Which element has a partially filled f orbital?

sm

os

ba

bi

Answers

Explanation:

1. Molarity is the ratio of the moles of a solute to the total liters of a solution. The solution includes both the solute and the solvent. Molality, on the other hand, is the ratio of the moles of a solute to the kilograms of a solvent.

2.Hund's rule: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

3. Atomic Number      Name of the ElementElectronic Configuration

        29                            Copper (Cu)                 [Ar] 3d10 4s1  

4. during ice phase of water , hydrogen bonds forms cage like structure . hence it occupies large volume, as temperature increases to 277K hydrogen bonds r broken hence volume decreases and density increases.. ... therefore water has it's maximum density at 277 K.

6. It increases with increase of pressure and decrease on lowering of pressure. At higher altitudes, the atmospheric pressure is low, and therefore water boils below 1000C. Hence, sufficient heat is not supplied for cooking the vegetables at hill stations. This difficulty may be overcome by using a pressure cooker.

7. According to Le Chatelier, if you increase the temperature of the water, the equilibrium will move to lower the temperature again. It will do that by absorbing the extra heat. That means that the forward reaction will be favoured, and more hydrogen ions and hydroxide ions will be formed.

Mark me as brianliest.

Sm is the answer

i hope i helped :)

Sm (Samarium) = [Xe] 4f6 6s2

Explanation:

Noble gas electron configuration of Sm (Samarium) = [Xe] 4f⁶ 6s²

Noble gas electron configuration of Os (Osmium) = [Xe] 4f¹⁴5d⁶6s²

Noble gas electron configuration of Ba (Barium) = [Xe] 6s²

Noble gas electron configuration of Bi (Bismuth) = [Xe] 4f¹⁴5d¹⁰6s²6p³

A completely filled s, p, d, and f orbitals have 2, 6, 10 and 14 electrons respectively. Half filled s, p, d, and f orbitals have 1, 3, 5 and 7 electrons respectively.  Bi and Os have completely filled f orbitals. Sm has partially filled f orbital.  

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:

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).

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:

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).

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 
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 


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