Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is 180 pm, the size of ammonium ion would be

Question

Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is 180 pm, the size of ammonium ion would be (A) 174 pm (B) 158 pm (C) 142 pm (D) 126 pm

Tardigrade Admin · Accepted Answer

For bcc structure Interionic distance r++r-=(√3/2)a (a = edge length) r+NH+4+r-Cl-=(√3/2)a r+NH+4+180=(√3/2)×390 r+NH+4 =338pm-180pm=158 pm

Q. Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is 180 pm, the size of ammonium ion would be

$174 p m$

$158 p m$

$142 p m$

$126 p m$

For bcc structure
Interionic distance $r^{+} + r^{-} = \frac{3}{2} a$
(a = edge length)
$r_{N H_{4}^{+}}^{+} + r_{C l^{-}}^{-} = \frac{3}{2} a$
$r_{N H_{4}^{+}}^{+} + 180 = \frac{3}{2} \times 390$
$r_{N H_{4}^{+}}^{+} = 338 p m - 180 p m = 158 p m$

Q. Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 pm. If the size of chloride ion is 180 pm, the size of ammonium ion would be

174pm

158pm

142pm

126pm

For bcc structure Interionic distance r++r−=23​​a (a = edge length) rNH4+​+​+rCl−−​=23​​a rNH4+​+​+180=23​​×390 rNH4+​+​=338pm−180pm=158pm

$174 p m$

$158 p m$

$142 p m$

$126 p m$

For bcc structure
Interionic distance $r^{+} + r^{-} = \frac{3}{2} a$
(a = edge length)
$r_{N H_{4}^{+}}^{+} + r_{C l^{-}}^{-} = \frac{3}{2} a$
$r_{N H_{4}^{+}}^{+} + 180 = \frac{3}{2} \times 390$
$r_{N H_{4}^{+}}^{+} = 338 p m - 180 p m = 158 p m$