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3 years ago

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How are ionic bonds formed and what is the attractive force within an ionic bond

1 answer:

Akimi4 [234]3 years ago

6 0

Answer and Explanation:

<u><em>How are ionic bonds formed?</em></u>

Ionic bonds are formed as a result of a redox reaction.

<em>Formation of cations:</em> In which atoms of an element (metal) with low ionization energy give some electrons to reach stable electronic configuration.

<em>Formation of anions:</em> in which atoms of another element with high electron affinity accepts electrons to achieve a stable electronic configuration.

<em>Example: NaCl</em>, the sodium atoms loss electron forming Na⁺ and the chlorine atoms gain electron forming Cl⁻, then the ions are attracted to each other with 1:1 ration forming the ionic compound NaCl.

Na + Cl → Na⁺ + Cl⁻ → NaCl

<u><em>What is the attractive force within the ionic bond?</em></u>

The attractive force within the ionic bond is the electrostatic attraction between the positively charged ions (cations, Na⁺) and the negatively charged ions (anions, Cl⁻).

<u><em> </em></u>

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Equal molar amounts of H2 and I2 are placed into an evacuated container and heated to 445°C, where the following equilibrium is

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<u>Answer:</u> The initial concentration of $H_2\text{ and }I_2$ are 0.0192 M and 0.0192 M respectively.

<u>Explanation:</u>

We are given:

Equilibrium concentration of HI = 0.030 M

Moles of hydrogen gas = Moles of iodine gas (concentration will also be the same)

For the given chemical equation:

$H_2+I_2\rightleftharpoons 2HI$

<u>Initial:</u> x x -

<u>At eqllm:</u> x-c x-c 2c

Calculating the value of 'c'

$2c=0.030\\\\c=\frac{0.030}{2}=0.015M$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HI]^2}{[H_2]\times [I_2]}$

We are given:

$K_{eq}=50.2$

$[H_2]=(x-c)=(x-0.015)$

$[I_2]=(x-c)=(x-0.015)$

Putting values in above equation, we get:

$50.2=\frac{(0.030)^2}{(x-0.015)\times (x-0.015)}\\\\x=0.0108,0.0192$

Neglecting the value of x = 0.0108 M, because the initial concentration cannot be less than the equilibrium concentration.

x = 0.0192 M

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A 3.25 L solution is prepared by dissolving 285 g of BaBr2 in water. Determine the molarity.

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Answer:

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Explanation:

Given data:

Volume of solution = 3.25 L

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Formula:

Molarity = number of moles of solute / L of solution

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