This set of Engineering Materials Science "Numerical Problems Questions & Answers" are based on "ATOMIC BONDING"
ATOMIC BONDING
Numerical Problems
Hello reader, here you can find some Numerical Problems with answers. But before reading this I will suggest you take a practice test on these questions given below. so you can understand your strength on this topic.
Click here to go Practice Test section on this topic namely Atomic Bonding.
But if you want to read this topic first and then practice it then read it carefully. The Numerical Problems based on Atomic Bonding are given below:
PART - I
1) The force of attraction between a divalent cation and a divalent anion is 1.67×10-8 N. If the ionic radius of the cation is 0.080nm, what is the anion radius in nm?
Correct Answer is "C"
We know that,
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ r2 = [(|Z1|e)(|Z2|e)] / 4πε0 × FA
⇒ r2 = [(|2|1.6 × 10-19 )(|-2|1.6 × 10-19)] / [4 × 3.14 × 8.85 × 10-12 × 1.67×10-8]
⇒ r2 = 5.53 × 10-20
⇒ r = 2.35 × 10-10 m = 0.235 nm
⇒ r+ + r- = 0.235 nm
⇒ r- = 0.235 - 0.080 = 0.155 nm
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ r2 = [(|Z1|e)(|Z2|e)] / 4πε0 × FA
⇒ r2 = [(|2|1.6 × 10-19 )(|-2|1.6 × 10-19)] / [4 × 3.14 × 8.85 × 10-12 × 1.67×10-8] ⇒ r2 = 5.53 × 10-20 ⇒ r = 2.35 × 10-10 m = 0.235 nm
⇒ r+ + r- = 0.235 nm
⇒ r- = 0.235 - 0.080 = 0.155 nm
2) The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively. What is the force of repulsion at their equilibrium interionic separation (i.e., when the ions just touch one another)?
Correct Answer is "B"
We know that,
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ FA = [1/4 × 3.14 × 8.85 × 10-12 × {(0.072 + 0.133)×10-9}2][(|2|1.6 × 10-19 )(|-1|1.6 × 10-19)]
⇒ FA = 1.096 ×10-8 N
At equilibrium interionic separation, FA + FR = 0
⇒ FR = - ⇒ FA = - 1.096 × 10-8 N
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ FA = [1/4 × 3.14 × 8.85 × 10-12 × {(0.072 + 0.133)×10-9}2][(|2|1.6 × 10-19 )(|-1|1.6 × 10-19)]
⇒ FA = 1.096 ×10-8 N
At equilibrium interionic separation, FA + FR = 0 ⇒ FR = - ⇒ FA = - 1.096 × 10-8 N
3) Compute the %IC of the interatomic bond for CsF. Given the electronegativity of Cs and F are 0.79 and 3.98 respectively.
Correct Answer is "D"
We know that,
% IC = {1 - exp [-(0.25)(XA -XB)2]} × 100
⇒ % IC = {1 - exp [-(0.25) (3.98 - 0.79)2]} × 100
⇒ % IC = 92.14
% IC = {1 - exp [-(0.25)(XA -XB)2]} × 100
⇒ % IC = {1 - exp [-(0.25) (3.98 - 0.79)2]} × 100
⇒ % IC = 92.14
4) The % IC of the bond in AB is 67.8%. If the electronegativity of B is 3.44 then what will be the electronegativity of A?
Correct Answer is "A"
We know that,
% IC = {1 - exp [-(0.25)(XA -XB)2]} × 100
⇒(XB -XA)2 = [ln (1 - %IC/100)] / (-0.25)
⇒(XB -XA)2 = [ln (1 - 67.8/100)] / (-0.25)
⇒(XB -XA) = √(4.533) = 2.13
⇒(XA = 3.44 - 2.13 = 1.31
% IC = {1 - exp [-(0.25)(XA -XB)2]} × 100
⇒(XB -XA)2 = [ln (1 - %IC/100)] / (-0.25)
⇒(XB -XA)2 = [ln (1 - 67.8/100)] / (-0.25)
⇒(XB -XA) = √(4.533) = 2.13
⇒(XA = 3.44 - 2.13 = 1.31
5) The chemical composition of the repeat unit for nylon 6,6 is given by the formula C12H22N2O2. Atomic weights for the constituent elements are AC = 12, AH = 1, AN = 14, and AO = 16. According to this chemical formula (for nylon 6,6), the percentage (by weight) of carbon in nylon 6,6 is most nearly
Correct Answer is "D"
We know that,
Wt% = (wt. of particular element / total weight ) × 100
⇒ Wt% of C = [(12 × 12) / {(12 × 12) + (22 × 1) + ( 2 × 14) + (2 × 16)}] × 100 = 63.7
Wt% = (wt. of particular element / total weight ) × 100 ⇒ Wt% of C = [(12 × 12) / {(12 × 12) + (22 × 1) + ( 2 × 14) + (2 × 16)}] × 100 = 63.7
6) If copper has bond energy of 56 kJ mol–1 of bonds, the enthalpy of atomization of copper in the same units is about
Correct Answer is "C"
Cu has face-centered cubic structure and each Cu atom has 12 nearest neighbors i.e. each Cu atom form 12 discrete bonds.
As two atoms are required to form a bond, the effective number of bonds per atom is 12/2 = 6.
Hence, enthalpy of atomization of copper = 56 × 6 = 336 kJ mol–1
As two atoms are required to form a bond, the effective number of bonds per atom is 12/2 = 6.
Hence, enthalpy of atomization of copper = 56 × 6 = 336 kJ mol–1
7) If the Fe–Fe bond length is 2.48 Ã…, the radius of the iron atom in Ã… is
Correct Answer is "D"
We know that,
Radius of atom = bond length / 2 = 2.48 / 2 = 1.24 Ã…
Radius of atom = bond length / 2 = 2.48 / 2 = 1.24 Ã…
8) The force of attraction between the two ions (have the same valency) in XY at their equilibrium interatomic separation distance (0.334 nm) is 2.07 × 10-9 N. The valency of the anion is
Correct Answer is "A"
We know that,
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ [|Z1||Z2|] = [{(FA) × (4πε0r2)} / e2]
⇒ [|Z1||Z2|] = [{2.07 × 10-9 × (4 × 3.14 × 8.85 × 10-12 × (0.334 × 10-9)2} / (1.6 × 10-19)2]
⇒ Z2 = 1 (where Z = |Z1| = |Z2| as both the ion have same valency)
⇒ Z = 1
Hence valency of cation = +1 and valency of anion = -1
Force of attraction, FA = (1/4πε0r2)[(|Z1|e)(|Z2|e)]
⇒ [|Z1||Z2|] = [{(FA) × (4πε0r2)} / e2]
⇒ [|Z1||Z2|] = [{2.07 × 10-9 × (4 × 3.14 × 8.85 × 10-12 × (0.334 × 10-9)2} / (1.6 × 10-19)2]
⇒ Z2 = 1 (where Z = |Z1| = |Z2| as both the ion have same valency)
⇒ Z = 1
Hence valency of cation = +1 and valency of anion = -1
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