Questions: Arrange the following elements in order of decreasing ionization energy. Rank from highest to lowest ionization energy. To rank items as equivalent, overlap them. silver phosphorus molybdenum strontium germanium

Solution Steps

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Generally, ionization energy increases across a period (from left to right on the periodic table) and decreases down a group (from top to bottom).

The elements given are:

• Silver (Ag)
• Phosphorus (P)
• Molybdenum (Mo)
• Strontium (Sr)
• Germanium (Ge)

Their positions on the periodic table are:

• Phosphorus (P) is in Group 15, Period 3.
• Germanium (Ge) is in Group 14, Period 4.
• Silver (Ag) is in Group 11, Period 5.
• Molybdenum (Mo) is in Group 6, Period 5.
• Strontium (Sr) is in Group 2, Period 5.

• Phosphorus (P): Being in the third period and a non-metal, it has a relatively high ionization energy.
• Germanium (Ge): Located in the fourth period, it has a lower ionization energy than phosphorus but higher than elements in the fifth period.
• Silver (Ag), Molybdenum (Mo), and Strontium (Sr): All are in the fifth period. Among these, strontium, being an alkaline earth metal, has the lowest ionization energy. Silver, being a transition metal, has a higher ionization energy than molybdenum and strontium.

Based on the trends:

1. Phosphorus (P): Highest ionization energy.
2. Germanium (Ge): Lower than phosphorus but higher than the fifth-period elements.
3. Silver (Ag): Higher ionization energy among the fifth-period elements.
4. Molybdenum (Mo): Lower than silver but higher than strontium.
5. Strontium (Sr): Lowest ionization energy.

Final Answer