NEET Practice Questions (MCQs) with Answers & Solutions

Schottky defect occurs mainly in electrovalent compound where positive and negative ions are of similar size.

The ratio Ni : O = 98:100

So, 100 atom of oxygen and 98 atom of Ni.

Let the number of atom Ni²⁺ = x

then number of Ni⁺³ =98-.x

(charge on Ni and O is equal)

So, then oxygen has charge =-2

3(98-x) + 2x = 2(100)

294-3x+2x=2(100)

-x=-94

x=94

% of Ni+2  = $\frac{atoms of N{i}^{+2}}{total number of atoms of Ni}\times 100$

94/98*100 = 96%

Percentage of Ni⁺³ = 100 - 96 =4%

d = Zm/a³N_A  d₁/d₂ = 4/4³ x 2³/2 = 0.25

$Density=\frac{mass}{volume}$

Mass of n molecules of H₂0 in hexagonal packing.

 $\frac{18n}{N_A}=\frac{18n}{6.023\times {10}^{23}}$         

^{$\therefore 0.92=\frac{18n}{6.023\times {10}^{23}\times 132\times {10}^{-24}}=4$}  

For BCC, 2^r+ + 2^r- =a$\sqrt{3}$

r+ + r- = a$\sqrt{3}$/2 = (387x1.732)/2 = 335pm

No. of A atom = 1/8 x 8=1
No. of B atom = 1/2 x 6-1/2=5/2

Hence, formula of compound is AB _5/2 or A₂B₅.

Occupied space in cubic close packed structure is 74% and in body centred packed structure is 68%.

ZnS crystallizes as a cubic close-packed array of S2− ions with Zn2+ ions in tetrahedral holes.
If S2− ions occupy the lattice points of a face-centered cubic unit cell and Zn2+ ions are packed into every other tetrahedral hole, we get the unit cell of $ZnS$

Suppose the number of oxide ions = N

The number of octahedral voids = N

Metal ions in octahedral voids = 2/3 N

Therefore, the ratio of M:O₂ = 2/3:1 or 2:3 or the formula is  M₂O₃