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Row 1: 2 elements

Row 2: 8 elements

Row 3: 8 elements

Row 4: 18 elements

Row 5: 18 elements

Row 6: 32 elements

Row 7: 32 elements

In other words: 2, 8, 8, 18, 18, 32, 32

Why does the first row have only 2 elements, but all the next rows have a pattern? There's a pattern of two rows having the same number of elements that only starts after the first row.

EDIT: I just noticed this alternate periodic table! It's first two rows have two elements each! Why does it work there? (I can't interpret the s, p, d, f part on the left). Which one is the right periodic table?

  • 2
    $\begingroup$ It's just because the order of filling goes 1s -> 2s -> 2p for the 1st and 2nd rows, but for the 4th row it's 4s -> 3d -> 4p. The difference is that the 3d only comes in after the 4s. If the order of filling was the apparently more "normal" 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f... then your 1st to 4th rows would have 2, 8, 18, and 32 elements. $\endgroup$ Commented Feb 3, 2016 at 10:05
  • 3
    $\begingroup$ So in a sense, the 1st and 2nd rows are not the "odd" part out. The 3rd row is the "odd" one out because if the order of filling was 3s 3p 3d it would have 18 elements. However, because the 3d orbitals are only filled after 4s, the d-block is delayed from row 3 to row 4. $\endgroup$ Commented Feb 3, 2016 at 10:07
  • $\begingroup$ @orthocresol I'm not sure I understand what you mean. Please turn it into a detailed answer if you can. $\endgroup$
    – DrZ214
    Commented Feb 17, 2016 at 10:06

5 Answers 5


The pattern is better expressed this way:

Row 1: 2 elements
Row 2: 2+6 elements
Row 3: 2+6 elements
Row 4: 2+6+10 elements
Row 5: 2+6+10 elements
Row 6: 2+6+10+14 elements
Row 7: 2+6+10+14 elements

The reason comes down to how the electrons fill the available energy levels. The thing that differentiates one element from another is the proton number, and each time a proton is added and a new element defined it requires one more electron to neutralise the charge. That electron naturally occupies the lowest available energy level in the atom.

The energy levels available are defined by the quantum state, including the quantum numbers $n, l,$ and $m_l$. $n$ is the principle quantum number and relates to the period the element is in, or the shell. $l$ is the angular momentum quantum number which defines the sub shell s, p, d, f, of which there are $n$ subshells whose values are $l=0,\dots {n-1}$. The magnetic quantum number $m_l$ further subdivides the subshell into orbitals, of which there are $2l+1$ orbitals whose values are $m_l=-l,\dots {+l}$.

subshell                 number of orbitals    subshell
label     l value       (number of ml values)  electron capacity
 s         0               1                   2 
 p         1               3                   6
 d         2               5                   10
 f         3               7                   14

Each orbital (i.e. each value of $m_l$) may contain 2 electrons The available quantum numbers are:

n    l values   (subshells)   ml values        total shell electron capacity
1    0           (s)           0                 2          
2    0,1         (s,p)        -1,0,1             2+6        
3    0,1,2       (s,p,d)      -2,-1,0,1,2        2+6+10     
4    0,1,2,3     (s,p,d,f)    -3,-2,-1,0,1,2,3   2+6+10+14  

The pattern is there, but it doesn't seem to match up because by row 7 you would have 98 electrons in the shell and might expect the row to contain 98 elements. This is not the case! The energy levels for the $l$ values with large deviations from $0$ (i.e. orbitals with high angular momentum) become increasingly far apart, so even though 3d orbitals exist in the third row, they are so much higher in energy than the 3p orbitals that they are higher even than the 4s orbitals. This happens again in row 4 where 4f orbitals are so much higher in energy than 4d orbitals, they are even higher than 5s.

The different in energy is not always so large in specific cases, so there are examples where bonding or ligands or symmetry cause the energy levels to switch around, but otherwise the trend is true. If they can produce (or discover) some elements in the next row, we might expect to see the 5g shell finally start to get filled and have a new block in the table, but realistically it doesn't look like they will be stable atoms.

I think I'm missing the reason why we arrange the elements into this table in the first place. The periodic table is arranged so that all elements in the same column have the same number of outer electrons. This is useful because elements with the same outer shell configuration of electrons react in similar ways, so they are grouped together in columns. This has the effect of creating 'periods' that begin with a reactive metal and end with an inert gas. Row 1 has two elements, and it's special in a way because it's the first in the series. Hydrogen has 1 electron, and a maximum capacity of 2 electrons. This means it behaves in a similar way as other elements that only have 1 electron in their outer shell (group 1), and also in a similar way to elements that only need 1 more electron to complete their outer shell (group 7). Where do we position it then? Often it is placed somewhere in the middle. Helium then has a full outer shell, which means it has many similarities to other elements with a full outer shell and goes in group 8 (or group 0) on the right.

