Why does losing two electrons to form a +2 Cation cause the greatest shrinkage in atomic radius of the following? People are telling me that's the answer but I don't understand.
a. Gaining one electron to form a 1+ cation
b. Losing two electrons to form a 2+ cation
c. Gaining two electrons to form a 2- anion
d. Gaining two electrons to form a 2+ cation
e. Gaining one electron to form a 1- anion
f. Losing one electron to form a 1+ cation
Losing electrons from an outside shell forms a + charge (cation) AND reduces the size (fewer electrons occupy less space).
Gaining electrons forms a negative charge (anion) AND increases the size (more electrons occupy a larger space).
a. Gaining one electron to form a 1+ cation Gaining one electron forms an anion.
b. Losing two electrons to form a 2+ cation OK
c. Gaining two electrons to form a 2- anion See above. Gaining electrons increases the size.
d. Gaining two electrons to form a 2+ cation Gaining two electrons forms a -2 charge
e. Gaining one electron to form a 1- anion True BUT gaining electrons increases the size.
f. Losing one electron to form a 1+ cation True and it decreases the size but not as much as losing two electrons.
Much obliged. I appreciate how you explain how to figure it out, as opposed to just telling me the answer.
How many electrons are transferred between the cation and anion to form the ionic bond in one formula unit of each compound? LIF, CaS, NaCI, KBr, BaO
To understand why losing two electrons to form a +2 cation causes the greatest shrinkage in atomic radius among the given options, we need to consider the concept of effective nuclear charge.
Effective nuclear charge is the net positive charge experienced by the valence electrons in an atom. It is determined by the number of protons in the nucleus and the shielding effect of other electrons in inner energy levels. The greater the effective nuclear charge, the stronger the attraction between the electrons and the nucleus, leading to a smaller atomic radius.
Now, let's analyze each option:
a. Gaining one electron to form a 1+ cation: When an atom gains an electron to form a +1 cation, the effective nuclear charge decreases because the number of protons remains the same while the number of electrons decreases. This leads to an increase in atomic radius.
b. Losing two electrons to form a 2+ cation: When an atom loses two electrons to form a +2 cation, the effective nuclear charge significantly increases. The reduction in the number of electrons is greater than the decrease in the number of protons, resulting in a stronger attraction between the remaining electrons and the nucleus. This leads to a significant shrinkage in atomic radius.
c. Gaining two electrons to form a 2- anion: When an atom gains two electrons to form a -2 anion, the effective nuclear charge decreases because the number of protons remains the same while the number of electrons increases. This leads to an increase in atomic radius.
d. Gaining two electrons to form a 2+ cation: This option is the same as option b, where losing two electrons to form a +2 cation results in a significant shrinkage in atomic radius due to the increase in effective nuclear charge.
e. Gaining one electron to form a 1- anion: When an atom gains one electron to form a -1 anion, the effective nuclear charge decreases because the number of protons remains the same while the number of electrons increases. This leads to an increase in atomic radius.
f. Losing one electron to form a 1+ cation: When an atom loses one electron to form a +1 cation, the effective nuclear charge increases but to a lesser extent compared to losing two electrons. The decrease in the number of electrons is less significant, resulting in a small shrinkage in atomic radius.
Therefore, among the given options, losing two electrons to form a +2 cation (option b) causes the greatest shrinkage in atomic radius due to the significant increase in effective nuclear charge.