Relation Between r and a
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- 0.134
- 0.293
- 0.26
- 0.32
Here
\(r \text{ is radius of the tetrahedral void}\)
\(R \text{ is radius of the atoms in close packing}\)
- h=√32×4r
- h=√23×4r
- h=√23×2r
- h=√32×2r
- R=0.732r
- r=0.732R
- R=0.414r
- r=0.414R
Take \(\sqrt{\dfrac{2}{3}}\approx0.82\)
- rCs⨁+rCI⊖=√32a
- rCs⨁+rCI⊖=√3a
- rCs⨁+rCI⊖=3a
- rCs⨁+rCI⊖=3a2
- 6
- 4
- 8
- 3
- LiF
- CsF
- CsI
- LiI
- LiF
- CsF
- CsI
- LiI
- 157
- 181
- 108
- 128
- 7.22×1023
- 21.68×1023
- 14.45×1023
- 10.83×1022
(JEE Main 2021)
Find the relation between r and a in the following figure.
if a = x ∗ r , type 100x rounded off to 2 decimal places.
Given: Molar mass of metal 'X' =78 g mol−1
The edge length of a F.C.C. cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is
288 pm
398 pm
618 pm
144 pm
KCl crystallizes in the same type of lattice as does NACl. Given that rNa+rCl−=0.55 and rK+rCl−=0.74. Calculate the ratio of the side of the unit cell for KCl to that of NaCl.
1.123
0.891
1.414
0.414
The edge length of a unit cell of a metal whose molecular weight is 75 g/mol is 5 angstrom. This crystallises in a cubic lattice. If the density is 2 g/cc, find the radius of the metal atom. Give the answer in pm.
NA=6×1023
457.8 pm
216.5 pm
789.5 pm
257.8 pm
(AIPMT -2009)
RbCl has NaCl type lattice and its unit cell length is 0.30∘A greater than that for KCl. If rk+=1.33∘A, the ionic radius of Rb+ is
1.48 ∘A
1.63 ∘A
1.03 ∘A
None of these
- 0.69
- 0.78
- 0.88
- 0.98
- 2.16A∘
- 3.62A∘
- 3.94A∘
- 4.15A∘
Sodium metal crystallises in a body-centred cubic lattice with the cell edge a = 4.29 Å . What is the radius of the sodium atom?
2.45 Å
1.86 Å
3.56 Å
1.46 Å
A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
108 pm
40 pm
127 pm
80 pm
- rCs+rCl−=√32 a
- rCl−+rCs=√3 a
- rCs+rCl−=3a
- rCs+rCl−=3a2
- 1.857×10−8
- 2.371×10−7
- 3.817×10−8
- 9.312×10−7
- 0.128 nm
- 0.091 nm
- 0.180 nm
- 0.64 nm