Question

write a ground state electron configuration for each neutral atom. ground state means that all of the lowest possible energy levels (up to the proper number of electrons for the element) are filled.
element atomic # # of electrons electron configuration abbreviated electron configuration
na 11 11
c 6 6
sr 38 38
ba 56 56
n 7 7
ca 20 20
sb 51 51
ne 10 10
cl 17 17
fe 26 26
he 2 2

Explanation:

Step1: Electron configuration rules

Electrons fill orbitals in order of increasing energy: \(1s < 2s<2p < 3s<3p<4s < 3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p\). Each \(s\) - orbital can hold 2 electrons, each \(p\) - orbital can hold 6 electrons, each \(d\) - orbital can hold 10 electrons and each \(f\) - orbital can hold 14 electrons. The abbreviated electron configuration uses the noble - gas core (the electron configuration of the previous noble gas).

Step2: Electron configuration of Na (Atomic number \(Z = 11\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{1}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Ne (\(1s^{2}2s^{2}2p^{6}\)). So, \([Ne]3s^{1}\)

Step3: Electron configuration of C (Atomic number \(Z = 6\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{2}\)
- **Abbreviated electron configuration**:
- The previous noble gas is He (\(1s^{2}\)). So, \([He]2s^{2}2p^{2}\)

Step4: Electron configuration of Sr (Atomic number \(Z = 38\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Kr (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\)). So, \([Kr]5s^{2}\)

Step5: Electron configuration of Ba (Atomic number \(Z = 56\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Xe (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}\)). So, \([Xe]6s^{2}\)

Step6: Electron configuration of N (Atomic number \(Z = 7\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{3}\)
- **Abbreviated electron configuration**:
- The previous noble gas is He (\(1s^{2}\)). So, \([He]2s^{2}2p^{3}\)

Step7: Electron configuration of Ca (Atomic number \(Z = 20\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Ar (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)). So, \([Ar]4s^{2}\)

Step8: Electron configuration of Sb (Atomic number \(Z = 51\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{3}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Kr (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}\)). So, \([Kr]5s^{2}4d^{10}5p^{3}\)

Step9: Electron configuration of Ne (Atomic number \(Z = 10\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}\)
- **Abbreviated electron configuration**:
- There is no noble gas before Ne in terms of core - electron - configuration simplification for Ne itself (since it is a noble gas). But if we follow the rule, we can write \([He]2s^{2}2p^{6}\)

Step10: Electron configuration of Cl (Atomic number \(Z = 17\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Ne (\(1s^{2}2s^{2}2p^{6}\)). So, \([Ne]3s^{2}3p^{5}\)

Step11: Electron configuration of Fe (Atomic number \(Z = 26\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\)
- **Abbreviated electron configuration**:
- The previous noble gas is Ar (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)). So, \([Ar]4s^{2}3d^{6}\)

Step12: Electron configuration of He (Atomic number \(Z = 2\))

- **Full electron configuration**:
- Fill orbitals: \(1s^{2}\)
- **Abbreviated electron configuration**:
- Since it is the first noble gas, there is no simpler noble - gas core. But if we consider the rule, we can write \([He]\) (but it's redundant as it is itself)

Answer:

Element	Electron configuration	Abbreviated electron configuration
C	\(1s^{2}2s^{2}2p^{2}\)	\([He]2s^{2}2p^{2}\)
Sr	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}\)	\([Kr]5s^{2}\)
Ba	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}\)	\([Xe]6s^{2}\)
N	\(1s^{2}2s^{2}2p^{3}\)	\([He]2s^{2}2p^{3}\)
Ca	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)	\([Ar]4s^{2}\)
Sb	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{3}\)	\([Kr]5s^{2}4d^{10}5p^{3}\)
Ne	\(1s^{2}2s^{2}2p^{6}\)	\([He]2s^{2}2p^{6}\)
Cl	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)	\([Ne]3s^{2}3p^{5}\)
Fe	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}\)	\([Ar]4s^{2}3d^{6}\)
He	\(1s^{2}\)	\([He]\)