Chemical Bonding Intex-1 Concise Class-10 ICSE Chemistry Selina Solutions. We Provide Step by Step Answer of Intext-1, Exercise-2, ICSE Class-10. The given Solutions is according to the Latest editions. Visit official Website CISCE for detail information about ICSE Board Class-10.

Chemical Bonding Intex-1 Concise Class-10 ICSE Chemistry Selina Solutions

Board	ICSE
Book / Publication	Concise / Selina
Subject	Chemistry
Class	10th
Writer	Dr SP Singh
Chapter-2	Chemical Bonding
Topics	Intext -1
Edition	2025-2026

Chemical Bonding Intex-1 ICSE Class-10 Concise Chemistry Selina Solutions Chapter-2

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Que-1: How do atoms attain noble gas configuration?

Ans: Atoms lose, gain or share electrons to attain noble gas configuration.

Que-2: Define:

(a) a chemical bond
(b) an electrovalent bond
(c) a covalent bond

Ans:
(a) A chemical bond may be defined as the force of attraction between any two atoms, in a molecule, to maintain stability.
(b) The chemical bond formed between two atoms by transfer of one or more electrons from the atom of a metallic electropositive element to an atom of a non-metallic electronegative element is called as electrovalent bond.
(c) The chemical bond formed due to mutual sharing of electrons between the given pairs of atoms of non-metallic elements is called as a covalent bond.

Que-3: What are the conditions for formation of an electrovalent bond?

Ans: Conditions for formation of Ionic bond are:
(i) The atom which changes into cation should possess 1, 2 or 3 valency electrons. The other atom which changes into anion should possess 5, 6 or 7 electrons in the valence shell.
(ii) A high difference of electronegativity of the two atoms is necessary for the formation of an Ionic bond.
(iii) There must be an overall decrease in energy i.e., energy must be released.
For this an atom should have low value of Ionisation potential and the other atom should have high value of electron affinity.
(iv) Higher the lattice energy, greater will be the case of forming an ionic compound.

Que-4: An atom X has three electrons more than noble gas configuration. What type of ion will it form? Write the formula of its (i) Sulphate (ii) Nitrate (iii) Phosphate (iv) carbonate (v) Hydroxide

Ans: It will form a cation: X³⁺
(i) X₂(SO₄)₃
(ii) X(NO₃)₃
(iii) XPO₄
(iv) X₂(CO₃)₃
(v) X(OH)₃

Que-5: Mention the basic tendency of an atom which makes it combine with other atoms.

Ans: Atoms combine with other atoms to attain stable octet or noble gas configuration.

Que-6: The element X has the electronic configuration 2, 8, 18, 8, 1. Without identifying x,

(a) predict the sigh and charge on a simple ion of X.
(B) write if X be an oxidizing agent or reducing agent and why.

Ans:
(a) X⁺
(b) X will be a strong reducing agent as it will have the tendency to donate its valence electron.

Que-7: In the formation of compound XY_2, an atom X gives one electron to each Y atom, what is the nature of bond in XY₂? Draw the electron dot structure of this compound?

Ans: X and Y form an ionic bond in XY₂.

Que-8: An atom has 2, 8, 7 electrons in its shell. It combines with Y having 1 electron in its outermost shell.

(a) What type of bond will be formed between X and Y?
(b) Write the formula of compound formed.

Ans:
(a) X has 7 electrons in its outermost shell and Y has only one electron in its outermost shell so Y loses its one electron and X gains that electron to form an ionic bond.
(b) The formula of the compound would be XY.

Que-9: Explain with the help of ionic equations Draw electron dot structural diagrams the formation of following electrovalent compound

(i) NaCl (ii) MgCl₂ (iii) CaO.

Ans: (i) Orbit structure and electron dot diagram of NaCl:
Sodium chloride is a compound formed from the ionic bonding of sodium and chloride. Sodium chloride is formed when sodium atoms interact with chlorine atoms. Sodium will donate an electron (which is a negatively-charged particle) to chlorine. This makes sodium slightly positive and chlorine slightly negative. Sodium becomes Na+ and Chlorine becomes Cl- Thus sodium is reduced in this process as it donates the electron and chlorine gain electron and is oxidized.

(ii )Orbit structure and electron dot diagram of MgCl₂:
The formation of magnesium chloride can be thought of as a reaction involving magnesium metal, Mg, and chlorine gas, Cl₂

1. A magnesium atom, which is a metal in Group II A, tends to lose its 2 outer-shell valence electrons to become a magnesium ion, (i.e. cation). The magnesium metal is said to be oxidized.
2. The covalently bonded Cl₂ molecule gains two electrons to become two chloride ions, (i.e. anions). Chlorine is said to be reduced. The two electrons lost by the magnesium is gained by the chlorine atom and thus produce magnesium ion and two chloride ions.
3. The oppositely charged of the magnesium and chloride ions attract each other and ionic bonds are formed. Each cation is surrounded by anions, and each anion is surrounded by cations. The simplest formula for this ionic compound is MgCl₂.

