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crimeas [40]
3 years ago
14

Which element has a partially filled F orbital? Sm Os Ba Bi

Chemistry
2 answers:
Sever21 [200]3 years ago
7 0
<span>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 </span>
uysha [10]3 years ago
7 0

<u>Answer:</u> The element having partially filled f-orbital is Samarium (Sm)

<u>Explanation:</u>

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:

  • <u>Option 1:</u> 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.

  • <u>Option 2:</u> 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.

  • <u>Option 3:</u> 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.

  • <u>Option 4:</u> 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).

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The required formula of hydrate is MgSO₃.6H₂O.

<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
n = W/M, where

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Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol

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So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

To know more about hydrate compound, visit the below link:

brainly.com/question/22411417

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