Edit: The alternative period table that you mention, the left-hand version, is laid out so as you read it from left to right you fill up the orbitals. This takes in to account that situation I describe where 3d orbitals are higher than 4s, and using this left-hand table you clearly see that. However this alternative table does not allow you to easily read the valance shell configuration like the Mendelev one does. Other types of table highlight different kinds of properties or relationships, which can be more useful in certain situations. The labels on the left (1s; 2s; 2p 3s; etc) relate to the outer subshell in that row, so Carbon is in the "2p" section of that chart while Magnesium is in the "3s" section on the same row.

  • $\begingroup$ Thanks but I'm still kinda confused. Can you explain exactly what are s, p, d, f shells and specifically how many electrons each of them can hold? $\endgroup$
    – DrZ214
    Commented Feb 17, 2016 at 14:07
  • $\begingroup$ I already added the notes to the table in my answer, s=2, p=6, d=10, f=14 electrons each. Row n=1 only has s subshell, n=2 has s and p subshells, n=3 has s, p, and d subshells, n=4 has s,p,d, and f subshells. $\endgroup$ Commented Feb 17, 2016 at 14:28
  • $\begingroup$ Does this mean, for example, all p shells always have capacity 6? So a 3p, 4p, or any p, always holds a max of 6 electrons? $\endgroup$
    – DrZ214
    Commented Feb 17, 2016 at 15:00
  • 2
    $\begingroup$ p orbitals always have have l=1 ('s' is the label for l=0, 'p' is the label for l=1, and so on) and ml=-1,0, or 1, each orbital can contain 2 electrons each , so that's 3 orbitals, 6 electrons in total for every p subshell $\endgroup$ Commented Feb 17, 2016 at 15:16


The general arrangement of The Periodic Table is a result of the orbitals available to electrons bound to a single, isolated nucleus.

Each orbital is described by a wave function, which is a regular function (similar to polynomials familiar to high school algebra) with respect to the $x$, $y$, and $z$ coordinates of space, and describes the behavior of the electrons occupying it.

We can find these wave functions (and equivalently: the orbitals) by solving the Schrödinger Equation, which is a differential equation describing the system: one electron orbiting one proton in the case of hydrogen. A system with more than one electron is actually impossible to solve exactly, so we usually just think about hydrogen orbitals. The set of orbitals for any other element is more or less the same as that for hydrogen--just with different energies--so your question can be answered by considering hydrogen alone.


Electrons in these orbitals have 4 characteristics, and no two electrons belonging to a single nucleus can have matching values for all 4 (The Pauli Exclusion Principle):

  1. Size
  2. Shape
  3. Orientation
  4. Spin

    1. For hydrogen: the size contains all of the information about the energy of the orbital. Denoted by the letter $n$.

    2. The shape reflects the angular momentum of the orbital. Denoted by $l$.

    3. The orientation reflects one (of three) components of the angular momentum, as a convention: the z-coordinate. Denoted by $m_l$.

    4. Unless you care about magnetism, the spin is only important because it allows you to put 2 electrons in one orbital (due to Pauli's principle). Denoted by $m_s$.

Since there is an energy associated with angular momentum, the possible values of $l$ for any electron is constrained by its value of $n$. For example: if $n=2$, only $l=1$ and $l=0$ are possible. In the case of the first row elements (in their ground states): $n=1$ so $l=0$ without exception.

The numbers of elements in the other rows are a result of rules governing possible orientations, $m_l$, of the orbitals (keeping in mind Pauli's principle, and that two spots are available per orbital due to spin, $m_s$).

In case you're curious: if an electron has angular momentum $l=1$, the z-component of its angular momentum, $m_l$ can only have the values $1$, $0$ or $-1$. For $l=2$, $m_l=\{2,1,0,-1,-2\}$, and for the first row: $l=0$ implies that $m_l=0$ without exception.

If you write out the number of possible orbitals following these rules: $$n=\{1,2,3,...\}$$ $$n \gt l \ge 0$$ $$+l \ge m_l \ge -l$$ $$m_s=\{-1/2,+1/2\}$$ ...where $n$ corresponds to the row number on The Periodic Table, you will find the pattern that you mentioned in your question.

