Question

If the radii of ${{{B}}^{{{3 + }}}}$ ions and ${{{O}}^{{{2 - }}}}$ ions are ${{27 pm and 140 pm}}$ respectively then which of the following should be correct crystal structure of boric anhydride.
(A)

(B)

(C)

(D) Hexagonal close packed(hcp) structure of oxide ions in which two-thirds of octahedral voids are occupied by boron atoms.

Answer 1

Hint: We calculate ionic radii ratio first to see the type of lattice or crystal formed by them. When we calculate the type of structure formed by the ions, we see the contribution of the ion in the lattice to figure out the chemical formula.

Complete step by step answer:
Firstly, we will refer to the table of radius ratio rules:
${{x = }}\dfrac{{{{{R}}^{{ + }}}}}{{{{{r}}^{{ - }}}}}$
Here R is radius of cation and r is the radius of anion. The ratio calculated shows a pattern or trend through which we predict the structure of lattice

x-value	Coordination number	Geometry
${{x < 0}}{{.155}}$	${{2}}$	Linear
${{0}}{{.155 }} \leqslant {{ x }} \leqslant {{ 0}}{{.225}}$	${{3}}$	Trigonal
${{0}}{{.225 }} \leqslant {{ x }} \leqslant {{ 0}}{{.414}}$	${{4}}$	Tetrahedral
	${{4}}$	Square planar
${{0}}{{.414 }} \leqslant {{ x }} \leqslant {{ 0}}{{.732}}$	${{6}}$	Octahedral
	${{12}}$	FCC/CCP or HCP
${{0}}{{.732 }} \leqslant {{ x }} \leqslant {{ 0}}{{.999}}$	${{8}}$	BCC

Now we calculate the ratio first,
Radius of cation ${{ = 27 pm}}$ Radius of anion ${{ = 140 pm}}$
${{x = }}\dfrac{{{{27}}}}{{{{140}}}}{{ pm = 0}}{{.193}}$
This ${{0}}{{.193}}$ on referring to the table above shows us coordination number ${{3}}$whose geometry is trigonal in shape. In opinions of Answer the D option is cancelled.
Now we determine the arrangement of the trigonal lattice by calculation the contribution of each ion in structure.
Number of boron ion per unit cell ${{ = 1}}$
Number of oxide ions per unit cell ${{ = 3 \times }}\dfrac{{{1}}}{{{2}}}{{ = }}\dfrac{{{3}}}{{{2}}}$
So the formula of the solid is ${{{B}}_{{1}}}{{{O}}_{\dfrac{{{3}}}{{{2}}}}}{{ = }}{{{B}}_{{2}}}{{{O}}_{{3}}}$
According to the calculations each corner of the triangle should be shared by two unit cells then only this formula can be possible.
Option A and C are eliminated because every corner is not shared by two unit cells whereas in option B every corner is shared by two unit cells.

So, the correct answer is Option B.

Additional information:
The simple cubic cell contains one atom as a whole at eight corners of the lattice
The body centered cubic cell contains two atoms as a whole at one at eight corners and one at body center.
The face centered cubic cell contains four atoms as whole, one at corners, 3 at face centers.

Note: The contribution of every atom should be calculated very carefully as it will be only predicting the right chemical formula. If we don’t know the correct contribution of atoms at corner position, body center, face center etc we cannot predict the lattice chemical formula.