JEE Advanced 2018 Paper 1 · Chemistry · Solutions, Solid State and Surface Chemistry · hard

Magnesium aluminate (spinel), MgAl2O4, has a cubic structure in which O^2- ions form a cubic close-packed lattice, Mg^2+ ions occupy 1/8 of the…

Numerical answer

Magnesium aluminate (spinel),

\mathrm{MgAl_2O_4}

, has a cubic structure in which

\mathrm{O^{2-}}

ions form a cubic close-packed lattice,

\mathrm{Mg^{2+}}

ions occupy

1/8

of the tetrahedral voids and

\mathrm{Al^{3+}}

ions occupy

1/2

of the octahedral voids. The unit cell edge is

808

pm. Take atomic masses Mg = 24.3, Al = 27.0, O = 16.0 and

N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}

. The density of the crystal in

\mathrm{g\,cm^{-3}}

is ______. (Give the answer correct to two decimal places.)

Show answer and solution

Answer: 3.58

Count ions per unit cell. The spinel unit cell contains 32 oxide ions in ccp arrangement; a ccp lattice has one octahedral and two tetrahedral voids per anion, so the cell has 32 octahedral and 64 tetrahedral voids. Mg

^{2+}

in 1/8 of 64 tetrahedral voids gives 8, and Al

^{3+}

in 1/2 of 32 octahedral voids gives 16, matching

8

formula units of

\mathrm{MgAl_2O_4}

(and 32 O). Mass per cell

=8\times(24.3+54.0+64.0)/N_A=8\times142.3/6.022\times10^{23}=1.890\times10^{-21}

g. Volume

=(8.08\times10^{-8}\ \mathrm{cm})^3=5.275\times10^{-22}\ \mathrm{cm^3}

. Density

=1.890\times10^{-21}/5.275\times10^{-22}=3.58\ \mathrm{g\,cm^{-3}}

, which agrees with the tabulated density of spinel.

Report a mistake

← Previous Solutions, Solid State and Surface Chemistry question Next Solutions, Solid State and Surface Chemistry question →

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake