Chemistry Chapter 1 Short Questions For Class 11 (with Answers)

The most important chemistry chapter 1 short questions for class 11. The 1st-year chapter 1 of chemistry is related to basic concepts. These questions are for the Punjab Textbook Board and can be used within all of Punjab where this syllabus is taught.

Students are advised to prepare these questions in order to perform the best in the board examination.

Chemistry Chapter 1 Short Questions for Class 11

Q.1: What is Dobereiner’s Law of Triads?
Ans: Importance of Moseley:
In 1829, Döbereiner grouped the elements into triads (a group of three) with similar properties, noting that the atomic weight of the middle element was roughly the average of the other two.
Example:
Examples of such triads include lithium, sodium, and potassium (⁷Li, ²³Na, ³⁹K).

Elements	Atomic Mass	Average
Lithium (Li)	6.9
Sodium (Na)	23.0	(6.9 + 39.0) / 2 = 22.95
Potassium (K)	39.0

Q.2: What is Mendeleev’s contribution in the making of periodic table?
Ans: Mendeleev’s contribution:
In 1869, Russian chemist Dmitri Mendeleev, considered the father of the Periodic Table, arranged 63 elements into eight vertical columns by increasing atomic mass, aligning elements with similar properties into vertical groups. The success of his table was hidden in leaving gaps for undiscovered elements and predicting their atomic mass and properties.

Q.3: What is the importance of Moseley in the development of periodic table?
Ans: Importance of Moseley:
In 1913, Moseley determined the exact atomic numbers of known elements using X-ray emission, resolving flaws and discrepancies in Mendeleev’s table by arranging the elements by atomic numbers instead of atomic masses. This significant breakthrough led Moseley to modify the Periodic Law to state that the properties of elements are periodic functions of their atomic numbers.

Q.4: What is modern periodic law?
Ans: Modern periodic law:
In 1911, Moseley noticed that elements could be classified more satisfactorily by using their atomic numbers, rather than their atomic masses.
The modern periodic law states that:
“If the elements are arranged in ascending order of their atomic numbers, their chemical properties repeat in a periodic manner”.

Q.5: Why elements of group 18 of the periodic table are called noble gases?
Ans: Noble Gases:
The noble gases are a group of unreactive elements present at the extreme right of the periodic table in Group 18.
Examples:
Examples include Helium, Neon, Argon, krypton, etc.
Due to their stable electron configuration (complete outermost shell), they are almost entirely unreactive under normal conditions and rarely form compounds with other elements.

Q.6: Why d and f-Block elements are called transition elements?
Ans: d and f block elements:
The d-block and the f-block elements are called transition elements because they are located between the s and p-block elements and their properties are in transition between the metallic elements of s-block and non metallic elements of the p-block.
The transition metals make up the largest family of elements in the middle of periodic table.

Examples:
They include four series of d-block elements, as well as the lanthanides and actinides (f-block elements) found in the two rows below.

Q.7: Why atomic radii decreases along a period and increases down the group?
Ans: Atomic radii:
“The average distance between the nucleus of the atom and its outermost electronic shell is called atomic radius”.
In Groups:
In the periodic table, the atomic radius increases from top to bottom within a group due to the addition of an extra shell of electrons in each period.
In Periods:
In a period, however, the atomic radius decreases. This gradual decrease in the radius is due to increase in the positive charge in the nucleus. As the positive nuclear charge increases, the negatively charged electrons in the shells are pulled closer to the nucleus. Thus, the size of the outermost shell becomes gradually smaller.

Q.8: How lanthanide contraction controls the atomic sizes, of the elements of 6th and 7th periods.
Ans: Lanthanide contraction:
“The gradual reduction in the size of the atoms of 6th and 7th periods is called as lanthanide contraction”.
Moving from left to right in a period of the periodic table, the atomic radius decreases, due to increase in the nuclear charge. This effect is quite remarkable in the elements of longer periods (6th and 7th) in which “d” and “f” subshells are involved.
Due to poor shielding of “d” and “f” orbitals the size of atoms of 6th and 7th periods decrease sharply and it is called as lanthanide contraction.

Q.9: What is shielding effect? How it varies in periodic table?
Ans: Shielding effect:
“The shielding effect is actually the repulsion due to the presence of electrons in between the nucleus and the outermost electronic shell”.
Variation in the periodic table:
In a period, the number of intervening electrons remains same, so shielding effect remain constant in a periodic.
In a group there is successive addition of electronic shells with increase in the atomic number from top to bottom. Hence more and more number of electrons shields the nucleus. Therefore, shielding effect increases in a group.

Q.10: Why the size of a cation is smaller than its parent atom?
Ans: Size of a cation:
Cations are formed, when a neutral atom loses one or more electrons. The size of cation is always smaller than its parent atom due to two reasons.
Na(g) ———> Na⁺(g) + 1e⁻
(187 pm) (95 pm)
Reasons:
(i) The removal of one or more electrons from a neutral atom usually results in the loss of the outer most shell.
(ii) The removal of electrons decreases the repulsion between the electronic clouds and hence the nucleus attracts the remaining electrons with greater force and draws them closer to the nucleus.

Q.11: Why the ionic radii of negative ions are larger than the size of their parent atoms?
Ans: Size of an anion:
When a neutral atom gains one or more electrons, it forms a negative ion. The size of the negative ion is always greater than its parent atom.
F(g) + 1e⁻ ———> F⁻(g)
(72 pm) (136 pm)
Reason:
With the addition of one or more electrons in the shell of a neutral atom, the repulsion between the electron increases. This increased repulsion causes the shell to expand. Thus the ionic radii of negative ion are larger than their parent atom.

Q.12: What are iso-electronic ions?
Ans: Iso-electronic ions:
“Ions having same number of electrons and electronic configuration are called as iso-electronic ions”.
Example:
Na(g) ———> Na⁺¹ + 1e⁻ | Mg(g) ———> Mg⁺² + 2e⁻ | Al(g) ———> Al⁺³ + 3e⁻
(2,8,1) (2,8) | (2,8,2) (2,8) | (2,8,3) (2,8)
Order:
N⁻³ > O⁻² > F⁻¹ > Ne > Na⁺¹ > Mg⁺² > Al⁺³
The size of the iso-electronic ions decreases with increase in the atomic number of the element.

Q.13: What is ionization energy? Write the factors affecting ionization energy.
Ans: Ionization energy:
“The ionization energy of an element is the minimum quantity of energy which is required to remove an electron from the outermost shell of its isolated gaseous atom in its ground state”.
Examples:
Na(g) ———> Na⁺(g) + e⁻ ΔH₁ = + 496 kJmol⁻¹
Factors affecting ionization energy:
The factors upon which the ionization energy of an atom mainly depends are

Magnitude of nuclear charge.
Size of the atom or ion.
Electronic arrangement.
Shielding effect.
Spin-pair repulsion.

Q.14: Why ionization energy decreases down the group and increases along a period?
Ans: Variation in the Group:
Going down in a group of periodic table, the nuclear charge increase but as the size of the atom and the number of electrons causing the shielding effect also increases the nuclear attraction become weak, therefore ionization energy decreases from top to bottom in a group.
Variation across the Period:
By moving from left to right in a period, the outer shell remains the same, while the nuclear charge increases effectively due to increase in atomic number. This increase in nuclear attraction makes the removal of an electron difficult and hence the ionization energy increases across a period.

Q.15: Why ionization energy of inert gases is maximum in every period of the periodic table?
Ans: Ionization energy of Noble gases:
By moving from left to right in a period, the outer shell remains the same, while the nuclear charge increases effectively that makes the removal of an electron difficult and hence the value of ionization energy increases.
Noble gases are present at the extreme right of the table in group VIIIA. Their trend reveals that inert gases have the highest values of ionization energy because due to complete outermost shell in them, the removal of electron is extremely difficult.

Q.16: Why second ionization energy of an element is greater than its first ionization energy?
Ans: First ionization energy:
“The energy needed to remove one electron from each atom in one mole of atoms of the element in the gaseous state to form one mole of gaseous 1+ ions is known as 1st ionization energy (ΔH1).”
Ca(g) ———> Ca⁺(g) + e⁻ ΔH₁ = + 590 kJ mol⁻¹
Second Ionization energy:
“If a second electron is removed from each ion in a mole of gaseous 1+ ions, we call it the 2nd ionization energy, ΔH₂.”
Ca⁺(g) ———> Ca⁺²(g) + e⁻ ΔH₂ = + 1150 kJ mol⁻¹
The second I.E of an element is always greater than its 1st I.E. It is because, with the loss of one electron the nuclear attraction on the remaining electrons increases hence more amount of energy is required to remove the second electron from a uni-positive ion than a neutral atom.

