Question

u2l14_exit ticket:
for each of the following elements, write the full electron configuration, the orbital notation, and the noble gas configurations.
element full e - configuration orbital notation noble gas config.
p (15 e - ) 1s²2s²2p⁶3s²3p³ ne3s²3p³
**si
cu
**v
n

Explanation:

Si (14 e⁻)

Full \( e^{-} \) configuration

- **Step1: Fill orbitals**
According to the Aufbau principle (\( 1s < 2s < 2p < 3s < 3p\)), fill electrons. \( 1s \) can hold 2 electrons, \( 2s \) can hold 2 electrons, \( 2p \) can hold 6 electrons, and \( 3s \) can hold 2 electrons. After filling these, we have \( 14-(2 + 2+6 + 2)=2 \) electrons left for the \( 3p \) orbital.
So the full electron configuration is \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2} \)

Orbital notation

- **Step2: Represent orbitals**
For \( 1s \): \( \frac{\uparrow\downarrow}{1s} \)
For \( 2s \): \( \frac{\uparrow\downarrow}{2s} \)
For \( 2p \): \( \frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p} \)
For \( 3s \): \( \frac{\uparrow\downarrow}{3s} \)
For \( 3p \): \( \frac{\uparrow}{3p}\frac{\uparrow}{3p}\frac{}{3p} \)

Noble gas config

- **Step3: Use noble gas core**
The noble gas before Si is Ne (\( 1s^{2}2s^{2}2p^{6} \)). So the noble - gas configuration is \( [Ne]3s^{2}3p^{2} \)

Cu (29 e⁻)

Full \( e^{-} \) configuration

- **Step4: Consider electron - filling exceptions**
The expected electron configuration is \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{9} \), but due to the stability of a half - filled (\( d^{5} \)) or fully - filled (\( d^{10} \)) \( d \) - subshell, one electron from the \( 4s \) orbital moves to the \( 3d \) orbital.
So the full electron configuration is \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10} \)

Orbital notation

- **Step5: Draw orbitals with electrons**
For \( 1s \): \( \frac{\uparrow\downarrow}{1s} \)
For \( 2s \): \( \frac{\uparrow\downarrow}{2s} \)
For \( 2p \): \( \frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p} \)
For \( 3s \): \( \frac{\uparrow\downarrow}{3s} \)
For \( 3p \): \( \frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p} \)
For \( 4s \): \( \frac{\uparrow}{4s} \)
For \( 3d \): \( \frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d} \)

Noble gas config

- **Step6: Use noble gas core**
The noble gas before Cu is Ar (\( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6} \)). So the noble - gas configuration is \( [Ar]4s^{1}3d^{10} \)

V (23 e⁻)

Full \( e^{-} \) configuration

- **Step7: Fill orbitals**
Using the Aufbau principle (\( 1s < 2s < 2p < 3s < 3p < 4s < 3d \))
\( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3} \)

Orbital notation

- **Step8: Represent orbitals**
For \( 1s \): \( \frac{\uparrow\downarrow}{1s} \)
For \( 2s \): \( \frac{\uparrow\downarrow}{2s} \)
For \( 2p \): \( \frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p} \)
For \( 3s \): \( \frac{\uparrow\downarrow}{3s} \)
For \( 3p \): \( \frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p} \)
For \( 4s \): \( \frac{\uparrow\downarrow}{4s} \)
For \( 3d \): \( \frac{\uparrow}{3d}\frac{\uparrow}{3d}\frac{\uparrow}{3d}\frac{}{3d}\frac{}{3d} \)

Noble gas config

- **Step9: Use noble gas core**
The noble gas before V is Ar (\( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6} \)). So the noble - gas configuration is \( [Ar]4s^{2}3d^{3} \)

N (7 e⁻)

Full \( e^{-} \) configuration

- **Step10: Fill orbitals**
Using the Aufbau principle (\( 1s < 2s < 2p \))
\( 1s^{2}2s^{2}2p^{3} \)

Orbital notation

- **Step11: Represent orbitals**
For \( 1s \): \( \frac{\uparrow\downarrow}{1s} \)
For \( 2s \): \( \frac{\uparrow\downarrow}{2s} \)
For \( 2p \): \( \frac{\uparrow}{2p}\frac{\uparrow}{2p}\frac{\uparrow}{2p} \)

Noble gas config

- **Step12: Use noble gas core**
There is no noble gas before N. But if we consider the closest noble - gas - like core (not a true noble - gas configuration in the strictest sense for such a small atom), we can't use a noble gas core in the traditional sense. But if we follow the pattern, we can write it as \( [He]2s^{2}2p^{3} \) (where He is \( 1s^{2} \))

Answer:

Element	Full \( e^{-} \) configuration	Orbital notation	Noble gas config
\( Cu(29e^{-}) \)	\( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10} \)	\( \frac{\uparrow\downarrow}{1s}\frac{\uparrow\downarrow}{2s}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{3s}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow}{4s}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d}\frac{\uparrow\downarrow}{3d} \)	\( [Ar]4s^{1}3d^{10} \)
\( V(23e^{-}) \)	\( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{3} \)	\( \frac{\uparrow\downarrow}{1s}\frac{\uparrow\downarrow}{2s}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{2p}\frac{\uparrow\downarrow}{3s}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{3p}\frac{\uparrow\downarrow}{4s}\frac{\uparrow}{3d}\frac{\uparrow}{3d}\frac{\uparrow}{3d}\frac{}{3d}\frac{}{3d} \)	\( [Ar]4s^{2}3d^{3} \)
\( N(7e^{-}) \)	\( 1s^{2}2s^{2}2p^{3} \)	\( \frac{\uparrow\downarrow}{1s}\frac{\uparrow\downarrow}{2s}\frac{\uparrow}{2p}\frac{\uparrow}{2p}\frac{\uparrow}{2p} \)	\( [He]2s^{2}2p^{3} \)