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The 1st ionisation order energy of Mg,S,...

The 1st ionisation order energy of Mg,S,P & Al is given by

A

`MggtAlgtPgtS`

B

`AlgtSgtPgtMg`

C

`MggtAlgtSgtP`

D

`AlgtMggtSgtP`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the order of first ionization energies for the elements Magnesium (Mg), Aluminum (Al), Phosphorus (P), and Sulfur (S), we will analyze their electronic configurations and the trends in ionization energy. ### Step-by-Step Solution: 1. **Identify the Elements and Their Atomic Numbers:** - Magnesium (Mg) - Atomic Number: 12 - Aluminum (Al) - Atomic Number: 13 - Phosphorus (P) - Atomic Number: 15 - Sulfur (S) - Atomic Number: 16 2. **Write the Electronic Configurations:** - Mg: 1s² 2s² 2p⁶ 3s² - Al: 1s² 2s² 2p⁶ 3s² 3p¹ - P: 1s² 2s² 2p⁶ 3s² 3p³ - S: 1s² 2s² 2p⁶ 3s² 3p⁴ 3. **Understand the Trend in Ionization Energy:** - Ionization energy generally increases across a period from left to right due to increasing effective nuclear charge (Z_eff) and decreasing atomic radius. - Fully filled and half-filled subshells provide extra stability, which affects ionization energy. 4. **Analyze Each Element:** - **Magnesium (Mg):** Has a stable configuration with a fully filled 3s² subshell. This makes it relatively stable but requires less energy to remove one electron compared to elements with more electrons in the p subshell. - **Aluminum (Al):** Has a configuration of 3s² 3p¹. The presence of one electron in the p subshell makes it easier to remove compared to Mg. - **Phosphorus (P):** Has a half-filled p subshell (3p³), which provides extra stability. Thus, it has a higher ionization energy than Al. - **Sulfur (S):** Has a configuration of 3s² 3p⁴. Although it has more electrons than P, the increased electron-electron repulsion in the p subshell makes it slightly easier to remove an electron compared to P. 5. **Establish the Order of Ionization Energies:** - From the analysis, we can conclude: - Highest Ionization Energy: Phosphorus (P) - Next: Sulfur (S) - Then: Magnesium (Mg) - Lowest: Aluminum (Al) 6. **Final Order:** - The correct order of first ionization energies is: \[ \text{P} > \text{S} > \text{Mg} > \text{Al} \]
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