Assign the position of the element having outer electronic configuration (i) ns² np⁴ for n=3 (ii) (n-1)d² ns²

Exercises · 3.30

Q. Assign the position of the element having outer electronic configuration

(i)

ns^2 np^4

for

n=3

(ii)

(n-1)d^2 ns^2

for

n=4

, and

(iii)

(n-2)f^7 (n-1)d^1 ns^2

for

n=6

, in the periodic table.

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Concept understanding — Periodic Trend Metallic Character

What "Metallic Character" Actually Means

Imagine you have a piece of copper wire and a lump of charcoal. The copper is shiny, you can hammer it into a thin sheet, and it conducts electricity. The charcoal is dull, brittle, and does not conduct electricity well. That difference — the set of properties that make a metal "metallic" — is what we call metallic character.

Metallic character is not a single number you can measure directly. It is a qualitative trend that describes how strongly an element behaves like a metal. The more metallic an element is, the more it shows these traits:

Shiny (lustrous) appearance
High electrical and thermal conductivity
Malleability (can be hammered into sheets) and ductility (can be drawn into wires)
Tendency to lose electrons and form positive ions (cations)

The last point is the key chemical reason behind the trend. Metals are electron-losers. Non-metals are electron-gainers.

The Periodic Trend: The Precise Statement

Important

Metallic character decreases from left to right across a period, and increases from top to bottom down a group.

Let's break that into two parts.

Across a Period (Left to Right)

Take Period 3: Na → Mg → Al → Si → P → S → Cl → Ar.

Sodium (Na) is a highly reactive metal — it loses its one valence electron very easily. As you move right, the elements become less willing to lose electrons. Magnesium loses two electrons but holds them a bit tighter. Aluminium still behaves like a metal but is less reactive. Silicon is a metalloid — it has some metallic and some non-metallic properties. Phosphorus, sulfur, chlorine, and argon are clearly non-metals.

Why? The nuclear charge (number of protons) increases across the period, pulling the electrons in tighter. The valence electrons are held more strongly, so the atom is less willing to give them away. Losing electrons becomes harder → metallic character decreases.

Down a Group (Top to Bottom)

Take Group 1: Li → Na → K → Rb → Cs → Fr.

Lithium is a metal, but it is relatively hard and has a high melting point for a metal. Caesium is so metallic that it melts in your hand and explodes on contact with water. The metallic character increases dramatically as you go down.

Why? The atomic radius increases down the group. The valence electron is farther from the nucleus and is shielded by more inner electron shells. The nucleus holds it much more loosely. Losing that electron becomes very easy → metallic character increases.

Note

The same logic applies to all groups. Even in Group 14, carbon (top) is a non-metal, silicon and germanium are metalloids, and tin and lead (bottom) are metals. The trend is consistent.

The One Reason Behind Both Trends

Both trends come down to a single idea: how easily an atom can lose an electron.

Direction	Change in electron loss ease	Effect on metallic character
Left → Right	Harder (higher ionization energy)	Decreases
Top → Bottom	Easier (lower ionization energy)	Increases

Tip

If you ever forget the trend, remember: Metals are electron-losers. The easier it is to lose an electron, the more metallic the element. Ionization energy (the energy needed to remove an electron) is your best friend here — lower ionization energy = higher metallic character.

A Quick Visual Summary

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The key is to decode the principal quantum number $n$ and the subshell block from the configuration, then locate the period, group, and block. (i) $3s^2 3p^4$ → Period 3, Group 16, p-block (Sulfur).

(ii) $3d^2 4s^2$ → Period 4, Group 4, d-block (Titanium).

(iii) $4f^7 5d^1 6s^2$ → Period 6, Group 3 (or Lanthanide series), f-block (Gadolinium).

Concept First: How to Place an Element from Its Electronic Configuration

The periodic table is organised by electronic configuration — specifically, the outermost electrons determine the block, period, and group. Here’s the logic:

Period = the highest principal quantum number $n$ that appears in the configuration (the outermost shell).
Block = the subshell that receives the last electron (s, p, d, or f).
Group = depends on the block:
- s-block: group = number of valence electrons (1 or 2).
- p-block: group = 10 + number of valence electrons (since s + p electrons together count).
- d-block: group = number of $(n-1)d$ electrons + number of $ns$ electrons (usually 2, but watch for exceptions).
- f-block: all elements in a given f-series (lanthanides or actinides) belong to Group 3, and the period is the $n$ of the $ns$ subshell.

Watch out

A common mistake: For d-block, the group is not simply the number of d-electrons. You must add the s-electrons too. For example, $3d^2 4s^2$ gives group $2+2=4$ , not 2.

Now let’s apply this step by step.

(i) $ns^2 np^4$ for $n=3$

Step 1: Identify the period.

$n=3$ means the outermost shell is the third shell. So the element lies in Period 3.

Step 2: Identify the block.

The last electron enters the $p$ subshell ( $3p^4$ ). Hence it is a p-block element.

Step 3: Find the group.

For p-block elements:

Group number = $10 + \text{(number of valence electrons)}$

Valence electrons = $ns^2 + np^4 = 2 + 4 = 6$

So group = $10 + 6 = 16$ .

Step 4: Name the element.

Period 3, Group 16 → Sulfur (S).

Tip

A quick check: The configuration $3s^2 3p^4$ is exactly that of sulfur. You can also recall that group 16 elements (chalcogens) have the general outer configuration $ns^2 np^4$ .

(ii) $(n-1)d^2 ns^2$ for $n=4$

Step 1: Identify the period.

Here $n=4$ , so the outermost shell is the fourth shell. The element is in Period 4.

Step 2: Identify the block.

The last electron enters the $(n-1)d$ subshell, i.e., $3d$ (since $n-1=3$ ). So it is a d-block element (transition metal).

Step 3: Find the group.

For d-block elements:

Group number = number of $(n-1)d$ electrons + number of $ns$ electrons

Here: $3d^2$ and $4s^2$ → $2 + 2 = 4$ .

So group = 4.

Step 4: Name the element.

Period 4, Group 4 → Titanium (Ti). …

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Q.Assign the position of the element having outer electronic configuration

Concept First: How to Place an Element from Its Electronic Configuration

(i) ns2np4ns^2 np^4ns2np4 for n=3n=3n=3

(ii) (n−1)d2ns2(n-1)d^2 ns^2(n−1)d2ns2 for n=4n=4n=4

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(i) $ns^2 np^4$ for $n=3$

(ii) $(n-1)d^2 ns^2$ for $n=4$

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