Question

The atomic radii of Mg2+ and Fions are 0.072 and 0.133 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another) b) What is the force of repulsion at this same separation distance

Answer 1

Answer:

$F_{attraction}=1.09923\times 10^{-8}\ N$

$F_{repulsion}=-1.09923\times 10^{-8}\ N$

Explanation:

The force of attraction can be written as:

$F=\frac {K\times |Z_{cation}|e\times |Z_{anion}|e}{r^2}$

Where,  

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

$Z_{cation}$  is the charge on the cation

$Z_{anion}$  is the charge on the anion

e is electronic charge = $1.602\times 10^{-19}\ C$

r is the distance between the cation and anion and in bond, it is the sum of the ionic radius of cation and anion

$r=r_{cation}+r_{anion}$

$r_{cation}$ = 0.072 nm

$r_{anion}$ = 0.133 nm

Thus,  

$r=0.072\ nm+0.133 nm$

$r$ = 0.205 nm

Also, 1 nm = 10⁻⁹ m

So, $r=2.05\times 10^{-10}\ m$

Given,

$Z_{cation}=2,\ Z_{anion}=1$

Applying in the equation,  

$F=\frac{9\times \:10^9\times \:2\times \:1.602\times \:10^{-19}\times \:1\times \:1.602\times \:10^{-19}}{\left(2.05\times \:\:10^{-10}\right)^2}$

$F=\frac{10^{-29}\times \:46.195272}{\left(10^{-10}\times \:2.05\right)^2}$

$F_{attraction}=1.09923\times 10^{-8}\ N$

Force of attraction = - Force of repulsion

So,

$F_{repulsion}=-1.09923\times 10^{-8}\ N$

Answer 2

Answer:

$\large \boxed{\text{(a) }-1.10 \times 10^{-8} \text{ N}; \text{(b) }+1.10 \times 10^{-8} \text{ N}}$

Explanation:

(a) Attractive force

To calculate the attractive force (F), we can use Coulomb's Law:

$F =k\dfrac{Q_{1}Q_{2}}{r^{2}}$

where  

Q₁ and Q₂ are the charges on the ions,

r is the distance between them, and

k = the Coulomb constant

Data:

Q₁ = 2+

Q₂ = 1-

r₊ = 0.072 nm

r₋ = 0.133 nm

k = 8.988 × 10⁹ N· m²C⁻²

Calculations:

Q₁ = (+2) × 1.602 × 10⁻¹⁹ C = +3.204 × 10⁻¹⁹ C

Q₂ = (-1) × 1.602 × 10⁻¹⁹ C = -1.602 × 10⁻¹⁹ C

r = r₊ + r₋ = 0.072 nm + 0.133 nm = 0.205 nm = 0.205 × 10⁻⁹ m

$\begin{array}{rcl}F & = & 8.988 \times 10^{9}\times\dfrac{3.204 \times 10^{-19} \times (-1.602) \times 10^{-19}}{(0.205 \times 10^{-9})^{2}}\\\\& = & \mathbf{-1.10 \times 10^{-8}} \textbf{ N}\\\end{array}\\\text{The attractive force between the two ions is \large \boxed{\mathbf{-1.10 \times 10^{-8}} \textbf{ N}} }$

(b) Repulsive force

The equilibrium position is reached when the ions just touch each other. If they come any closer, the nuclear repulsions will outweigh the Coulomb attraction.

Thus, the repulsive force is equal and opposite to the attractive force.

$\text{The repulsive force is \large \boxed{\mathbf{+1.10 \times 10^{-8}} \textbf{ N}} }$