Electron densities for orbitals of various $l$ (vertical) and $m_l$ (horizontal) values

Image Ref: https://en.wikipedia.org/wiki/Spherical_harmonics


Have you gone to a library?

You might have, and might have observed that the books on same topic are in same stack.

Let chemists be the librarian of the library of elements. And he has got to arrange his elements into stacks. Okay?

Chemist 1: How do I reduce the disorder in the arrangement of the these elements?

Chemist 2: Umm ... we need some criteria. According to that criteria, we will stack the elements.

Chemist 3: Let the criterion be electronic configuration.

Chemist 4 (looks at you): Do you know about subshells and orbitals?

Welcome to the world — sorry, shell — of electrons!

Periodic Table of Elements Showing Electron Shells

An electron shell may be thought of as an orbit followed by electrons around an atom’s nucleus.

The closest shell to the nucleus is called the “1 shell” (also called “K shell”), followed by the “2 shell” (or “L shell”), then the “3 shell” (or “M shell”), and so on farther and farther from the nucleus. The shells correspond with the principal quantum numbers (n = 1, 2, 3, 4 ...) or are labelled alphabetically with letters used in the X-ray notation (K, L, M, …).

Each shell can contain only a fixed number of electrons.

The electron shells are labelled K, L, M, N, O, P, and Q; or 1, 2, 3, 4, 5, 6, and 7; going from innermost shell outwards. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called 1s; the second (L) shell has two subshells, called 2s and 2p; the third shell has 3s, 3p, and 3d; the fourth shell has 4s, 4p, 4d and 4f.

Each s subshell holds at most 2 electrons.
Each p subshell holds at most 6 electrons.
Each d subshell holds at most 10 electrons.
Each f subshell holds at most 14 electrons.

Row 1 contains only the elements that have only the K shell occupied*.
Row 2 contains only the elements that have only the K and L shells occupied*.
Row 3 contains only the elements that have only the K, L and M shells occupied*.
And so on.

* Occupied means that shell may or may not be filled, but there certainly are some electrons.


the number of elements in a particular period is equal to the number of vacant orbitals available to be filled in the particular shell this also follows the n+l rule according to the energies of the orbitals.

Coming to first period in the first shell only one sub shell is there which is 1s and two electrons can be successively filled hence two elements are present in the first period.

Second shell has two sub shells 2s and 2p which have a total of four orbitals hence eight electrons can be successively filled in the four orbitals and hence 2nd period has eight elements.

The third shell has an exception due to energy considerations because of the n+l rule though the third period has three sub shells 3s , 3p and 3d and total of 9 orbitals i.e one 3s orbital three 3p orbitals and five 3d orbitals but because 3d orbitals can be filled only after 4s orbitals due to the n+l rule only four orbitals are available which are the one 3s and three 3p orbitalshence eight electrons can be successively filled and hence eight elements are present in the third period.

the fourth shell has similar story with four sub shells 4s 4p 4d and 4f but because of the n+l rule or the madelung energy ordering rule

4d orbitals are filled after 5s orbitals and 4f orbitals after the 6s orbitals but here the 3d orbitals are also available to be filled up so a total of 9 orbitals i.e one 4s three 4p and five 3d orbitals are available the next periods follow suit and things happen in a similar manner . Hope it helps .


Simple Answer:

The elements are ordered into periods based on which electron shells are being filled (from left to right). In the first period, the first electron shell is being filled. In the second period, the second shell is being filled. And so on.

There are 8 elements in period 2 because all those elements have electrons in the second shell and no electrons in the third shell.


Electron configuration describes the electrons of an atom in more detail. The elements in the same period all have the same thing at the beginning of their electron configuration. Here are some examples so you can see the pattern:

Li - Period 2 - [He] 2s1
Be - Period 2 - [He] 2s2
Ne - Period 2 - [He] 2s2 2p6
Na - Period 3 - [Ne] 3s1
Cl - Period 3 - [Ne] 3s2 3p5

This is the only way I can think of explaining it without going into detail about what the electron configuration means and how it can be used. But as to why there are 8 elements in period 2, all of those 8 elements have an electron configuration that starts like this: [He] 2s

Alternate Table:

Which one is the right periodic table?

There is no right table. These tables are just organizational tools, and you can organize the elements based on a different criteria if you want to. No matter how you organize the table of elements, the elements are still going to have the same properties and react the same way anyway.

  • $\begingroup$ There is no "right table", but is there a "right periodic table"? If the elements were in a different arrange, wouldn't the periodic properties become disarranged? There would not be "periods" of properties as density, e.g. $\endgroup$
    – mguima
    Commented Jun 5, 2023 at 15:59

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