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Classification of Elements and Periodicity in Properties - s-, p-, d- and f-Block Elements

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The elements are classified into four blocks: ss-block, pp-block, dd-block, and ff-block, depending on the type of atomic orbitals that are being filled with electrons.

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The ss-block elements (Groups 1 and 2) have the general outer electronic configuration ns1−2ns^{1-2}. They are all reactive metals with low ionization enthalpies.

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The pp-block elements (Groups 13 to 18) have the general outer electronic configuration ns2np1−6ns^2 np^{1-6}. These elements, together with ss-block elements, are called Representative Elements or Main Group Elements.

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The dd-block elements (Groups 3 to 12) are known as Transition Elements. Their general outer electronic configuration is (n−1)d1−10ns0−2(n-1)d^{1-10} ns^{0-2}. They are characterized by variable oxidation states and the formation of colored ions.

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The ff-block elements (Lanthanoids and Actinoids) are known as Inner Transition Elements. Their general outer electronic configuration is (n−2)f1−14(n−1)d0−1ns2(n-2)f^{1-14} (n-1)d^{0-1} ns^2.

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Across a period, the effective nuclear charge (ZeffZ_{eff}) increases, leading to a decrease in atomic radii and an increase in ionization enthalpy.

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Down a group, the addition of new shells increases the atomic radii and decreases the ionization enthalpy due to increased screening effects.

📐Formulae

s-block: ns1−2s\text{-block: } ns^{1-2}

p-block: ns2np1−6p\text{-block: } ns^2 np^{1-6}

d-block: (n−1)d1−10ns0−2d\text{-block: } (n-1)d^{1-10} ns^{0-2}

f-block: (n−2)f1−14(n−1)d0−1ns2f\text{-block: } (n-2)f^{1-14} (n-1)d^{0-1} ns^2

Zeff=Z−σZ_{eff} = Z - \sigma

ΔegH=Electron Gain Enthalpy\Delta_{eg}H = \text{Electron Gain Enthalpy}

💡Examples

Problem 1:

Predict the position of the element with atomic number Z=24Z = 24 in the periodic table.

Solution:

The electronic configuration of Z=24Z = 24 (Chromium) is [Ar]3d54s1[Ar] 3d^5 4s^1.

Explanation:

Since the last electron enters the dd-orbital, it belongs to the dd-block. The period is determined by the highest principal quantum number (n=4n = 4), so it is in Period 4. For dd-block, Group number = (number of electrons in (n−1)d(n-1)d subshell + number of electrons in nsns subshell) = 5+1=65 + 1 = 6. Thus, it is in Group 6.

Problem 2:

Which of the following elements has the highest negative electron gain enthalpy: FF, ClCl, BrBr, or II?

Solution:

ClCl (Chlorine) has the highest negative electron gain enthalpy.

Explanation:

Although fluorine (FF) is more electronegative, its small size leads to strong inter-electronic repulsions in the relatively compact 2p2p subshell. This makes the addition of an electron less favorable compared to chlorine (ClCl), where the 3p3p subshell is larger and can accommodate the incoming electron more easily.

Problem 3:

Calculate the effective nuclear charge for the valence electron of Nitrogen (Z=7Z=7) if the shielding constant σ\sigma is 3.103.10.

Solution:

Zeff=7−3.10=3.90Z_{eff} = 7 - 3.10 = 3.90

Explanation:

The effective nuclear charge is calculated using the formula Zeff=Z−σZ_{eff} = Z - \sigma, where ZZ is the atomic number and σ\sigma is the shielding (screening) constant.