Classification of Elements and Periodicity in Properties • Topic 1 of 3

Development & Modern Periodic Law

Chemists faced a flood of new elements in the nineteenth century and needed a way to organise them. The story of the periodic table is really a story of pattern-hunting — spotting that certain elements behave like each other and refusing to treat that as a coincidence.

Early attempts

Döbereiner's Triads (1829): Johann Döbereiner grouped elements in threes where the atomic mass of the middle element was nearly the arithmetic mean of the other two. For the triad Li (7), Na (23), K (39), the mean of Li and K is $(7+39)/2 = 23$, exactly Na. The idea only worked for a handful of triads, so it was abandoned.

Newlands' Law of Octaves (1865): John Newlands arranged the known elements in order of increasing atomic mass and noticed that every eighth element resembled the first, like the octaves in music. It held up to calcium but collapsed beyond it, and the Chemical Society of London refused to publish his work.

Mendeleev's Periodic Table (1869): Dmitri Mendeleev ordered elements by increasing atomic mass and stated his Periodic Law: the properties of the elements are a periodic function of their atomic masses. His genius lay in two bold moves — he left gaps for undiscovered elements and even predicted their properties. He named them eka-aluminium, eka-boron and eka-silicon; these turned out to be gallium, scandium and germanium. He also reversed a few mass-based positions (e.g. Te before I) to keep chemically similar elements together, hinting that mass was not the deepest cause.

The Modern Periodic Law

Henry Moseley (1913), studying X-ray spectra, showed that the atomic number $Z$ — not atomic mass — is the fundamental property of an element. This resolved Mendeleev's anomalies (Te, $Z=52$, genuinely comes before I, $Z=53$). The Modern Periodic Law states: the physical and chemical properties of the elements are a periodic function of their atomic numbers. Periodicity arises because similar outer (valence) electronic configurations recur at regular intervals as $Z$ increases.

Long form of the periodic table

The modern long form has 7 horizontal periods and 18 vertical groups. A period number equals the principal quantum number $n$ of the outermost shell being filled. Down a group, elements share the same valence-shell configuration, which is why they show similar chemistry. The number of elements in each period (2, 8, 8, 18, 18, 32, 32) follows directly from the order in which orbitals are filled.

s-, p-, d- and f-blocks

The table is divided into four blocks by the subshell that receives the last electron:

s-block (Groups 1 and 2): outer configuration $ns^{1-2}$ — reactive metals. p-block (Groups 13–18): $ns^2np^{1-6}$ — includes metals, metalloids, non-metals and the noble gases ($ns^2np^6$). d-block (Groups 3–12, the transition elements): $(n-1)d^{1-10}\,ns^{0-2}$. f-block (lanthanoids and actinoids): $(n-2)f^{1-14}$ — the inner-transition elements, shown separately at the foot of the table.

IUPAC nomenclature for $Z>100$

To name newly discovered super-heavy elements provisionally, IUPAC uses numerical roots for each digit of $Z$: nil(0), un(1), bi(2), tri(3), quad(4), pent(5), hex(6), sept(7), oct(8), enn(9). The roots are joined in order, followed by the suffix -ium. For example $Z=104$ is un-nil-quad-ium = unnilquadium (Unq), later named rutherfordium. This system gives every element a name even before a permanent one is agreed.

Long form periodic table showing s, p, d and f blocksspdf (lanthanoids and actinoids)Gp 1-2Gp 13-18Gp 3-12periods 1-7
1
Worked Example
Verify Döbereiner's triad rule for Cl (35.5), Br (80) and I (127).
Solution
  1. The middle element by mass is Br.
  2. Take the mean of the lightest and heaviest: $\dfrac{35.5 + 127}{2} = \dfrac{162.5}{2} = 81.25$.
  3. Compare with the actual mass of Br, which is $80$.
  4. The predicted and actual values agree closely.

Answer: The mean $81.25$ is very close to the observed atomic mass of bromine ($80$), so Cl, Br, I form a valid Döbereiner triad.

