Question

a solid compound (XY) has NaCl type structure. If the radius of the cation is 120 pm, the edge length of unit cell of XY will be
(1) 819.71 pm
(2) 772.25 pm
(3) 901.55 pm
(4) 715.68 pm​

Answer 1

Answer:

819.71 pm

Explanation:

Nacl is ccp therefore √2a=4r

r= 120

a=4/√2

a = 2√2×120

819.71 pm

Answer 2

Answer:

The unit cell's edge length, a, measured is $819.7pm$.

Therefore, option 1) $819.71pm$ is correct.

Explanation:

Given,

The radius of the cation, r₊ = $120pm$

The unit cell's edge length, a =?

As we know,

• $\frac{r_{+} }{r_{-} } = 0.414$

Here,

• r₊ = Radius of the cation
• r₋ = Radius of the anion

After putting the given values in the equation, we get:

• $\frac{120 }{r_{-} } = 0.414$
• r₋ = $289.85pm$

Now, as we know,

• Edge length, a = $2( r_{+} + r_{-} )$
• Edge length, a = $2( 120 + 289.85 )$ = $819.7pm$

Hence, the edge length of the unit cell, a = $819.7pm$.