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The effective radius of the iron atom is 1.42A°. It has FCC structure. Calculate its density (Fe=56amu).

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Important Questions on Solid State

MEDIUM
JEE Main/Advance
IMPORTANT
If the anions (A) form hexagonal closest packing and cations (C) occupy only 23 of the octahedral voids in it, then the general formula of the compound would be
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JEE Main/Advance
IMPORTANT
NH4Cl crystallizes in a body-centered cubic type lattice with a unit cell edge length of 387pm. The distance between the oppositively charged ions in the lattice is
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JEE Main/Advance
IMPORTANT

The density of a pure substance 'A' whose atoms are packed in a cubic-close packed arrangement is 1 g cc-1. If B atoms can occupy tetrahedral void and all the tetrahedral voids are occupied by 'B' atoms, what is the density of resulting solid in g cc-1? [Atomic mass (A)=30 g mol-1 and atomic mass (B)=50 g mol-1].

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JEE Main/Advance
IMPORTANT
In the body-centered cubic unit cell \& face centered cubic unit cell, the radius of atom in terms of edge lengh(a) of the unit cell is respectively:
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JEE Main/Advance
IMPORTANT
In FCC unit cell, what fraction of edge length is not covered by atoms?
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JEE Main/Advance
IMPORTANT
Diamond structure can be considered as ZnS (Zinc blend) structure in which each Zn2+ in alternate tetrahedral void and S2 - in cubic close pack arrangement is replaced by one carbon atom. If C-C covalent bond length in diamond is 1.5A°, what is the edge length of diamond unit cell (z=8).
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JEE Main/Advance
IMPORTANT
Copper metal crystallizes in FCC lattice. Edge length of unit cell is 362pm. The radius of largest atom that can fit into the voids of copper lattice without disturbing it.
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JEE Main/Advance
IMPORTANT
Packing fraction in 2-D hexagonal arrangement is