Anand Classes provides detailed NCERT Solutions for Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties (Questions 3.26 to 3.30). These step-by-step solutions help students understand the concepts of periodic table, periodic trends, electronic configuration, ionization enthalpy, reactivity, and properties of s, p, d, and f block elements. This study material is designed for CBSE and competitive exam preparation like NEET and JEE. Click the print button to download study material and notes.
NCERT Solutions of Problems 3.26 – 3.30 of Chapter Classification of Elements and Periodicity in Properties Chemistry Class 11 pdf Download
NCERT 3.26 : What are the major differences between metals and non-metals?
Answer.
Elements which have a strong tendency to lose electrons to form cations are called metals, while those which have a strong tendency to accept electrons to form anions are called non-metals.
- Metals
- Strong reducing agents
- Low ionization enthalpies
- Less negative electron gain enthalpies
- Low electronegativity
- Form basic oxides
- Generally form ionic compounds
- Non-metals
- Strong oxidizing agents
- High ionization enthalpies
- Highly negative electron gain enthalpies
- High electronegativity
- Form acidic oxides
- Generally form covalent compounds
Comparison of Metals and Non-Metals
| Property | Metals | Non-Metals |
|---|---|---|
| Tendency | Strong tendency to lose electrons (form cations) | Strong tendency to gain electrons (form anions) |
| Redox nature | Strong reducing agents | Strong oxidizing agents |
| Ionization enthalpy | Low | High |
| Electron gain enthalpy | Less negative | Highly negative |
| Electronegativity | Low | High |
| Oxides | Form basic oxides | Form acidic oxides |
| Compounds | Generally form ionic compounds | Generally form covalent compounds |
NCERT 3.27 : Use the periodic table to answer the following questions
(a) Identify an element with five electrons in the outer subshell
(b) Identify an element that would tend to lose two electrons
(c) Identify an element that would tend to gain two electrons
(d) Identify the group having metal, non-metal, liquid as well as gas at room temperature
Answer :
(a) Identify an element with five electrons in the outer subshell
The general electronic configuration of the elements having five electrons in the outer subshell is:
$$ns^2 \, np^5$$
This electronic configuration is characteristic of elements of Group 17, i.e., halogens.
Examples: F, Cl, Br, I, At.
(b) Identify an element that would tend to lose two electrons
The elements which have a tendency to lose two electrons must have two electrons in the valence shell.
Therefore, their general configuration should be:
$$ns^2$$
This electronic configuration is characteristic of Group 2 elements, i.e., alkaline earth metals.
Examples: Mg, Ca, Sr, Ba.
(c) Identify an element that would tend to gain two electrons
The elements which have a tendency to accept two electrons must have six electrons in the valence shell.
Therefore, their general electronic configuration is:
$$ns^2 \, np^4$$
This electronic configuration is characteristic of Group 16 elements.
Examples: O, S.
(d) Identify the group having metal, non-metal, liquid as well as gas at room temperature
- A metal which is liquid at room temperature is Mercury (Hg). It is a transition metal and belongs to Group 12.
- A non–metal which is a gas at room temperature: Nitrogen (Group 15), Oxygen (Group 16), Fluorine, Chlorine (Group 17), Inert gases (Group 18).
- A non–metal which is a liquid at room temperature: Bromine (Group 17).
NCERT 3.28 : The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer
The elements of Group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron.
The tendency to lose electrons, in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group, therefore, the reactivity of Group 1 elements increases in the same order:
$$\text{Li < Na < K < Rb < Cs}$$
In contrast, the elements of Group 17 have seven electrons in their respective valence shells and thus have a strong tendency to accept one more electron.
The tendency to accept electrons, in turn, depends upon their electrode potentials. Since the electrode potentials of Group 17 elements decrease in the order:
$$\text{F (+2.87 V) > Cl (+1.36 V) > Br (+1.08 V) > I (+0.53 V)}$$
Therefore, their reactivity also decreases in the same order:
$$\text{F > Cl > Br > I}$$
NCERT 3.29 : Write the general outer electronic configuration of s–, p–, d– and f–block elements
Answer
(i) s–Block elements:
$$ns^{1-2}, \quad n = 2 \text{ to } 7$$
(ii) p–Block elements:
$$ns^2 \, np^{1-6}, \quad n = 2 \text{ to } 6$$
(iii) d–Block elements:
$$(n-1)d^{1-10} \, ns^{0-2}, \quad n = 4 \text{ to } 7$$
(iv) f–Block elements:
$$(n-2)f^{0-14} \, (n-1)d^{0-1} \, ns^2, \quad n = 6 \text{ to } 7$$
NCERT 3.30 : 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 $n = 4$
(iii) $(n – 2)f^7 \, (n – 1)d^1 \, ns^2$ for $n = 6$
Answer :
(i) $ns^2 \, np^4$ for $n = 3$
When $n = 3$, it suggests that the element belongs to the third period.
Since the last electron enters the p–orbital, therefore, the given element is a p–block element.
Valence shell contains $2 + 4 = 6$ electrons.
Group number of the element:
$$10 + \text{no. of valence electrons} = 10 + 6 = 16$$
The complete electronic configuration is:
$$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^4$$
Hence, the element is Sulphur (S).
(ii) $(n – 1)d^2 \, ns^2$ for $n = 4$
Here $n = 4$ suggests that the element lies in the 4th period.
Since the d–orbitals are incomplete, it is a d–block element.
Group number of the element:
$$\text{No. of d–electrons} + \text{No. of s–electrons} = 2 + 2 = 4$$
Thus, the element lies in Group 4 and 4th period.
The complete electronic configuration is:
$$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^2 \, 4s^2$$
Hence, the element is Titanium (Ti).
(iii) $(n – 2)f^7 \, (n – 1)d^1 \, ns^2$ for $n = 6$
Here $n = 6$ means the element lies in the 6th period.
Since the last electron goes to the f–orbital, the element is an f–block element.
All f–block elements lie in Group 3.
The complete electronic configuration is:
$$[Xe] \, 4f^7 \, 5d^1 \, 6s^2$$
Atomic number of the element:
$$54 + 7 + 1 + 2 = 64$$
Hence, the element is Gadolinium (Gd).
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