If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is

Question

If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is (A) r = 0.414R (B) R = 0.414r (C) r=2R (D) r=√2R

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f6ad0a8ddeffd635a4020abeffcbcd1b-.jpeg" /> (2R)2=(R+r)2+(R+r)2 √2 R=R+r (√2-1)R=r 0.414R=r

Q. If the radius of an octahedral void is $r$ and radius of atoms in close packing is $R$ , the relation between $r$ and $R$ is

$r = 0.414 R$

$R = 0.414 r$

$r = 2 R$

$r = 2 R$

$(2 R)^{2} = (R + r)^{2} + (R + r)^{2}$
$2 R = R + r$
$(2 - 1) R = r$
$0.414 R = r$

Q. If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is

r=0.414R

R=0.414r

r=2R

r=2​R

(2R)2=(R+r)2+(R+r)2 2​R=R+r (2​−1)R=r 0.414R=r

Q. If the radius of an octahedral void is $r$ and radius of atoms in close packing is $R$ , the relation between $r$ and $R$ is

$r = 0.414 R$

$R = 0.414 r$

$r = 2 R$

$r = 2 R$

$(2 R)^{2} = (R + r)^{2} + (R + r)^{2}$
$2 R = R + r$
$(2 - 1) R = r$
$0.414 R = r$