Question

in which of the following sets are the charges given correctly for all the ions?
○ rb-, ba²-, p³+
○ na+, mg+, al+
○ n-, o²-, f³-
○ k+, sr²+, o²-

Explanation:

Step1: Analyze each option based on periodic table trends

• Option 1:
• \(Rb\) (rubidium) is in Group 1A, so it should form \(Rb^{+}\) (loses 1 electron), not \(Rb^{-}\).
• \(Ba\) (barium) is in Group 2A, so it should form \(Ba^{2 +}\) (loses 2 electrons), not \(Ba^{2-}\).
• \(P\) (phosphorus) is in Group 5A, it usually gains 3 electrons to form \(P^{3-}\), not \(P^{3+}\).
• Option 2:
• \(Na\) (sodium) in Group 1A forms \(Na^{+}\).
• \(Mg\) (magnesium) in Group 2A should form \(Mg^{2+}\) (loses 2 electrons), not \(Mg^{+}\).
• \(Al\) (aluminum) in Group 3A should form \(Al^{3+}\) (loses 3 electrons), not \(Al^{+}\).
• Option 3:
• \(N\) (nitrogen) in Group 5A usually gains 3 electrons to form \(N^{3-}\), not \(N^{-}\).
• \(O\) (oxygen) in Group 6A gains 2 electrons to form \(O^{2-}\).
• \(F\) (fluorine) in Group 7A gains 1 electron to form \(F^{-}\), not \(F^{3-}\).
• Option 4:
• \(K\) (potassium) in Group 1A loses 1 electron to form \(K^{+}\).
• \(Sr\) (strontium) in Group 2A loses 2 electrons to form \(Sr^{2+}\).
• \(O\) (oxygen) in Group 6A gains 2 electrons to form \(O^{2-}\).

Answer:

\(K^{+}, Sr^{2 +}, O^{2-}\) (the fourth option)