Respuesta :
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
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
Answer: The element having partially filled f-orbital is Samarium (Sm)
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:
[tex]Sm:[Xe]4f^66s^2[/tex]
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:
[tex]Os:[Xe]5d^56s^2[/tex]
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:
[tex]Ba:[Xe]6s^2[/tex]
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:
[tex]Bi:[Xe]6s^26p^3[/tex]
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).