Question

q1 chemistry: u2l12 electron config day 2
prepwork part 1:
follow the exemplar answer below and write the full electron configuration and orbital diagram for the given elements.

#	element	# e-	full e- configuration	orbital diagram
ex	oxygen	8 e-	1s²2s²2p⁴	1↓ 1↓ 1↓ 1 1

1s 2s 2p |

1**	lithium
2**	aluminum
3*	magnesium
4**	phosphorous

Explanation:

Step1: Determine electrons of Lithium

Lithium has atomic number 3, so 3 electrons.
$1s^{2}2s^{1}$

Step2: Draw orbital diagram for Lithium

$\frac{1\downarrow}{1s}\frac{1}{2s}$

Step3: Determine electrons of Aluminum

Aluminum has atomic number 13, so 13 electrons.
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$

Step4: Draw orbital diagram for Aluminum

$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}\frac{1}{3p}$

Step5: Determine electrons of Magnesium

Magnesium has atomic number 12, so 12 electrons.
$1s^{2}2s^{2}2p^{6}3s^{2}$

Step6: Draw orbital diagram for Magnesium

$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}$

Step7: Determine electrons of Phosphorus

Phosphorus has atomic number 15, so 15 electrons.
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$

Step8: Draw orbital diagram for Phosphorus

$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}\frac{1}{3p}\frac{1}{3p}\frac{1}{3p}$

Answer:

Element	# e-	Full e- configuration	Orbital diagram
Aluminum	13 e-	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$	$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}\frac{1}{3p}$
Magnesium	12 e-	$1s^{2}2s^{2}2p^{6}3s^{2}$	$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}$
Phosphorus	15 e-	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$	$\frac{1\downarrow}{1s}\frac{1\downarrow}{2s}\frac{1\downarrow1\downarrow1\downarrow}{2p}\frac{1\downarrow}{3s}\frac{1}{3p}\frac{1}{3p}\frac{1}{3p}$