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Multiple Choice
Which of the following is the correct electron configuration for nobelium (No, atomic number 102)?
A
[Rn] 5f^{14} 7s^{2}
B
[Rn] 5f^{14} 6d^{2}
C
[Rn] 5f^{12} 7s^{2}
D
[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{6} 7s^{2}
Verified step by step guidance
1
Step 1: Identify the atomic number of nobelium (No), which is 102. This tells us the total number of electrons that need to be arranged in the electron configuration.
Step 2: Recognize the noble gas core that precedes nobelium in the periodic table. Nobelium is an actinide, so its electron configuration builds upon the electron configuration of radon (Rn), which has 86 electrons.
Step 3: After the radon core, electrons fill the 5f, 6d, and 7s orbitals. According to the Aufbau principle, electrons fill orbitals in order of increasing energy, generally filling 5f orbitals before 6d orbitals in actinides.
Step 4: Determine the number of electrons that fill the 5f and 7s orbitals for nobelium. Since nobelium has 102 electrons, subtract the 86 electrons of radon to find the number of electrons to place in the 5f, 6d, and 7s orbitals (102 - 86 = 16 electrons).
Step 5: Use the known electron filling pattern for nobelium, which fills the 5f orbital completely with 14 electrons and places 2 electrons in the 7s orbital, resulting in the configuration [Rn] 5f^{14} 7s^{2}.