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Which is the electron configuration for boron?

A. [tex]\(1s^2 2s^3\)[/tex]
B. [tex]\(1s^2 2s^2 3s^1\)[/tex]
C. [tex]\(1s^1 2s^2 2p^2\)[/tex]
D. [tex]\(1s^2 2s^2 2p^1\)[/tex]


Sagot :

To determine the correct electron configuration for boron, we need to consider its atomic number, which is 5. The atomic number tells us that boron has 5 protons and, assuming neutrality, 5 electrons.

The electron configuration of an element describes the distribution of its electrons among the different atomic orbitals. The order of filling the orbitals follows the Aufbau principle, which states that electrons fill orbitals starting from the lowest energy level moving to higher levels. The general order of filling is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s...

Following this order, we distribute the 5 electrons of boron:

1. Fill the 1s orbital:
- The 1s orbital can hold up to 2 electrons.
- After filling, we have: [tex]\(1s^2\)[/tex]
- Remaining electrons: 5 - 2 = 3 electrons

2. Fill the 2s orbital:
- The 2s orbital can hold up to 2 electrons.
- After filling, we have: [tex]\(1s^2 2s^2\)[/tex]
- Remaining electrons: 3 - 2 = 1 electron

3. Place the remaining electrons in the 2p orbital:
- The 2p orbital can hold up to 6 electrons, but we only have 1 electron left.
- After placement, we have: [tex]\(1s^2 2s^2 2p^1\)[/tex]

So, the correct electron configuration for boron is [tex]\(1s^2 2s^2 2p^1\)[/tex].

Upon checking the given options:

- [tex]\(1s^2 2s^3\)[/tex]: This configuration incorrectly places 3 electrons in the 2s orbital, which can only hold 2 electrons.
- [tex]\(1s^2 2s^2 3s^1\)[/tex]: This configuration incorrectly places an electron in the 3s orbital before filling the 2p orbital.
- [tex]\(1s^1 2s^2 2p^2\)[/tex]: This configuration incorrectly places only 1 electron in the 1s orbital and places too many electrons overall (6 instead of 5).
- [tex]\(1s^2 2s^2 2p^1\)[/tex]: This matches the correct electron configuration we've derived.

Therefore, the electron configuration for boron is [tex]\(1s^2 2s^2 2p^1\)[/tex], which corresponds to the fourth option:

[tex]\(\boxed{4}\)[/tex]