JEE Advanced · Chemistry · 15. Solid State

The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is (are)

A The number of the nearest neighbours of an atom present in the topmost layer is 12
B The efficiency of atom packing is $74 %$
C The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively
D The unit cell edge length is $2 \sqrt{2}$ times the radius of the atom

Verified Solution

Answer & Solution

Correct Answer

(B) The efficiency of atom packing is $74 %$

Step-by-step Solution

Detailed explanation

CCP is ABC ABC ..... type packing
A] In topmost layer, each atom is in contact with 6 atoms in same layer and 3 atoms below this layer.
B] Packing fraction

= \frac{4 \times \frac{4}{3} π r^{3}}{{(\frac{4 r}{\sqrt{2}})}^{3}} = (0.74)

C] Each FCC unit has effective no of atoms = 4
Octahedral void = 4
Tetrahedral void = 8
D]

4 r = a \sqrt{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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	List- $I$		List- $II$
$(I)$	Probability of $(X_{2} \geq Y_{2})$ is	$(P)$	$\frac{3}{8}$
$(II)$	Probability of $(X_{2} > Y_{2})$ is	$(Q)$	$\frac{11}{16}$
$(III)$	Probability of $(X_{3} = Y_{3})$ is	$(R)$	$\frac{5}{16}$
$(IV)$	Probability of $(X_{3} > Y_{3})$ is	$(S)$	$\frac{355}{864}$
		$(T)$	$\frac{77}{432}$