Choose the element in each of the sets you would expect to have the highest IE2

AI Thread Summary
The discussion centers on identifying the element with the highest second ionization energy (IE2) from a given set, with potassium (K) being the correct answer. The reasoning highlights that after the first ionization, K achieves the electron configuration of argon, making it highly stable and thus requiring more energy for the second ionization. In contrast, elements like beryllium (Be), magnesium (Mg), and calcium (Ca) have outer electron configurations that allow for easier ionization to achieve stability. The trends in ionization energy indicate that it increases across a period and decreases down a group, but exceptions exist. Overall, K's stability after losing one electron explains its higher IE2 compared to the other elements listed.
CMATT
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Homework Statement


Choose the element in each of the sets you would expect to have the highest IE2.

a. K
b. Be
c. Mg
d. Ca
e. Al

Homework Equations


The correct answer is K

The Attempt at a Solution


I do not understand why it is K ...I kind of guessed by using my Ionization Energy diagram that shows the trend on the periodic table as:

--------> IE increases across a period/row
|
|
|
IE decreases down a group

Can someone please explain why K is correct?
 
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What do you know about number of valence electrons in elements? How does it depend on the the element position in the periodic table? Why do the valence electrons are valence electrons?
 
That's a general trend.But there are exceptions. Besides you have to compare IE2.So this can't be used.
After 1st Ionisation K attains the EC of nearest noble gas i.e. Argon which means it is highly stable.So IE2 of K is highest.
On the other hand if you see Be,Mg,Ca then the outer shell EC of these elements after 1st ionisation is ns1.So they can easily ionise to gain stability.

CMATT said:

Homework Statement


Choose the element in each of the sets you would expect to have the highest IE2.

a. K
b. Be
c. Mg
d. Ca
e. Al

Homework Equations


The correct answer is K

The Attempt at a Solution


I do not understand why it is K ...I kind of guessed by using my Ionization Energy diagram that shows the trend on the periodic table as:

--------> IE increases across a period/row
|
|
|
IE decreases down a group

Can someone please explain why K is correct?
 

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