A compound formed by elements X and Y crystallizes in the cubic structures where Y atoms are at the corner of the cube and X atoms are at alternate edge centres. What is the formula of the compound?
Answer
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Hint:The regular arrangement of the constituents particles such as atoms, ions and molecules in a definite pattern in three-dimensional space is known as the crystal lattice or space lattice.
Complete step by step answer:
Here it is given that the atoms Y are present at the corners and we know there are eight corners in a simple unit cell and a simple cubic unit cell has eight atoms on the corner. So, the atom at the corner is shared by eight unit cells. Hence the contribution of each atom present at the corner is 1/8. As a result, the number of Y atoms presents In the unit cell is:
$8 \times \dfrac{1}{8} = 1$
Since there are twelve edges and each atom on the edge centre shared by four unit cells. Hence the contribution of each atom present at the edge centre is 1/4. Now it is given is the atom are present at alternate edge centre that means a number of an atom of X is $\dfrac{1}{2} \times 12 = 6$ and the contribution is \[\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 4$} \]. Hence the number of X atom present at alternate edge centre in a unit cell is:
$6 \times \dfrac{1}{4} = \dfrac{3}{2}$. Hence the formula of a compound is given as:
$X:Y$
$ \Rightarrow 1:\dfrac{3}{2}$
$ \Rightarrow 2:3$
The formula of a compound is ${X_2}{Y_3}$.
Note:
The number of particles present in a unit cell is called unit cell constant and is represented by “z”. The general expression for calculation of unit cell constant is given as:
$z = \dfrac{{{N_c}}}{8} + \dfrac{{{N_f}}}{2} + \dfrac{{{N_e}}}{4} + \dfrac{{{N_i}}}{1}$
Here subscript c, f, e and I represent corner, face centre, edge centre and inside. It is important to remember that in the case of compounds, the number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence it helps to predict the formula of a compound.
Complete step by step answer:
Here it is given that the atoms Y are present at the corners and we know there are eight corners in a simple unit cell and a simple cubic unit cell has eight atoms on the corner. So, the atom at the corner is shared by eight unit cells. Hence the contribution of each atom present at the corner is 1/8. As a result, the number of Y atoms presents In the unit cell is:
$8 \times \dfrac{1}{8} = 1$
Since there are twelve edges and each atom on the edge centre shared by four unit cells. Hence the contribution of each atom present at the edge centre is 1/4. Now it is given is the atom are present at alternate edge centre that means a number of an atom of X is $\dfrac{1}{2} \times 12 = 6$ and the contribution is \[\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
\kern-.15em\lower.25ex\hbox{$\scriptstyle 4$} \]. Hence the number of X atom present at alternate edge centre in a unit cell is:
$6 \times \dfrac{1}{4} = \dfrac{3}{2}$. Hence the formula of a compound is given as:
$X:Y$
$ \Rightarrow 1:\dfrac{3}{2}$
$ \Rightarrow 2:3$
The formula of a compound is ${X_2}{Y_3}$.
Note:
The number of particles present in a unit cell is called unit cell constant and is represented by “z”. The general expression for calculation of unit cell constant is given as:
$z = \dfrac{{{N_c}}}{8} + \dfrac{{{N_f}}}{2} + \dfrac{{{N_e}}}{4} + \dfrac{{{N_i}}}{1}$
Here subscript c, f, e and I represent corner, face centre, edge centre and inside. It is important to remember that in the case of compounds, the number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence it helps to predict the formula of a compound.
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