Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of A+ ion is 110 pm, the radius of B- ion in pm will be

Question

Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of A+ ion is 110 pm, the radius of B- ion in pm will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Edge length =2rA++2rB- 508=2× 110+2rB- 2rB-=508-220=288 rB-=144 pm

Q. Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of $A^{+}$ ion is 110 pm, the radius of $B^{-}$ ion in pm will be

Edge length $= 2 r_{A^{+}} + 2 r_{B^{-}}$

$508 = 2 \times 110 + 2 r_{B^{-}}$

$2 r_{B^{-}} = 508 - 220 = 288$

$r_{B^{-}} = 144$ pm

Q. Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of A+ ion is 110 pm, the radius of B− ion in pm will be

Edge length =2rA+​+2rB−​ 508=2×110+2rB−​ 2rB−​=508−220=288 rB−​=144 pm

Q. Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of $A^{+}$ ion is 110 pm, the radius of $B^{-}$ ion in pm will be

Edge length $= 2 r_{A^{+}} + 2 r_{B^{-}}$

$508 = 2 \times 110 + 2 r_{B^{-}}$

$2 r_{B^{-}} = 508 - 220 = 288$

$r_{B^{-}} = 144$ pm