Question

section 2: electron configuration practice
instructions: write full electron configuration and noble gas shorthand for each element.
oxygen (o) 8
calcium (ca) 20
iron (fe) 26
chlorine (cl) 17
krypton (kr) 36
copper (cu) 29
uranium (u) 92

Explanation:

Step1: Determine electron configuration rules

Electrons fill orbitals in order of increasing energy (Aufbau principle). Orbitals are filled as follows: \(1s < 2s<2p < 3s<3p<4s < 3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p\). Each \(s\) - orbital holds 2 electrons, each \(p\) - orbital holds 6 electrons, each \(d\) - orbital holds 10 electrons and each \(f\) - orbital holds 14 electrons. For noble - gas shorthand, we use the noble gas from the previous period.

Step2: For Oxygen (O, \(Z = 8\))

• Full configuration:
• Fill orbitals: \(1s^{2}2s^{2}2p^{4}\)
• Noble - gas shorthand:
• The previous noble gas is \(He\) (\(Z = 2\)). So, \([He]2s^{2}2p^{4}\)

Step3: For Calcium (Ca, \(Z = 20\))

• Full configuration:
• Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)
• Noble - gas shorthand:
• The previous noble gas is \(Ar\) (\(Z = 18\)). So, \([Ar]4s^{2}\)

Step4: For Iron (Fe, \(Z = 26\))

• Full configuration:
• Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\)
• Noble - gas shorthand:
• The previous noble gas is \(Ar\) (\(Z = 18\)). So, \([Ar]4s^{2}3d^{6}\)

Step5: For Chlorine (Cl, \(Z = 17\))

• Full configuration:
• Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
• Noble - gas shorthand:
• The previous noble gas is \(Ne\) (\(Z = 10\)). So, \([Ne]3s^{2}3p^{5}\)

Step6: For Krypton (Kr, \(Z = 36\))

• Full configuration:
• Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\)
• Noble - gas shorthand:
• Since it is a noble gas, we can also write it in terms of the previous noble gas \(Ar\) (\(Z = 18\)): \([Ar]4s^{2}3d^{10}4p^{6}\)

Step7: For Copper (Cu, \(Z = 29\))

• Full configuration:
• Due to the stability of half - filled and fully - filled orbitals, the electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\) (instead of \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{9}\))
• Noble - gas shorthand:
• The previous noble gas is \(Ar\) (\(Z = 18\)). So, \([Ar]4s^{1}3d^{10}\)

Step8: For Uranium (U, \(Z = 92\))

• Full configuration:
• \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{3}6d^{1}\)
• Noble - gas shorthand:
• The previous noble gas is \(Rn\) (\(Z = 86\)). So, \([Rn]7s^{2}5f^{3}6d^{1}\)

Answer:

Element	Atomic Number	Full Configuration	Shorthand Configuration
Calcium (Ca)	20	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)	\([Ar]4s^{2}\)
Iron (Fe)	26	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\)	\([Ar]4s^{2}3d^{6}\)
Chlorine (Cl)	17	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)	\([Ne]3s^{2}3p^{5}\)
Krypton (Kr)	36	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\)	\([Ar]4s^{2}3d^{10}4p^{6}\)
Copper (Cu)	29	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}\)	\([Ar]4s^{1}3d^{10}\)
Uranium (U)	92	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^{2}5f^{3}6d^{1}\)	\([Rn]7s^{2}5f^{3}6d^{1}\)