Question

name: _____________ pd: ____ chemistry u3l14_q2 ct module day 3 hw

** 9. which of the following is the ground - state electron configuration for the $\ce{na+}$ ion?
(a) $1s^2 2s^2 2p^5$
(b) $1s^2 2s^2 2p^6$
(c) $1s^2 2s^2 2p^6 3s^1$
(d) $1s^2 2s^2 2p^6 3s^2$

• 10. which of the following ground - state electron configurations represents the atom that has the lowest first - ionization energy?

(a) $1s^2 2s^1$
(b) $1s^2 2s^2 2p^2$
(c) $1s^2 2s^2 2p^6$
(d) $1s^2 2s^2 2p^6 3s^1$

1. $1s^2 2s^2 2p^6 3s^2 3p^6$

how many unpaired electrons are in the atom represented by the electron configuration above?
(a) 0
(b) 1
(c) 2
(d) 3

• 12. which of the following is the electron configuration of an excited atom that is likely to emit a quantum of energy?

(a) $1s^2 2s^2 2p^6 3s^2 3p^1$
(b) $1s^2 2s^2 2p^6 3s^2 3p^5$
(c) $1s^2 2s^2 2p^6 3s^1$
(d) $1s^2 2s^2 2p^6 3s^1 3p^1$

** 13. which of the following represents the electron configuration of an oxygen atom in the ground state?
(a)
\\\

$$\begin{array}{c}\ \\frac{\\uparrow\\downarrow}{1s}\\quad\\frac{\\uparrow\\downarrow}{2s}\\quad\\frac{\\uparrow}{2p}\\quad\\frac{\\uparrow}{2p}\\quad\\frac{}{2p}\\\ \\end{array}$$

\\
(b)
\\\

$$\begin{array}{c}\ \\frac{\\uparrow\\downarrow}{1s}\\quad\\frac{\\uparrow\\downarrow}{2s}\\quad\\frac{\\uparrow}{2p}\\quad\\frac{\\uparrow}{2p}\\quad\\frac{\\uparrow}{2p}\\\ \\end{array}$$

\\
(c)
\\\

$$\begin{array}{c}\ \\frac{\\uparrow\\downarrow}{1s}\\quad\\frac{\\uparrow\\downarrow}{2s}\\quad\\frac{\\uparrow\\downarrow}{2p}\\quad\\frac{\\uparrow}{2p}\\quad\\frac{\\uparrow}{2p}\\\ \\end{array}$$

\\
(d)
\\\

$$\begin{array}{c}\ \\frac{\\uparrow\\downarrow}{1s}\\quad\\frac{\\uparrow\\downarrow}{2s}\\quad\\frac{\\uparrow\\downarrow}{2p}\\quad\\frac{\\uparrow\\downarrow}{2p}\\quad\\frac{}{2p}\\\ \\end{array}$$

\\

Explanation:

Step1: Analyze Na⁺ electron count

Na atom has 11 electrons; Na⁺ loses 1 → 10 electrons.

Step2: Match electron configuration

10 electrons correspond to $1s^22s^22p^6$.

Step1: Relate ionization energy to electron position

First ionization energy is lowest for valence electrons farthest from nucleus.

Step2: Identify outermost electron

Option D ($1s^22s^22p^63s^1$) has a 3s electron (farthest).

Step1: Check orbital filling

$1s^22s^22p^63s^23p^6$: all orbitals are fully paired (s holds 2, p holds 6).

Step2: Count unpaired electrons

No unpaired electrons → 0.

Step1: Define excited state

Excited state: electron in higher energy orbital than ground state.

Step2: Identify excited configuration

Option D ($1s^22s^22p^63s^13p^1$): 3p has an electron before 3s is full (ground state 3s²).

Answer:

1. B. $1s^22s^22p^6$
2. D. $1s^2 2s^22p^6 3s^1$
3. A. 0
4. D. $1s^22s^22p^63s^13p^1$
5. C. (1s: ↑↓, 2s: ↑↓, 2p: ↑↓, ↑, ↑)