. Then value of $(a + b + c + d)$ is $E$ the value $\frac{E}{15}$ is (considering unit cell is electrically neutral).

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Detailed Solution

Rank of anion $- A = \frac{6}{8} = \frac{3}{4}$
Rank of anion $- B = \frac{2}{8} = \frac{1}{4}$
Rank of cation $- z = 1 + \frac{6}{2} = 4$
Hence, ionic compound empirical formula is $A_{3} B_{1} C_{16}$
$d = \frac{(\frac{6}{8} \times 20) + (\frac{2}{8} \times 40) + (4 \times 15)}{{(1)}^{3}} \frac{a m u}{A^{3}}$
$d = 85$
So, $a + b + c + d = 3 + 1 + 16 + 85 = 105$