- Not quite sure why this is D...please help!
- wouldn't Cl- become Ar with the extra e- which would decrease its atomic radius?-
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Periodic table trends...confusion
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- Not quite sure why this is D...please help! - wouldn't Cl- become Ar with the extra e- which would decrease its atomic radius?First ionization energy is the energy required to remove a (valence) electron from an atom. Looking at the outermost shell of Element 'D', there is only one valence electron. The next closest shell has a full octet. Remember that elements desire to have a full octet that fills all the orbitals. With that said, Element D is very willing to give up that one electron, so it won't take much energy to remove. Therefore, it has the lowest first ionization energy. The answer is not Element B because Element B has a smaller radius, so the attractive forces between the nucleus and the electrons are stronger. This makes the first ionization energy for Element B higher than Element D.
For your second question:
First of all, adding electrons NEVER CHANGES THE ELEMENT!!!!!!!!! Adding electrons makes the ION of the element!!!!!!!! Elements in the period table DIFFER FROM THE NUMBER OF PROTONS IN THE NUCLEUS! Don't make that mistake!!!
Anyway, adding an electron to Cl would make it's anion, Cl-. Adding an electron makes the anion bigger than it's uncharged counterpart. Since the number of protons remains unchanged, the nucleus can't hold the extra electron(s) as tightly. Therefore, atomic radius becomes larger for anions, and smaller for cations
First of all, adding electrons NEVER CHANGES THE ELEMENT!!!!!!!!! Adding electrons makes the ION of the element!!!!!!!! Elements in the period table DIFFER FROM THE NUMBER OF PROTONS IN THE NUCLEUS! Don't make that mistake!!!
Anyway, adding an electron to Cl would make it's anion, Cl-. Adding an electron makes the anion bigger than it's uncharged counterpart. Since the number of protons remains unchanged, the nucleus can't hold the extra electron(s) as tightly. Therefore, atomic radius becomes larger for anions, and smaller for cations
1) First ionization energy decrease as we goes towards the bottom left. Looking at the choices, you can immediately eliminate A and C because A is halogen(7 valence e) while C is noble gas. That would leave you with B and D, eliminate B because D is bigger (valence electrons of larger atoms are more loosely held), answer is D
2)Atom -> cations decreases in atomic radius, vice versa for anion. Think of it as positive-negative attractions. The positive nucleus is holding the negative electron in its orbital because of this + - attraction. But with the addition of an extra negative charge, this force is "weakened", allowing the atomic radius to expand.
Chad explains this very thoroughly in his video.
2)Atom -> cations decreases in atomic radius, vice versa for anion. Think of it as positive-negative attractions. The positive nucleus is holding the negative electron in its orbital because of this + - attraction. But with the addition of an extra negative charge, this force is "weakened", allowing the atomic radius to expand.
Chad explains this very thoroughly in his video.
Got it. Thank you all for thorough and quick responses.
