This set of Ceramics Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ionic Bonding in Ceramics”.

1. What is ionization energy?

a) Energy required for removing loosely bound electron

b) Energy released when an electron is added to neutral atom

c) Energy required for forming bonds between two molecules

d) Delocalization of electrons

View Answer

Explanation: During ionic bond formation, electron is either removed or added in the valence shell. When there is an extra electron in the outermost shell, the atom tends to lose it to form ionic bond. Energy is required to remove this loosely bound electron. This is called ionization energy.

2. What is electron affinity?

a) Energy released when an electron is added to neutral atom

b) Energy required for removing loosely bound electron

c) Energy required for forming bonds between two molecules

d) Delocalization of electrons

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Explanation: When an electron is added to a neutral atom, there is some release of energy as the atom forms a negative ion. Electron affinity can also be defined as an atom’s ability to form a negative ion.

3. Ionic bonds are non-directional.

a) True

b) False

View Answer

Explanation: Ionic bonds are formed by strong electrostatic force of attractions between unlike charges. Each ion is surrounded by oppositely charged ions so there is no definite direction. Therefore ionic bonds are non-directional.

4. To obtain equilibrium separation, there must be repulsion to _________

a) Shield the charge

b) Form ionic bonds

c) Balance the attraction

d) Prevent formation of bonds

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Explanation: Equilibrium separation between means balance of both attractive and repulsive forces between ions and electrons. If there is only repulsion between like charges or attraction between electrons and positive ion cores, bond formation is not possible. Therefore, to obtain equilibrium separation, there must be repulsion to balance the attraction.

5. Find the electrostatic attractive energy between magnesium and oxygen in the ceramic magnesium oxide if the radius of cation is 72 pm and that of anion is 140 pm.

a) -12.55 x 10^{20} J

b) -9.56 x 10^{20} J

c) -12.55 x 10^{21} J

d) -9.56 x 10^{21} J

View Answer

Explanation: By formula,

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\)

Where, Z

_{u}and Z

_{v}are charges of cation and anion respectively.

Charges of cation and anion are -2 and +2 respectively.

R = radius of cation + radius of anion

R = 72 + 140 = 212 pm, Putting value of R in the formula, we get,

E = -12.55 x 10

^{20}J

6. Find the electrostatic attractive energy between barium and oxygen in the ceramic barium oxide if the radius of cation is 135 pm and that of anion is 140 pm.

a) -2.56 x 10^{20} J

b) -9.68 x 10^{20} J

c) -1.56 x 10^{15} J

d) -5.55 x 10^{15} J

View Answer

Explanation: By formula,

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\)

Where, Z

_{u}and Z

_{v}are charges of cation and anion respectively.

Charges of cation and anion are -2 and +2 respectively.

R = radius of cation + radius of anion

R = 135 + 140 = 275 pm, Putting value of R in the formula, we get,

E = -9.68 x 10

^{20}J

7. Find the electrostatic attractive energy between aluminium and oxygen in alumina if the radius of cation is 53 pm and that of anion is 140 pm.

a) -9.68 x 10^{20} J

b) -1.11 x 10^{20} j

c) -12.41 x 10^{19} J

d) -3.33 x 10^{19} J

View Answer

Explanation: By formula,

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\)

Where, Z

_{u}and Z

_{v}are charges of cation and anion respectively.

Charges of cation and anion are -6 and +6 respectively.

R = radius of cation + radius of anion

R = 53 + 140 = 193 pm, Putting value of R in the formula, we get,

E = -12.41 x 10

^{19}J

8. Find the electrostatic attractive energy between magnesium and chlorine in magnesium chloride if the radius of the cation is 72 pm and that of chlorine anion is 184 pm.

a) -1.22 x 10^{22} J

b) -5.56 x 10^{22} J

c) -30.5 x 10^{21} J

d) -10.39 x 10^{20} J

View Answer

Explanation: By formula,

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\)

Where, Z

_{u}and Z

_{v}are charges of cation and anion respectively.

Charges of cation and anion are -2 and +2 respectively.

R = radius of cation + radius of anion

R = 72 + 184 = 256 pm, Putting value of R in the formula, we get,

E = -10.39 x 10

^{20}J

9. What is the formula of total energy of an ion pair?

a) E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R} + \frac{B}{r^n}\)

b) E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\)

c) E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R} + \frac{B}{r^3}\)

d) E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R} + \frac{B}{r}\)

View Answer

Explanation: The formula of electrostatic attractive energy is given by

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R}\). Repulsive energy is given by

E

_{r}= \(\frac{B}{r^n}\). Here B is a constant and n is known as the Born exponent. Therefore, total energy is

E = \(-\frac{Z_u Z_v e^2}{4 \pi \epsilon_0 R} + \frac{B}{r^n}\)

10. What is the ionic radius of silica if the radius of the cation and anion is 118 pm and 140 pm respectively?

a) 129

b) 230

c) 258

d) 315

View Answer

Explanation: Ionic radius of silica is given by,

R = radius of action + radius of anion

R = 118 + 140 = 258 pm.

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