Question

The first ionisation potential (in eV) of Be and B, respectively are

• A. 8.29, 9.32
• B. 9.32, 9.32
• C. 8.29, 8.29
• D. 9.32, 8.29

Answer 1

The Correct Option is D

 The question asks us to determine the first ionization potentials of Beryllium (Be) and Boron (B), and choose the correct pair from the given options.

First, let's understand the concept of ionization potential:

• The ionization potential (IP) is the energy required to remove the outermost electron from an isolated gaseous atom to form a cation.
• The first ionization potential refers to the energy required to remove the first electron.

Now, considering the periodic trends and the electron configurations:

• Beryllium (Be) has an atomic number of 4, with the electron configuration: \(1s^2\, 2s^2\). The removal of an electron from the filled 2s orbital requires significant energy because of the stable configuration.
• Boron (B) has an atomic number of 5, with the electron configuration: \(1s^2\, 2s^2\, 2p^1\). Removing an electron from this partially filled 2p orbital requires less energy compared to Be.

Therefore, typically, Beryllium will have a higher ionization energy compared to Boron because:

• Beryllium's electron is removed from a 2s orbital that is more stable and closer to the nucleus than the 2p orbital in Boron.
• Ionic removal from a fully-filled s orbital requires more energy than from a partially filled p orbital.

Hence, the ionization potentials (in eV) for Be and B are as follows:

Element	Ionization Potential (eV)
Beryllium (Be)	9.32
Boron (B)	8.29

Based on this information, the correct option that lists the first ionization potentials for Be and B is: 9.32, 8.29.

This corresponds to option: 9.32, 8.29.