Question

write the complete ground - state electron configuration of cl^-.
for multi - digit superscripts or coefficients, use each number in succession.
1s^22s^22p^63s^23p^5
incorrect, 3 attempts remaining

• 0.1 points for incorrect attempt

you have provided the complete ground - state electron configuration.
consider how the magnitude of the charge of the ion will result in the addition
or removal of electrons. cations are formed by removing electrons from the
outermost shell. anions are formed by the addition of electrons to the
outermost shell.

Explanation:

Step1: Determine the number of electrons in Cl⁻

Chlorine (Cl) has an atomic number of 17, so it has 17 electrons. Cl⁻ has a charge of - 1, which means it has 1 more electron than Cl. So, Cl⁻ has \(17 + 1=18\) electrons.

Step2: Write the electron configuration

Using the Aufbau principle (filling orbitals in order of increasing energy: \(1s<2s<2p<3s<3p<4s<3d<4p\cdots\))

• The \(1s\) orbital can hold 2 electrons: \(1s^{2}\)
• The \(2s\) orbital can hold 2 electrons: \(2s^{2}\)
• The \(2p\) orbital can hold 6 electrons: \(2p^{6}\)
• The \(3s\) orbital can hold 2 electrons: \(3s^{2}\)
• The \(3p\) orbital can hold 6 electrons. After filling the previous orbitals (\(1s^{2}2s^{2}2p^{6}3s^{2}\)), we have \(18-(2 + 2+6 + 2)=6\) electrons left for the \(3p\) orbital: \(3p^{6}\)

Answer:

\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)