Question

qauyvb! m7 FUQz13MkJaWkxYRVC1SzBHUDHRTT9357 Potassium iodide (KI) is a white, odorless solid that is typically used in its powder form. Kl is added to table salt as an important source of dietary iodine and exhibits predominantly ionic bonding. 1. Using your periodic table, determine the number of neutrons, protons and electrons of each element (K and I). 2. Write the full electron configurations of this two elements. 3. The K+ and I-ions have electron structures that are identical to which of the two inert gases? 4. Compute the percent ionic character of the interatomic bonds. 5. Calculate the force of attraction between K+ and I-ions whose centers are separated by a distance of 1.25 nm. IT Problem 2 I Entrez votre réponse Bonus Exercise

Answer 1

(1)

No of Electrons = Number of Protons = Atomic Number

No of Neutrons = Atomic Mass - Atomic Number

For Potassium :-

Atomic Number = 19

Atomic Mass = 39

No of Electrons = Number of Protons = 19

No of Neutrons = Atomic Mass - Atomic Number = 39-19 = 20

For Iodine :-

Atomic Number = 53

Atomic Mass = 127

No of Electrons = Number of Protons = 53

No of Neutrons = Atomic Mass - Atomic Number = 127-53 = 74

(2)

Electronic Configuration of Potassium having Atomic Number = 19 :-

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$

Electronic Configuration of Iodine having Atomic Number = 53 :-

1s 2s 2p 3s 3p 3d1045-4p4d105s-5p

(3)

has atomic configuration as Argon (Ar)

-

$I^{-}$ has atomic configuration as Xenon (Xe)

(4)

Percentage Ionic Character of Interatomic Bonds = %

100 1-e-(92)

Electronegativity of Iodine = $x_{I}$ = 2.66

Electronegativity of Potassium = $x_{K}$ = 0.82

Percentage Ionic Character of Interatomic Bonds = $100\left [ 1-e^{-(\frac{2.66-0.82}{2})^{2}} \right ]$

Percentage Ionic Character of Interatomic Bonds = 57.10 %

(5)

Force of Attraction = F

$F = k\frac{q_{1}q_{2}}{d^{2}}$

where k is a constant and its value is 8.99 x  $10^{9}$  $\frac{Nm^{2}}{C^{2}}$

d is the distance between the centres in metres.

$q_{1}$ is the charge on Potassium Ion i.e +1 electron

$q_{2}$ is the charge on Iodine Ion i.e. -1 electron

$q_{1}$ = + 1.6 x $10^{-19}$ C

$q_{2}$ = - 1.6 x $10^{-19}$ C

Substituting the values , Therefore the Magnitude of force of attraction is :-

$F =\frac{ 8.99\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(1.25\times 10^{-9})^{2}}$

F = 1.4729216 x $10^{-10}$ Newtons