Q.17: What is first electron affinity?
Ans: 1st Electron affinity, ΔH°ea1
“The first electron affinity is the enthalpy change involved when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous uni-negative ions under standard conditions.”
Example:
Electron affinity of chlorine atom.
Cl(g) + e⁻ ———> Cl⁻(g) ΔH°ea1 = -348.8 kJ mol⁻¹
This is the amount of energy released when 6.02 x 10²³ atoms of chlorine in the gaseous state are converted into Cl⁻ ions. Since, energy is released, so first electron affinity carries negative sign.

Q.18: Discuss the factors affecting electron affinity?
Ans: Factors affecting electron affinity:
Important factors affecting the magnitude of electron affinity values of elements are as follows:
(i) Size of atom:
For small sized atoms the attraction of the nucleus for the incoming electron is stronger. Thus, smaller is the size of the atom, greater is its electron affinity.
(ii) Nuclear charge:
Greater the magnitude of nuclear charge of an element stronger is the attraction of its nucleus for the incoming electron. Thus, with the increase in the magnitude of nuclear charge, electron affinity also increases.
(iii) Electronic configuration of atom:
The electron affinity is low when the electron is added to a half filled sub-shell than that for partially filled one. The second value of electron affinity of an element is usually shown with a positive sign?

Q.19: Why the second electron affinity, ΔH°ea2 “The second electron affinity, is the amount of energy required to add electrons to 1 mole of uni-negative gaseous ions to form 1 mole of gaseous 2- ions under standard conditions.”
Example:
When first electron is added to a neutral oxygen atom, 141 kJ mol⁻¹ energy is released.
O(g) + e⁻ ———> O⁻(g) ΔH°ea1 = -141 kJ mol⁻¹
But, when a second electron is added to a uni-negative ion, the incoming electron is repelled by the already present negative charge. So, 798 kJ mol⁻¹ of energy is absorbed on adding second electron to a uni-negative ion (O⁻) ion.
O⁻(g) + e⁻ ———> O⁻²(g) ΔH°ea2 = +798 kJ mol⁻¹

Q.20: Why group 15 (VA) elements show low electron affinity?
Ans: Electron affinity values of ‘N’ and ‘p’ group-15:
Electron affinity values of ‘N’ and ‘p’ group-15 (V-A), atoms are very low. This is because of the presence of half-filled ‘np’ orbitals in their valence shell.
N = 1s² 2s² 2px¹ 2py¹ 2pz¹ | ΔH°ea1 = -7 kJ mol⁻¹
1s² 2s² 2p⁶ 3s² 3px¹ 3py¹ 3pz¹ | ΔH°ea1 = -71.7 kJ mol⁻¹
These half-filled p-subshells, being very stable, have very little tendency to accept any extra electron to be added to them.

Q.21: Why the electron affinity of fluorine is less than chlorine?
Ans: Electron affinity of F is less than Cl:
The electron affinity of the elements increases left or right in a period and decreases down the group. But exceptionally fluorine has electron affinity less than that of chlorine.
Reason:
Fluorine has very small size and seven electrons in 2s and 2p sub shells have thick electronic cloud. This thick cloud repels the incoming electron. So initially energy is absorbed to add the electron and then energy is released. Therefore, the E.A of fluorine is less than that of Chlorine.
Example:
F(g) + 1e⁻ ———> F⁻(g) E.A = -328 kJ/mole
Cl(g) + 1e⁻ ———> Cl⁻(g) E.A = -349 kJ/mole

Q.22: Define electronegativity. How it varies in periods and groups?
Ans: Electronegativity:
“The power of an atom to attract shared electron pair towards itself is called its electronegativity.”
Variation of electronegativities in periodic table:
A comparison of electronegativities shows that the values increase in a period with the decrease in atomic size and increase in the nuclear charge.
While these values decrease in a group due to the increase in the size of the atoms and shielding effect of the electronic shells.

Q.23: How the criteria of electronegativity help us to understand the nature of a bond? / How the difference of electronegativity decides the nature of a chemical bond?
Ans: Nature of Chemical Bond:
The difference in the electronegativity value of the bonded atoms determines the nature of the bond.
(1) If the difference is (0 – 0.4), the bond between the two atoms is non-polar. e.g., H₂, Cl₂, Br₂ etc.
(2) If the difference is (0.5 – 1.6), the bond between the atoms is polar covalent e.g., HCl, HBr etc.
(3) A difference of (1.7) units shows equal contribution of ionic and covalent bond e.g., AlCl₃.
(4) If the difference is greater than 1.7 units, the bond will be ionic in nature e.g., NaCl KBr etc.

