Question

Which element has a partially filled F orbital?

Sm

Os

Ba

Bi

Answer 1

Answer: The element having partially filled f-orbital is Samarium (Sm)

Step-by-step explanation:

Maximum number of electrons that can be filled in f-orbital are 14. So, for the element having 14 electrons in the f-orbital will be considered as fully-filled and the element having 7 electrons in the f-orbital will be considered as half-filled.

It will be judged by electronic configuration of the elements.

For the given options:

• Option 1: Samarium (Sm)

The atomic number of samarium is 62 and number of electrons that are present are 62. The electronic configuration of samarium is:

`Sm:[Xe]4f^66s^2`

As, this element has electrons in f-orbital. Thus, this has partially filled f-orbital.

• Option 2: Osmium (Os)

The atomic number of osmium is 76 and number of electrons that are present are 76. The electronic configuration of osmium is:

`Os:[Xe]5d^56s^2`

As, this element does not have electrons in f-orbital. Thus, this does not has partially filled f-orbital.

• Option 3: Barium (Ba)

The atomic number of barium is 56 and number of electrons that are present are 56. The electronic configuration of barium is:

`Ba:[Xe]6s^2`

As, this element does not have electrons in f-orbital. Thus, this does not has partially filled f-orbital.

• Option 4: Bismuth (Bi)

The atomic number of bismuth is 83 and number of electrons that are present are 83. The electronic configuration of bismuth is:

`Bi:[Xe]6s^26p^3`

As, this element does not have electrons in f-orbital. Thus, this does not has partially filled f-orbital.

Hence, the correct answer is samarium (Sm).

Answer 2

Sm

For an element to have a partially filled f orbital, it will have to have an f orbital in the first place, this cancels barium, as it is the lightest of the elements listed:


Barium does not have an f orbital: [Xe]6s^2 or 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2


Sm: [Xe] 4f6 6s2 Does have an f orbital AND they are partially filled (the F subshell has the potential to hold 14 electrons, but Sm only holds 6 electrons on its F subshell, therefore the electrons, by the rule of maximum multiplicity, in which the electrons will try to occupy orbitals by themselves first (the F subshell has 7 orbitals because 14/2 = 7), it leaves the f subshell with partially filled orbitals.

Os: Xe 4f14 5d6 6s2 all occupied f orbitals

Bi: Xe 4f14 5d10 6s2 6p3 Has full F orbitals