19,000+ solved questions for JEE Advanced, JEE Mains, NEET & IChO — with answers and expert explanations.
A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :
A sample of electrically neutral NaCl crystal is analysed for its density which has some unoccupied sites. Two readings where taken.
Assertion : If density of CsCl (cubic structure) is 3.99 g/cm3, the distance between Cs⁺ and Cl⁻ ions will be 357 pm [atomic mass of Cs = 133]. Reason : CsCl has bcc lattice.
In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is
Assertion : If three elements A, B and C crystallize in a cubic solid lattice with A atoms at the corners, B atoms at the cube centre and C atoms at the centre of the faces of the cube, then the formula of the compoun…
The unit cell of a binary alloy composed of A and B metals, has ccp. Structure with A atoms occupying the corners and B atoms occupying centres of each face of the cube. If during the crystallisation of this alloy, in…
Assertion : Graphite is an example of tetragonal crystal system. Reason : For a tetragonal system, a = b c, α = β = γ = 90º.
Ag crystallises as fcc. If radius of Ag is 144 pm then its density will be (a)₁₀ g cm–3
A metallic element has a cubic lattice. Each edge of the unit cell is 2 Å. The density of the metal is 2.5 g cm–3 . The unit cells in 200 g of the metal are
A compound contains two types of atoms X and Y. It crystallises in a cubic lattice with atoms X at the corners of the unit cell and atoms Y at the body centre. The simplest possible formula of this compound is
Density of NaCl crystal by not considering the unoccupied sites but only the occupied sites = 2.165 × 103 kg m–3. The percentage of unoccupied sites in NaCl crystal is
Total volume of atoms present in a face centred cubic unit cell of a metal is (r = radius of atom) (a) r (b) r (c) r (d) r
Edge length of a cube is 400 pm, its body diagonal would be
Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately:
In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the corners and face centres of the cubic unit cell but all the face-centred atoms along one of the axes are removed and ‘B’ atoms occupy octahedral voids, t…
In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be
Assertion : Glass is an amorphous solid. Reason : In glass, there is only short range order.
Number of unit cells in 4 g of X (atomic mass=40). Which crystallises in bcc pattern is (NA=Avogadro number)
Assertion : In caesium chloride structure, Cl⁻ ions are at the corners of a primitive cubic array and Cs⁺ ions fits into big central empty region in each Cl⁻ array. Reason : The radius ratio r Cs r Cl is greater than …
The density of KBr is 2.75 g cm –3. The length of the unit cell is 654 pm. Atomic mass of K = 39, Br = 80. Then the solid is
Percentage of free space in cubic close packed structure and in body centred packed structure are respectively
Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom is
Density of NaCl crystal assuming all sites are occupied = 2.178 × 103 kg m–3
Assertion : If density of CsCl (cubic structure) is 3.99 g/cm3, the distance between Cs⁺ and Cl⁻ ions will be 357 pm [atomic mass of Cs = 133]. Reason : CsCl has bcc lattice.
In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be