
Define metallic bond. Iron has a body centred cubic unit cell with a cell dimension of $286.65pm$. Calculate the density of iron (atomic mass of iron is $56g/mol$).
Answer
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Hint: The constituent particles of metallic solids are metal atoms which are held together by metallic bond. To find density use formula $d = \dfrac{{zM}} {{{a^3} {N_A}}} $ where, $z$ is the number of atoms present in one unit cell, $M$ is the molar mass of iron, $a$ is the edge length and ${N_A} $ is the avogadro number.
Complete Step by Step answer:
In order to explain the nature of metallic bonds, Lorentz proposed a simple theory known as electron gas model or electron sea model. A metal atom is supposed to consist of two parts, valence electrons and the remaining part (the nucleus and inner shells) are called kernels. Lattice sites are occupied by the kernels of metal atoms while the space between the kernel is occupied by valence electrons. Due to small ionisation energy the valence electrons are not held by the nucleus firmly. Therefore, the electrons leave the field of influence of one kernel and come under the influence of another kernel. Thus the electrons are not localized but are mobile. The simultaneous attraction between the kernels and the mobile electron which hold the kernel together is known as metallic bond.
To calculate the density of iron we will use the formula $d = \dfrac{{zM}} {{{a^3} {N_A}}} $
where, $z$ is the number of atoms present in one unit cell, $M$ is the molar mass of iron, $a$ is the edge length and ${N_A} $is the avogadro number.
Given information: $M = 56g/mol$, $a = 286.65pm$, $z = 2$ and ${N_A} = 6.022 \times {10^ {23}} g/mol$.
Putting all these values in the above formula we get, $d = \dfrac {2 \times 56} {(286.65 \times {10}^ {- 12})^3}$
$ \Rightarrow d = 0.00000079 \times {10^7} g/c{m^3} d = 0.00000079 \times {10^7} g/c{m^3} $
$ \Rightarrow d = 7.9g/c{m^3} $
Hence the density of iron is $7.9g/c{m^3} $.
Note:The metal possesses various properties like lustrous, high conductivity, malleability and ductility, alloy formation, cohesive forces in lattice, has close packed structures.
Complete Step by Step answer:
In order to explain the nature of metallic bonds, Lorentz proposed a simple theory known as electron gas model or electron sea model. A metal atom is supposed to consist of two parts, valence electrons and the remaining part (the nucleus and inner shells) are called kernels. Lattice sites are occupied by the kernels of metal atoms while the space between the kernel is occupied by valence electrons. Due to small ionisation energy the valence electrons are not held by the nucleus firmly. Therefore, the electrons leave the field of influence of one kernel and come under the influence of another kernel. Thus the electrons are not localized but are mobile. The simultaneous attraction between the kernels and the mobile electron which hold the kernel together is known as metallic bond.
To calculate the density of iron we will use the formula $d = \dfrac{{zM}} {{{a^3} {N_A}}} $
where, $z$ is the number of atoms present in one unit cell, $M$ is the molar mass of iron, $a$ is the edge length and ${N_A} $is the avogadro number.
Given information: $M = 56g/mol$, $a = 286.65pm$, $z = 2$ and ${N_A} = 6.022 \times {10^ {23}} g/mol$.
Putting all these values in the above formula we get, $d = \dfrac {2 \times 56} {(286.65 \times {10}^ {- 12})^3}$
$ \Rightarrow d = 0.00000079 \times {10^7} g/c{m^3} d = 0.00000079 \times {10^7} g/c{m^3} $
$ \Rightarrow d = 7.9g/c{m^3} $
Hence the density of iron is $7.9g/c{m^3} $.
Note:The metal possesses various properties like lustrous, high conductivity, malleability and ductility, alloy formation, cohesive forces in lattice, has close packed structures.
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