2
Worked Example
Why did Mendeleev place tellurium (atomic mass $127.6$) before iodine (atomic mass $126.9$) even though Te is heavier?
Solution
  1. Mendeleev's law ordered elements by atomic mass, which would put I before Te.
  2. But iodine resembles the halogens (F, Cl, Br) and tellurium resembles the oxygen-group elements (O, S, Se).
  3. To keep chemically similar elements in the same group, Mendeleev reversed the order.
  4. Moseley later showed Te has $Z=52$ and I has $Z=53$, so by atomic number the order Te then I is actually correct.

Answer: Mendeleev prioritised chemical similarity over mass; the modern law (ordering by $Z$) confirms his arrangement.

3
Worked Example
Predict the block, period and group of the element with $Z = 26$.
Solution
  1. Write the configuration: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2$.
  2. The last electron enters a $d$ subshell, so it is a d-block element.
  3. The highest principal quantum number is $n=4$, so the period is 4.
  4. For d-block, group number $= 8 + (\text{number of } (n-1)d \text{ electrons beyond filled} ) $; here $ns^2 + (n-1)d^6$ gives group $= 2 + 6 = 8$.

Answer: $Z=26$ (iron) lies in the d-block, period 4, group 8.

4
Worked Example
Give the IUPAC provisional name and symbol for the element with $Z = 119$.
Solution
  1. Split $Z$ into digits: 1, 1, 9.
  2. Assign roots: 1 = un, 1 = un, 9 = enn.
  3. Join the roots and add the suffix -ium: un + un + enn + ium.
  4. The symbol takes the first letter of each root: U, U, E.

Answer: Name = ununennium; symbol = Uue.

5
Worked Example
An element has the valence-shell configuration $ns^2np^3$ in the third period. Identify the element, its group and block.
Solution
  1. Third period means $n = 3$, so the configuration is $3s^2\,3p^3$.
  2. The full configuration is $1s^2\,2s^2\,2p^6\,3s^2\,3p^3$, giving $Z = 15$.
  3. The last electron enters a $p$ subshell, so it is a p-block element.
  4. Group number for p-block $= 10 + (\text{valence electrons}) = 10 + 5 = 15$.

Answer: The element is phosphorus ($Z=15$), period 3, group 15, p-block.

6
Worked Example
How many elements are present in the 4th period of the long form of the periodic table, and why?
Solution
  1. In period 4 the subshells filled are $4s$, $3d$ and $4p$ (in order of increasing energy).
  2. $4s$ holds 2 electrons, $3d$ holds 10 electrons and $4p$ holds 6 electrons.
  3. Total electrons added $= 2 + 10 + 6 = 18$.
  4. Each new electron corresponds to one new element, so the period contains 18 elements.

Answer: The 4th period has 18 elements (from K, $Z=19$, to Kr, $Z=36$).

Key Points

  • Döbereiner's triads, Newlands' octaves and Mendeleev's table (with gaps and predictions for eka-elements) were the key early classifications.
  • Moseley established atomic number $Z$ as the basis; the Modern Periodic Law: properties are a periodic function of atomic number.
  • The long form has 7 periods (period no. = outermost $n$) and 18 groups (same valence configuration down a group).
  • Blocks are named after the subshell filled last: s ($ns^{1-2}$), p ($ns^2np^{1-6}$), d ($(n-1)d^{1-10}ns^{0-2}$), f ($(n-2)f^{1-14}$).
  • IUPAC names elements with $Z>100$ using digit roots (nil, un, bi, tri...) plus the suffix -ium.
Tap an option to check your answer0 / 4
Q1.Who first arranged elements by atomic number rather than atomic mass as the fundamental property?
Explanation: Moseley's X-ray work (1913) showed atomic number, not mass, is the fundamental property, leading to the Modern Periodic Law.
Q2.Mendeleev's 'eka-silicon' was later discovered and named:
Explanation: Eka-silicon corresponds to germanium; eka-aluminium is gallium and eka-boron is scandium.
Q3.The element with $Z=24$ belongs to which block?
Explanation: $Z=24$ (chromium) has configuration $[\text{Ar}]3d^5\,4s^1$; the last electron is in a $d$ subshell, so it is a d-block element.
Q4.The IUPAC provisional symbol for the element $Z=120$ is:
Explanation: Digits 1, 2, 0 give roots un, bi, nil; the name is unbinilium and the symbol Ubn.