NH_4Cl crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl^(-) ion is 181 pm, radius…

Question

NH_4Cl crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl^(-) ion is 181 pm, radius for NH4+ ion is

MARKS Content Team · Accepted Answer

Correct answer: (A) 154.1 pm. In body centered cubic lattice, oppositely charged ions touch each other along the cube 2r_c+2r_a=√3a r_c+r_a=(√3)/2a =(√3)/2×387=335.15pm given r_a=181pm r_c=335.15-181.0=154.15pm

$N H_{4} C l$ crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the $C l^{-}$ ion is 181 pm, radius for $N H_{4}^{+}$ ion is

Solution

More Solid State Questions

NH4Cl crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl- ion is 181 pm, radius for NH4+ ion is

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More Solid State Questions

$N H_{4} C l$ crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the $C l^{-}$ ion is 181 pm, radius for $N H_{4}^{+}$ ion is