In which sublevels the valence electrons are located for the following atoms? Sr: Te: 2) What is the number f unpaired electrons found in the following atoms? Ni: Ca: 3) Label the following atoms as paramagnetic or diamagnetic; a) NI b) Se c) C
The Correct Answer and Explanation is:
Correct Answers:
1) Sublevels of valence electrons:
- Sr (Strontium): 5s
- Te (Tellurium): 5s and 5p
2) Number of unpaired electrons:
- Ni (Nickel): 2 unpaired electrons
- Ca (Calcium): 0 unpaired electrons
3) Paramagnetic or Diamagnetic:
- a) Ni: Paramagnetic
- b) Se: Diamagnetic
- c) C: Paramagnetic
Explanation:
To understand where valence electrons are located, one must write the electron configuration of the atoms. Strontium (Sr) has an atomic number of 38. Its electron configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
The outermost electrons are in the 5s sublevel. Hence, Sr has its valence electrons in the 5s sublevel.
Tellurium (Te) has an atomic number of 52. Its electron configuration is:
[Kr] 5s² 4d¹⁰ 5p⁴
Its valence electrons are in both the 5s and 5p sublevels.
To determine the number of unpaired electrons, examine the orbital diagram based on Hund’s rule and the Aufbau principle. Nickel (Ni) has an atomic number of 28, and its electron configuration is:
[Ar] 4s² 3d⁸
In the 3d sublevel, 8 electrons fill five orbitals, with two of them remaining unpaired.
Calcium (Ca), atomic number 20, has the configuration:
[Ar] 4s²
Both electrons in the 4s orbital are paired. Therefore, calcium has 0 unpaired electrons.
An atom is paramagnetic if it has at least one unpaired electron, and diamagnetic if all electrons are paired.
- Nickel, with two unpaired electrons in 3d, is paramagnetic.
- Selenium (Se), atomic number 34, has a configuration ending in 4p⁴. All electrons are paired in its orbitals, making it diamagnetic.
- Carbon (C), atomic number 6, ends in 2p², with two unpaired electrons, making it paramagnetic.