(iii)Electron dot Orbit structure diagram of CaO:
The Calcium ion is an Alkaline earth metal and wants to give up the 2 s orbital elections and become a +2 cation.
Oxygen has six valence electrons and is looking to gain two electrons to complete the octet (8) electron count in the valence shell making it a -2 anion.
When the charges of the Calcium +2 and the Oxygen -2 are equal and opposite, the ions for an electrical attraction. (Remember Paula Abdul told us “Opposites Attract”)
This one to one ratio of charges makes the formula CaO.

Que-10: Compare :

(a) Sodium atom and sodium ion
(b) Chlorine atom and chlorine ion With respect to
(i) Atomic structure
(ii) Electrical state
(iii) Chemical action and
(iv) toxicity

Ans:
(a) Sodium atom and sodium ion
(i) Sodium atom has one electron in M shell while sodium ion has 8 electrons in L shell.
(ii) Sodium atom is neutral while sodium ion is positively charged.
(iii) Sodium atom is highly reactive while its ion is inert.
(iv)Sodium atom is poisonous while sodium ion is non-poisonous.
(b) Chlorine atom and chlorine ion
(i) Chlorine atom has 7 electrons in its M shell while Chloride ion has 8 electrons in the same shell.
(ii) Chlorine atom is neutral while chloride ion is negatively charged.
(iii) Chlorine atom is highly reactive while its ion is inert.
(iv)Chlorine gas is poisonous while chloride ion is non-poisonous.

Que-11: The electronic configuration of fluoride ion is the same as that of a neon atom. What is the difference between two?

Ans: Fluoride ion is negatively charged while neon atom is neutral.

Que-12: State Which of the following is reduction reaction and Which is oxidation reactions .

(i) Pb → Pb²⁺ + 2e^–
(ii) Fe²⁺ – e^– → Fe³⁺
(iii) A³⁺ + e^-1 → A²⁺ 
(iv)  Cu → Cu²⁺ 2e^–

Ans: 
(i) Pb → Pb²⁺ + 2e^–  = Oxidation Reaction
(ii) Fe²⁺ – e^– → Fe³⁺ = Oxidation Reaction
(iii) A³⁺ + e^-1 → A²⁺ = Reduction Reaction
(iv) Cu → Cu²⁺ 2e^– = Oxidation Reaction

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Que-13: What do you understand by redox reactions? Explain oxidation and reduction in terms of loss or gain of electrons.

Ans: Transfer of electron(s) is involved in the formation of an electrovalent bond. The electropositive atom undergoes oxidation, while the electronegative atom undergoes reduction. This is known as a redox process.
Oxidation: In the electronic concept, oxidation is a process in which an atom or ion loses electron(s).
Zn → Zn²⁺ + 2e^–
Reduction: In the electronic concept, the reduction is a process in which an atom or ion accepts electron(s).
Cu²⁺ + 2e^–→ Cu

Que-14: 

Ans: 

Que-15: Divide the following redox reactions into oxidation and reduction half reactions.

(i) Zn + Pb²⁺ → Zn²⁺ + Pb
(ii) Zn + Cu²⁺ → Zn²⁺ + Cu
(iii) Cl₂ + 2Br^– → Br₂ + 2Cl^–
(iv) Sn²⁺ + 2Hg²⁺ → Sn⁴⁺ + Hg²⁺
(v) 2Cu⁺ → Cu + Cu²⁺

Ans:
(i) Zn→ Zn²⁺ + 2e^– (Oxidation)
Pb²⁺ + 2e^–  → Pb (Reduction)
(ii) Zn→ Zn²⁺ + 2e^– (Oxidation)
Cu²⁺ + 2e^–→ Cu (Reduction)
(iii) Cl₂ + 2e^–→ 2Cl^– (Reduction)
2Br^–→ Br₂ + 2e^– (Oxidation)
(iv) Sn²⁺→ Sn⁴⁺ + 2e^– (Oxidation)
2Hg²⁺ + 2e^–→ Hg₂ (Reduction)
(v) Cu⁺→ Cu²⁺ + e^– (Oxidation)
Cu⁺ + e^– → Cu (Reduction)

Que-16: Potassium (Atomic No. 19) and chlorine (Atomic No. 17) react to form a compound. On the basis of electronic concept, explain

(i) oxidation
(ii) reduction
(iii) oxidising agent
(iv) reducing agent

Ans: Potassium (Atomic No. 19) and chlorine (Atomic No. 17) react to form a compound. On the basis of electronic concept, explain
(i) oxidation
(ii) reduction
(iii) oxidising agent
(iv) reducing agent

—:  End of Chemical Bonding Intex-1 Concise Class-10 ICSE Chemistry Selina Solutions Ch-2 : –

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