WB JEE 2021 | Solid State Question 4 | Chemistry

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

An element crystallises in a body centered cubic lattice. The edge length of the unit cell is 200 pm and the density of the element is 5.0 g cm^$$-$$3. Calculate the number of atoms in 100 g of this element.

2.5 $$\times$$ 10²³

2.5 $$\times$$ 10²⁴

5.0 $$\times$$ 10²³

5.0 $$\times$$ 10²⁴

WB JEE 2020

MCQ (Single Correct Answer)

-0.25

In the face centered cubic lattice structure of gold the closest distance between gold atoms is ('a' being the edge length of the cubic unit cell)

$$a\sqrt 2 $$

$${a \over {\sqrt 2 }}$$

$${a \over {2\sqrt 2 }}$$

$$2\sqrt 2 a$$

WB JEE 2018

MCQ (Single Correct Answer)

-0.33

A compound formed by elements X and Y crystallises in the cubic structure, where X atoms are at the corners of a cube and Y atoms are at the centre of the body. The formula of the compounds is

XY₂

X₂Y₃

XY₃

WB JEE 2017

MCQ (Single Correct Answer)

-0.5

In a close-packed body-centred cubic lattice of potassium, the correct relation between the atomic radius (r) of potassium and the edge-length (a) of the cube is

$$r = {a \over {\sqrt 2 }}$$

$$r = {a \over {\sqrt 3 }}$$

$$r = {{\sqrt 3 } \over 2}a$$

$$r = {{\sqrt 3 } \over 4}a$$

WB JEE Subjects