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4 years ago

Which of the following elements would you expect to have the highest ionization energy value, and why?

Chemistry

2 answers:

Alik [6]4 years ago

7 0

Neon, Ne, would have the highest ionization energy because it has a high effective nuclear charge and small radius.

Whitepunk [10]4 years ago

4 0

Answer:

D. Neon (Ne), because it has a high effective nuclear charge and small radius

Explanation:

Ionization energy is the amount of energy required to remove an electron from the valence orbital.

Chlorine (Cl) and fluorine (F) are halogens which have a general valence configuration of ns²np⁵ and are highly electronegative

Lithium (Li) is an alkali metal with a valence configuration of ns¹ and has a tendency to lose electrons readily

Neon (Ne) is a noble gas with a valence configuration of ns²np⁶. Here the outer shells have a complete octet and stable configuration which makes it difficult to remove electrons. In addition the high nuclear charge and small radius of Ne results in a strong force of attraction between the nucleus and the valence electrons. Therefore, Ne would have the highest ionization energy value.

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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

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