Question No:1
A compound PQ has NaCl type structure. The
face diagonal of the unit cell is 240 √2 pm.
The nearest inter ionic distance between P+
and Q- is (in pm)
1. 240 √2
2. 120/2
3. 60
4. 120
Answers
answer : option (3) 60 pm
It has given that a compound PQ has NaCl type structure. the face diagonal of unit cell is 240/√2 pm.
we have to find the nearest inter ionic distance between P⁺ and Q¯ .
we NaCl crystal structure is similar to face centered cubic lattice. where Na⁺ ions are located at octahedral voids (i.e., edge centres and centre of cube ) and Cl¯ ions are located at all corners of cubic lattice and face centres.
here face diagonal of unit cell = 240/√2
so edge length of unit cell = (240/√2)/√2 = 120 pm
see figure , it is clear that edge length of unit cell = 2 × radius of anions + 2 × captions
= 2Q¯ + 2P⁺
= 2(P⁺ + Q¯) = 120
⇒P⁺ + Q¯ = 60 pm
therefore nearest inter ionic distance between P⁺ and Q¯ is 60 pm.
Answer:
answer : option (3) 60 pm
It has given that a compound PQ has NaCl type structure. the face diagonal of unit cell is 240/√2 pm.
we have to find the nearest inter ionic distance between P⁺ and Q¯ .
we NaCl crystal structure is similar to face centered cubic lattice. where Na⁺ ions are located at octahedral voids (i.e., edge centres and centre of cube ) and Cl¯ ions are located at all corners of cubic lattice and face centres.
here face diagonal of unit cell = 240/√2
so edge length of unit cell = (240/√2)/√2 = 120 pm
see figure , it is clear that edge length of unit cell = 2 × radius of anions + 2 × captions
= 2Q¯ + 2P⁺
= 2(P⁺ + Q¯) = 120
⇒P⁺ + Q¯ = 60 pm
therefore nearest inter ionic distance between P⁺ and Q¯ is 60 pm.