Question

1 multiple choice
. select the best answer.
(i) (5 points) the electron configuration for ne+ is: 1s²2s²2p⁶ 1s²2s²2p⁵ 1s²2s²2p⁶3s¹ 2s²2p⁶ 1s²2s²2p³
figure 1: the periodic table
table 1: useful information
conversion factors
2.54 cm = 1 in
1 ml = 1 cm³
1 cal = 4.184 j/g°c
mᵢvᵢ = mբvբ
δe=-2.178x10⁻¹⁸j(1/n²_final - 1/n²_initial)
ke = hν - φ
δe = q + w = δke + δpe
c = 2.998 x 10⁸ m/s
equations
f=(1.8×°c)+32
k = °c + 273.15
a = εbc
1/coeffacidmacidvacid = 1/coeffbasembasevbase
e = hν = hc/λ
c = λν
w=-pδv
h = 6.626 x 10⁻³⁴j s

Explanation:

Step1: Recall Ne atom electron - configuration

Neon (Ne) has an atomic number of 10, so its electron - configuration is $1s^{2}2s^{2}2p^{6}$.

Step2: Consider the positive charge

$Ne^{+}$ has lost one electron. The electron is removed from the highest - energy orbital. The highest - energy orbital in Ne is the 2p orbital. So, the electron - configuration of $Ne^{+}$ is $1s^{2}2s^{2}2p^{5}$.

Answer:

$1s^{2}2s^{2}2p^{5}$