JEE Advanced · Chemistry · 15. Solid State
A compound \(\mathrm{M}_{\mathrm{p}} \mathrm{X}_{\mathrm{q}}\) has cubic close packing (ccp) arrangement of \(\mathrm{X}\). Its unit cell structure is shown below. The empirical formula of the compound is
- A \(\mathrm{MX}\)
- B \(\mathrm{MX}_{2}\)
- C \(\mathrm{M}_{2} \mathrm{X}\)
- D \(\mathrm{M}_{5} \mathrm{X}_{14}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{MX}_{2}\)
Step-by-step Solution
Detailed explanation
No. of \(M\) atoms \(=\frac{1}{4} \times 4+1=1+1=2\)
No. of \(X\) atoms \(=\frac{1}{2} \times 6+\frac{1}{8} \times 8=3+1=4\)
So, formula \(=M_{2} X_{4}=M X_{2}\)
No. of \(X\) atoms \(=\frac{1}{2} \times 6+\frac{1}{8} \times 8=3+1=4\)
So, formula \(=M_{2} X_{4}=M X_{2}\)
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