Question

A solid compound (XY) has NaCl type structure. If the radius of cation is 120 pm, the edge length of unit cell of XY will be

Answer 1

Answer:

819.71 pm

Explanation:

Nacl is ccp therefore √2a=4r

Put the value of radius r in the above equation, i.e. 120 pm

U will get the value of a which is the answer

Mark me as brainliest !!

Answer 2

Answer:

The edge length (a) of unit cell of XY is equal  to $819.71pm$.

Explanation:

Consider that $r_{+} =$ radius of cation

and, $r_{-}=$  radius of an anion

Now $XY$ has $NaCl$ type structure,

Therefore the radius ratio:

$\frac{r_{+}}{r_{-}}=0.414$

$r_{-}=\frac{r_{+}}{0.414}$                                                      ................(1)

Given, $r_{+}=120pm$  put in eq.(1)

∴  $r_{-}=\frac{120}{0.414} =289.85pm$

Now, Edge length $a=2(r_{+}+r_{-})$

$a=2(120+289.85)\\a=2(409.85)\\a=819.71pm$

Therefore the edge length (a) of given unit cell will be equal to 819.71pm.