How many electrons can have the value n=2, I=1 and s - ½ in the configuration 1S2?, 2S2?,2p3?

Explanation

This is the correct answer. Let's break down the quantum numbers:n = 2: This indicates that the electrons are in the second principal energy level. l = 1: This indicates that the electrons are in a p subshell.s = -1/2: This indicates that the electrons have a spin of -1/2. NNow, let's analyze each configuration:1s2: This configuration has 2 electrons in the 1s orbital. Since the 1s orbital has l = 0, none of these electrons can have l = 1. 2s2: This configuration has 2 electrons in the 2s orbital. Since the 2s orbital has l = 0, none of these electrons can have l = 1. 2p3: This configuration has 3 electrons in the 2p subshell. Each p orbital can hold a maximum of 2 electrons, and each electron can have either s = +1/2 or s = -1/2. Therefore, there can be a maximum of 1 electron in the 2p subshell with n = 2, l = 1, and s = -1/2. Adding up the electrons from each configuration, we find that a total of 3 electrons can have the values n = 2, l = 1, and s = -1/2 in the given configurations.