Which of the following has the lowest second ionization energy?
a. Ca b. Ne c. K d. Ar e. Na
Arrange the following set of ions in order of increasing atomic radii.
Ca2+, Cl-, K+, P3-, S2-
a. Ca2+ < K+ < P3- < S2- < Cl-
b. K+ < Cl- < Ca2+ < S2- < P3-
c. P3- < S2- < Cl- < K+ < Ca2+
d. Ca2+ < K+ < Cl- < S2- < P3-
e. Cl- < S2- < P3- < Ca2+ < K+
To determine which element has the lowest second ionization energy, we need to analyze their electron configurations. The second ionization energy refers to the energy required to remove a second electron from a positively charged ion.
Looking at the given options:
a. Ca: Ca has an atomic number of 20. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². The first ionization energy involves removing an electron from the 4s orbital, and the second ionization energy involves removing an electron from the 3p orbital.
b. Ne: Ne has an atomic number of 10. Its electron configuration is 1s² 2s² 2p⁶. Since it is a noble gas, Ne already has a full valence shell, and its second ionization energy is significantly higher than the other elements.
c. K: K has an atomic number of 19. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The first ionization energy involves removing an electron from the 4s orbital, and the second ionization energy involves removing an electron from the 3p orbital.
d. Ar: Ar has an atomic number of 18. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶. Similar to Ne, Ar is also a noble gas, so its second ionization energy is significantly higher.
e. Na: Na has an atomic number of 11. Its electron configuration is 1s² 2s² 2p⁶ 3s¹. The first ionization energy involves removing an electron from the 3s orbital. If we consider the second ionization energy, it would involve removing an electron from the 2p orbital.
Based on the electron configurations, we can conclude that the lowest second ionization energy among the given options is for Ca (option a).
To arrange the given set of ions in order of increasing atomic radii, we need to consider the effective nuclear charge and the number of electron shells.
Looking at the given options:
a. Ca2+, Cl-, K+, P3-, S2-
b. K+, Cl-, Ca2+, S2-, P3-
c. P3-, S2-, Cl-, K+, Ca2-
d. Ca2+, K+, Cl-, S2-, P3-
e. Cl-, S2-, P3-, Ca2+, K+
To determine the order of increasing atomic radii, we need to consider the following factors:
1. Effective nuclear charge: The greater the positive charge of the nucleus, the stronger it pulls the electrons closer, resulting in a smaller atomic radius.
2. Number of electron shells: The greater the number of electron shells, the larger the atomic radius.
Based on these factors, the order of increasing atomic radii among the given ions is:
a. Ca2+ < K+ < P3- < S2- < Cl- (option a)