Question

Which atoms are indicated by the following configurations? (a) |He| $2s^{1}$ (b) |Ne| $3s^{2}3p^{3}$ (c) |Ar| $4s^{2}3d^{1}$

Answer 1

a. HELIUM

b. NEON

c. ARGON

Answer 2

Electrons are added one at a time to the lowest energy orbitals available until all the orbitals are filled.

According to the Aufbau principle, electrons fill lower energy orbitals first.

In the electronic configuration, Octet electronic configuration of an element is replaced by nearest halogen.

a) $[He]2{ s }^{ 1 }$

“He” atomic number = 4

Electronic configuration = $1{ s }^{ 2 }2{ s }^{ 2 }$

Therefore, total electrons = 1 + 4

Total electrons = 5

Therefore, the given electronic configuration of element is Boron.

b) $[Ne]3{ s }^{ 2 }3{ p }^{ 3 }$

Ne atomic number = 10

Electronic configuration = $1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }$

Therefore, total electrons = 10 + 2 + 3  

Total electrons = 15

Therefore, the given electronic configuration of element is Phosphorous.

c) $[Ar]4{ s }^{ 2 }3{ d }^{ 1 }$

Ar atomic number = 18

Electronic configuration = $1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }^{ 2 }3{ s }^{ 2 }3{ p }^{ 6 }$Therefore, total electrons = 18 + 2 + 1 = 21

Therefore, the given electronic configuration of element is Scan.