The question is incomplete, here is the complete question:
Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in 
The image is attached below.
<u>Answer:</u> The density of CsI is 
<u>Explanation:</u>
To calculate the density of metal, we use the equation:
where,
= density
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of CsI = 259.8 g/mol
= Avogadro's number = 
a = edge length of unit cell = 
(Conversion factor: 
)
Putting values in above equation, we get:
Hence, the density of CsI is
Explanation :
As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.
He placed the elements with similar nature in the same group.
According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.
However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.
As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127 and tellurium isotopes are tellurium-128 and tellurium-130.
Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.
