Chemistry, asked by Rohantan07, 9 months ago

how many subshells are possible if arbitrary period number (10) is established

Answers

Answered by sathwikgracy
1

Answer:

Well, this is kind of open to interpretation... but if I interpreted it correctly, I get

27

elements, compared to the original

18

elements, in the 4th period of the periodic table.

(If you already knew that the number of electrons allowed in a given orbital is derived from the properties of electrons, and NOT of the orbitals themselves, the answer quickly follows.)

PRELIMINARY THINGS

I think there's a typo in the question... I looked this up elsewhere, and it's probably...

If each orbital can hold a maximum of 3 electrons, the number of elements in the 4th period of the periodic table (long form) is...?

Also, note that this is entirely theoretical, as all electrons only have two possible spins (

m

s

=

±

1

2

) in a given orbital, and no two electrons can share the same quantum state (Pauli Exclusion Principle); that restricts each orbital to contain only two electrons in real life.

That aside, when we suppose three electrons are "allowed" in a single orbital (assuming the other three quantum numbers are as normal), we suddenly "allow"

50

%

more elements in a given quantum level.

(It's not really why the periodic table was arranged historically, but... I suppose that's what the intent of this question was...)

EXPANDING THE PERIOD...

An electron configuration for the fourth period in a generalized manner is written as:

4

s

x

3

d

x

4

p

x

where:

x

=

(

2

l

+

1

)

N

m

s

is the total number of electrons in all the orbitals in a given subshell.

l

is the angular momentum quantum number.

l

=

0

,

1

,

2

,

3

,

.

.

.

corresponds to

s

,

p

,

d

,

f

,

.

.

.

orbitals.

2

l

+

1

is known as the degeneracy of the subshell; it is how many orbitals are in that subshell.

N

m

s

is an arbitrary number of spins the electron could have, as it also then gives the maximum number of electrons per orbital. In this case we SUPPOSE that

N

m

s

=

3

, but in real life it is just

2

.

Now, the number of allowed

m

s

values derives from the properties of the electron, not of the orbitals themselves, so having more spins allowed does NOT change the orbital shapes or relative energies.

For the

4

s

orbital:

l

=

0

,

2

l

+

1

=

1

For the

3

d

orbitals:

l

=

2

,

2

l

+

1

=

5

For the

4

p

orbitals:

l

=

1

,

2

l

+

1

=

3

Thus, the hypothetical electron configuration we would then write is...

4

s

1

3

3

d

5

3

4

p

3

3

=

4

s

3

3

d

15

4

p

9

And that would apparently expand the fourth period of the periodic table from

18

elements to

27

elements.

Answered by Theopekaaleader
0

</p><p></p><p></p><p></p><p>\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}</p><p></p><p>\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered} </p><p>⟶ </p><p>−10</p><p>1</p><p>	</p><p> = </p><p>v</p><p>1</p><p>	</p><p> − </p><p>−30</p><p>1</p><p>	</p><p> </p><p>⟶ </p><p>v</p><p>1</p><p>	</p><p> = </p><p>−10</p><p>1</p><p>	</p><p> + </p><p>30</p><p>1</p><p>	</p><p> </p><p>⟶ </p><p>v</p><p>1</p><p>	</p><p> = </p><p>30</p><p>−3+1</p><p>	</p><p> </p><p>⟶ </p><p>v</p><p>1</p><p>	</p><p> = </p><p>30</p><p>−2</p><p>	</p><p> </p><p>	</p><p> </p><p>⟶ </p><p>v</p><p>1</p><p>	</p><p> = </p><p>−15</p><p>1</p><p>	</p><p> </p><p>⟶v=−15cm</p><p>	</p><p> </p><p></p><p></p><p>\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}} </p><p>MAGNIFICATION:</p><p>	</p><p> </p><p></p><p>\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered} </p><p>↦m= </p><p>Heightofobject(O)</p><p>Heightofimage(I)</p><p>	</p><p> = </p><p>Distanceofobject</p><p>Distanceofimage</p><p>	</p><p> = </p><p>u</p><p>v</p><p>	</p><p> </p><p>↦m= </p><p>−15</p><p>−30</p><p>	</p><p> </p><p>	</p><p> </p><p>↦m=2cm</p><p>	</p><p> </p><p></p><p>Thus;</p><p></p><p>The magnification will be 2 cm .</p><p></p><p>

Similar questions
Math, 9 months ago