Question

what is the noble gas electron configuration of bismuth (bi)?
a. kr 6s² 6p³
b. xe 6s² 6p³
c. xe 4f¹⁴ 5d¹⁰ 6s² 6p³
d. kr 5s² 5d¹⁰ 5p⁶ 6s² 6d¹⁰ 6p³
e. xe 6s² 5d¹⁰ 6p³

Explanation:

Step1: Determine Bismuth's Atomic Number

Bismuth (Bi) has an atomic number of 83.

Step2: Identify the Noble Gas before Bi

The noble gas with an atomic number less than 83 and closest to it is Xenon (Xe), which has an atomic number of 54.

Step3: Calculate Electrons after Xe

Electrons in Bi after Xe: \( 83 - 54 = 29 \).

Step4: Fill Orbitals for Remaining Electrons

The electron configuration for the remaining 29 electrons:

• \( 4f^{14} \) (14 electrons),
• \( 5d^{10} \) (10 electrons),
• \( 6s^{2} \) (2 electrons),
• \( 6p^{3} \) (3 electrons).

Summing these: \( 14 + 10 + 2 + 3 = 29 \), which matches.

So the noble gas configuration is \([Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{3}\).

Answer:

C. \([Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{3}\)