Write both the complete electron- configuration notation and the noble-gas notation for iodine, I. How many inner-shell electrons does an iodine atom contain?
The correct answer and explanation is:
Complete Electron Configuration for Iodine (I):
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁵
Noble-Gas Notation for Iodine (I):
[Kr] 5s² 4d¹⁰ 5p⁵
Number of Inner-Shell Electrons in Iodine:
46 inner-shell electrons
Explanation:
Iodine (I) has an atomic number of 53, which means a neutral iodine atom contains 53 protons and 53 electrons. To write its electron configuration, the electrons are placed into orbitals following the order of increasing energy levels using the Aufbau principle.
The complete electron configuration starts with the lowest energy orbital and fills upward:
- 1s² (2 electrons)
- 2s², 2p⁶ (8 electrons)
- 3s², 3p⁶ (8 electrons)
- 4s², 3d¹⁰, 4p⁶ (18 electrons)
- 5s², 4d¹⁰, 5p⁵ (17 electrons)
Adding up the electrons confirms that this configuration accounts for all 53 electrons.
The noble-gas notation simplifies this by replacing the configuration of krypton (atomic number 36) with [Kr], since krypton has 36 electrons. So, starting from [Kr], the remaining 17 electrons are written as 5s² 4d¹⁰ 5p⁵.
Inner-shell electrons are those not in the outermost shell (valence shell). For iodine, the electrons in the n=5 shell (5s² 5p⁵) are the valence electrons (7 in total). Subtracting these from the total gives the number of inner-shell electrons:
53 total electrons − 7 valence electrons = 46 inner-shell electrons.
These inner-shell electrons occupy the 1s through 4d orbitals and help shield the nucleus, playing a role in the chemical reactivity and bonding behavior of iodine.