(B) For an $fcc$ lattice, the atoms touch along the face diagonal.The relation between the edge length $(a)$ and the radius $(r)$ is given by $4r = a

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(B) For an $fcc$ lattice, the atoms touch along the face diagonal.The relation between the edge length $(a)$ and the radius $(r)$ is given by $4r = a \sqrt{2}$.The diameter $(d)$ of the atom is $2r$.Therefore, $d = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.Given $a = 407 \ pm$, we have $d = \frac{407}{1.414} \approx 287.8 \ pm$.

Class 12 Chemistry Solid State Crystal structure and Coordination number

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Gold has a face-centred cubic $(fcc)$ lattice with an edge length of the unit cube of $407 \ pm$. Assuming the closest packing, the diameter of the gold atom is .............. $pm$.

AIIMS 2011 All AIIMS

A
$576.6$
B
$287.8$
C
$352.5$
D
$704.9$

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Solid State

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Crystal structure and Coordination number

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AIIMS 2011 Paper All AIIMS years →

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An element has a body-centered cubic $(BCC)$ structure with an edge length of $288 \ pm$. The density of the element is $7.2 \ g/cm^3$. How many atoms are present in $208 \ g$ of the element?

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$A$ compound is formed by elements $A$ and $B$. This crystallizes in a cubic structure where atoms $A$ are at the corners of the cube and atoms $B$ are at the body center. The simplest formula of the compound is:

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Gold has a face-centred cubic $(fcc)$ lattice with an edge length of the unit cube of $407 \ pm$. Assuming the closest packing, the diameter of the gold atom is .............. $pm$.

Similar Questions

If the radius of an anion is double that of a cation,the coordination number of the cation and the type of hole occupied are respectively:

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