Question

q1 chemistry: u2l14 electron config day 4
prepwork part 1: electron configuration practice
for each of the following elements, write the full electron - configuration, the orbital notation, and the noble gas configurations. calculate the number of electron first.
element full e - configuration orbital notation noble gas config.
p (15 e -) 1s²2s²2p⁶3s²3p³ 1↑↓ 1s 1↑↓ 2s 1↑↓ 1↑↓ 1↑↓ 2p 1↑↓ 3s 1↑ 1↑ 1↑ 3p ne3s²3p³
*c
*as
*y
tc
*sr

Explanation:

Step1: Determine electron - count for Carbon (C)

Carbon has atomic number 6, so 6 electrons.

Step2: Write full electron - configuration

The full electron - configuration is $1s^{2}2s^{2}2p^{2}$.

Step3: Write orbital notation

The orbital notation:
$\frac{\uparrow\downarrow}{1s}\frac{\uparrow\downarrow}{2s}\frac{\uparrow}{2p_x}\frac{\uparrow}{2p_y}\frac{}{2p_z}$

Step4: Write noble - gas configuration

The noble - gas before carbon is helium (He). So the noble - gas configuration is $[He]2s^{2}2p^{2}$.

Step5: Determine electron - count for Arsenic (As)

Arsenic has atomic number 33, so 33 electrons.

Step6: Write full electron - configuration

The full electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}$.

Step7: Write orbital notation

For the outer - shell electrons (after $3p^{6}$): $\frac{\uparrow\downarrow}{4s}\frac{\uparrow\downarrow}{3d_{xy}}\frac{\uparrow\downarrow}{3d_{xz}}\frac{\uparrow\downarrow}{3d_{yz}}\frac{\uparrow\downarrow}{3d_{x^{2}-y^{2}}}\frac{\uparrow\downarrow}{3d_{z^{2}}}\frac{\uparrow}{4p_x}\frac{\uparrow}{4p_y}\frac{\uparrow}{4p_z}$

Step8: Write noble - gas configuration

The noble - gas before arsenic is argon (Ar). So the noble - gas configuration is $[Ar]4s^{2}3d^{10}4p^{3}$.

Step9: Determine electron - count for Technetium (Tc)

Technetium has atomic number 43, so 43 electrons.

Step10: Write full electron - configuration

The full electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{5}$.

Step11: Write orbital notation

For the outer - shell electrons (after $4p^{6}$): $\frac{\uparrow\downarrow}{5s}\frac{\uparrow}{4d_{xy}}\frac{\uparrow}{4d_{xz}}\frac{\uparrow}{4d_{yz}}\frac{\uparrow}{4d_{x^{2}-y^{2}}}\frac{\uparrow}{4d_{z^{2}}}$

Step12: Write noble - gas configuration

The noble - gas before technetium is krypton (Kr). So the noble - gas configuration is $[Kr]5s^{2}4d^{5}$.

Step13: Determine electron - count for Strontium (Sr)

Strontium has atomic number 38, so 38 electrons.

Step14: Write full electron - configuration

The full electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}$.

Step15: Write orbital notation

For the outer - shell electrons (after $4p^{6}$): $\frac{\uparrow\downarrow}{5s}$

Step16: Write noble - gas configuration

The noble - gas before strontium is krypton (Kr). So the noble - gas configuration is $[Kr]5s^{2}$.

Answer:

Element	Full $e^{-}$ configuration	Orbital notation	Noble gas config.
As	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}$	$\frac{\uparrow\downarrow}{4s}\frac{\uparrow\downarrow}{3d_{xy}}\frac{\uparrow\downarrow}{3d_{xz}}\frac{\uparrow\downarrow}{3d_{yz}}\frac{\uparrow\downarrow}{3d_{x^{2}-y^{2}}}\frac{\uparrow\downarrow}{3d_{z^{2}}}\frac{\uparrow}{4p_x}\frac{\uparrow}{4p_y}\frac{\uparrow}{4p_z}$	$[Ar]4s^{2}3d^{10}4p^{3}$
Tc	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{5}$	$\frac{\uparrow\downarrow}{5s}\frac{\uparrow}{4d_{xy}}\frac{\uparrow}{4d_{xz}}\frac{\uparrow}{4d_{yz}}\frac{\uparrow}{4d_{x^{2}-y^{2}}}\frac{\uparrow}{4d_{z^{2}}}$	$[Kr]5s^{2}4d^{5}$
Sr	$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}$	$\frac{\uparrow\downarrow}{5s}$	$[Kr]5s^{2}$