Which of the following pairs is correct (for the correct reason) regarding the ionization energy of the atoms? *
A.Ca < Ba; fewer electron shells in Ca
B.Na < K; increasing effective nuclear charge in K
C.Ca < Co; increasing effective nuclear charge in Co
D.F > C ; higher electron repulsion forces in F
To determine which of the given pairs is correct regarding the ionization energy of the atoms, let's examine the explanations provided for each option:
A. Ca < Ba; fewer electron shells in Ca: This explanation is incorrect. Ionization energy generally increases as you move across a period from left to right, which means that the element in the rightmost position usually has the highest ionization energy. In this case, Ba is to the right of Ca, so the correct order would be Ba < Ca.
B. Na < K; increasing effective nuclear charge in K: This explanation is correct. Ionization energy generally increases as you move across a period from left to right due to the increasing effective nuclear charge. Since K is to the right of Na in the periodic table, K would have a higher ionization energy compared to Na.
C. Ca < Co; increasing effective nuclear charge in Co: This explanation is incorrect. Co (Cobalt) is located to the right of Ca (Calcium) in the periodic table, so Co should have a higher ionization energy compared to Ca. The given explanation does not correctly justify the ordering.
D. F > C; higher electron repulsion forces in F: This explanation is incorrect. Ionization energy generally increases as you move across a period from left to right due to the increasing effective nuclear charge. In this case, while there may be higher electron repulsion forces in F (Fluorine) compared to C (Carbon), the primary factor determining ionization energy is the effective nuclear charge, which increases from left to right.
Therefore, the correct answer is option B - Na < K; increasing effective nuclear charge in K.