Question

The edge length of fee cell is 508 pm. If radius of cation is 110pm,then the radius of the anion would be ___.

a) 120pm
b) 110pm
c) 144pm
d) 130pm​

Answer 1

• c) 144pm ✓

Step-by-step explanation :

We know that,

For any ionic substance in F.C.C arrangement,

• 
  `2(r {}^( + ) + r {}^( - ) ) = edge \: length`

Where,

• r+ = radius of the cation
• r- = radius of the anion

Plugging in the values,

`2(110pm+ r {}^( - ) ) = 508pm \\ 220pm \: + 2r {}^( - ) = 508pm \\ 2r {}^( - ) = 508pm - 220pm \\ 2r {}^( - ) = 288pm \\ r {}^( - ) = 288pm / 2 \\ r {}^( - ) = 144pm`

Thus, the radius of the anion would be equal to 144pm.