NEET Practice Questions (MCQs) with Answers & Solutions

When all particle along are body diagonal one removed, these 2 X atoms from corner are removed, one Y particle removed & 2 Z particle removed.

Hence new arrangement, X particle = $\frac{1}{8}\times 6+\frac{1}{2}\times 6=\frac{15}{4}$; Y particle = 6 ; Z particle = 3

Hence formula = X_15/4Y₃Z₆ = X_5/4YZ₆ = X₅Y₄Z₈

d_c-c = $\sqrt{\frac{1}{16}+\frac{1}{16}+\frac{1}{16}}=\sqrt{\frac{3}{16}}$ and a_FCC = 1 

In 1 mole of M_xO, mole of M²⁺ ion = y

∴ mole of M³⁺ ion = x – y.

Applying charge balance, 2y + 3(x – y) – 2 = 0

⇒ y = 3x – 2

∴ $\frac{{\text{M}}^{\text{3+}}}{{\text{M}}^{\text{2+}}}=\frac{\text{x\hspace{0.17em}}-\text{\hspace{0.17em}y}}{\text{y}}=\frac{\text{x\hspace{0.17em}}-(3\text{x\hspace{0.17em}}-2)}{3\text{x\hspace{0.17em}}-2}=\frac{2(\text{x\hspace{0.17em}}-1)}{3\text{x\hspace{0.17em}}-2}$

d⁸ → Ni²⁺, d⁷ → Ni³⁺.

Total charge of nickel.

(0.96 × 2) + (0.04 × 3) = 2.04.

No. of O^2- ion = $\frac{2.04}{2}\hspace{0.17em}=\hspace{0.17em}1.02.$= 1.02. 

Formula of solid = NiO_1.02 = Ni_0.98O.

At room temperature, thermal energy causes some electrons of the Si-Si bonds to leave. The electron deficient bond becomes a positive hole. Both the positive holes and electrons released account for the conduction of electricity.

We have, $d=\frac{a}{\sqrt{h^2+k^2+{\ell }^2}}$; $d_{111}=\frac{0.556}{\sqrt{1^2+1^2+1^2}}=0.321nm$ 

The separation of the 111 planes is twice as great as that of 222 planes.

The hexagonal base consists of six equilateral triangles, each with side 2r and altitude 2r sin 60°.

Hence, area of base = $\text{6\hspace{0.17em}}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\left(2r\right)\hspace{0.17em}\left(2rsin{60}^o\right)\right)=6\sqrt{3}.r^2$

The height of the hexagonal is twice the distance between closest packed layers.

The latter can be determined to a face centred cubic lattice with unit cell length a. In such a lattice, the distance between closest packed layers is one third of the body diagonal, i.e. $\frac{\sqrt{3}a}{3}$

Hence $\text{ Height }\left(h\right)=2\left[\frac{{\displaystyle \sqrt{3}a}}{{\displaystyle 3}}\right]=\frac{2a}{\sqrt{3}}$

Now, in the face centred lattice, atoms touch one another along the face diagonal,

Thus, $4r=\sqrt{2}.a$

With this, the height of hexagonal becomes : $\text{ Height }\left(h\right)=\frac{2}{\sqrt{3}}\left[\frac{{\displaystyle 4r}}{{\displaystyle \sqrt{2}}}\right]=\left[4\frac{{\displaystyle \sqrt{2}}}{{\displaystyle 3}}\right].r$

Volume of hexagonal unit is, V = (base area) × (height) $=\left(6\sqrt{3}{r}^2\right)\left[\frac{{\displaystyle 4\sqrt{2}}}{{\displaystyle \sqrt{3}}}.r\right]=24\sqrt{2}.r^3$

In one hexagonal unit cell, there are 6 atoms as described below :

• 3 atoms in the central layer which exclusively belong to the unit cell.

• 1 atom from the centre of the base. There are two atoms of this type and each is shared between two hexagonal unit cells.

• 2 atoms from the corners. There are 12 such atoms and each is shared amongst six hexagonal unit cells.

Now, the volume occupied by atoms = $6\left[\frac{{\displaystyle 4}}{{\displaystyle 3}}\pi r^3\right]$

Fraction of volume occupied by atoms $=\frac{\text{Volume occupied by atoms}}{\text{Volume of hexagonal unit cell}}$$=\frac{6\left(\frac{4}{3}\pi r^3\right)}{24\sqrt{2}.r^3}=\pi /3\sqrt{2}=0.74.$

Fraction of empty space = $(1.00-0.74)=0.26$

Percentage of empty space = 26%   

In (bcc) crystals, atoms touch each other along the cross diagonal.

Hence, Atomic radius $\left(R\right)=\frac{a\sqrt{3}}{4}=351\times \frac{\sqrt{3}}{4}=151.98 pm$