Question #32109

If each orbital can hold a maximum of three electrons, then the number of elements in the 9th period of the Modern Periodic Table are?

Expert's answer

According to Klechkovski rule

1) Filling the energetic layers with electrons begins from levels with lowest values of n and l and continues in ascending order of n+l total.

2) If n+l total is same for several orbitals, the one with lower n is filled first.

It can be easily shown that in 9-th period next orbitals would be filled: 9s6g7f8d9p.

s-level consists of 1 orbital

p-level consists of 3 orbitals

d-level consists of 5 orbitals

f-level consists of 7 orbitals

g-level consists of 9 orbitals

So we have 1 + 3 + 5 + 7 + 9 = 36 orbitals total. If each orbital can be filled with 3 electrons, then we have 36 ∙ 3 = 108 elements.

1) Filling the energetic layers with electrons begins from levels with lowest values of n and l and continues in ascending order of n+l total.

2) If n+l total is same for several orbitals, the one with lower n is filled first.

It can be easily shown that in 9-th period next orbitals would be filled: 9s6g7f8d9p.

s-level consists of 1 orbital

p-level consists of 3 orbitals

d-level consists of 5 orbitals

f-level consists of 7 orbitals

g-level consists of 9 orbitals

So we have 1 + 3 + 5 + 7 + 9 = 36 orbitals total. If each orbital can be filled with 3 electrons, then we have 36 ∙ 3 = 108 elements.

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