Q.24: What is electro-positivity or metallic character? How it varies in the periodic table?
Ans: Electropositivity or Metallic character:
“The tendency of an atom to lose electron and form positive ions is called metallic character or electropositivity”.
Variation in periodic table:
As it becomes easier to remove the electron of an atom bigger in size, therefore metallic character increases from top to bottom in a given group of elements. On the contrary, it decreases from left to right across a period.

Q.25: Oxides of metals are basic in nature. Justify?
Ans: Basic oxides:
The oxides of strong electropositive metals of group 1 and group 2 are basic oxides.
They produce strong alkalis when combine with water.
Na₂O(s) + H₂O(l) ———> 2NaOH(aq)
They react with acids to give neutralization reaction.
Na₂O(s) + H₂SO₄(l) ———> 2NaSO₄(aq) + H₂O(l)

Q.26: Write mechanism of the reaction of metal oxide with water?
Ans: Mechanism:
The oxides of alkali and alkaline earth metals produce O⁻² ions in water. The O⁻² ions have high affinity for proton (H⁺) and cannot exist alone in an aqueous solution. Therefore, it immediately takes proton from water and forms OH⁻¹ ion and hence the solution become alkaline in nature.
Na₂O(s) ———> 2Na⁺¹(aq) + O⁻²(aq)
O⁻²(aq) + H₂O(l) ———> 2OH⁻¹(aq)

Q.27: Justify that Al₂O₃ is amphoteric in nature? / Define amphoteric oxides? Give one example.
Ans: Amphoteric oxides:
Amphoteric oxides are oxides that can react with both acids and bases. This means they have the ability to behave as either an acid or a base, depending on the conditions.
Examples:
Aluminum oxide (Al₂O₃) is an amphoteric oxide. It is insoluble in water but reacts with hydrochloric acid to form aluminium chloride and water, and with sodium hydroxide to form sodium aluminate and water.
Al₂O₃(s) + 6HCl(aq) ———> 2AlCl₃(s) + 3H₂O(l) (Aluminium chloride)
Al₂O₃(s) + 2NaOH(aq) ———> Na₂Al₂O₄(s) + H₂O(l) (Sodium aluminate)

Q.28: What are acidic oxides?
Ans: Acidic oxides:
“An acidic oxide is an oxide that when combined with water gives off an acid.”
Non-metals react with oxygen to form acidic oxides which are held together by covalent bonds.
Examples:
Examples of acidic oxides in period 3 are, P₂O₃, P₂O₅, SO₃, SO₂, etc.
P₂O₃(s) + H₂O(l) ———> H₃PO₃(aq) (Phosphorous acid)
Reactions of theses oxides with bases are given below,
P₂O₃(s) + 6NaOH(aq) ———> 2Na₃PO₃(aq) + 3H₂O(l) (Sodium phosphite)

Q.29: Although both sodium and phosphorus are present in the same period of the periodic table, yet their oxides are different in nature, Na₂O is basic while P₂O₅ is acidic in character.
Ans: Na₂O and P₂O₅:
The acidic and basic character of an oxide depend upon the metallic and non-metallic character of the element. Generally oxides of metals are basic, while that of non-metals are acidic in nature.
Sodium is a metal belonging to group 1, hence its oxide is basic in nature.
Na₂O(s) + H₂O(l) ———> 2NaOH(aq)
While phosphorous is a non-metal of group 15, hence oxide of phosphorous (P₂O₅) is acidic in nature.
P₂O₅(s) + 3H₂O(l) ———> 2H₃PO₄(aq)

Q.30: What are Neutral chlorides? / Justify that chlorides of group 1 and 2 are neutral in nature?
Ans: Neutral Chlorides:
Neutral chlorides are salts that, when dissolved in water, produces a neutral solution with a pH close to 7.
At the start of period 3, sodium and magnesium chlorides do not react with water. Their aqueous solutions contain the positive metal ions and negative chloride ions surrounded by water molecules.
For example,
NaCl(s) ———> Na⁺(aq) + Cl⁻(aq) (pH = 7)
MgCl₂(s) ———> Mg⁺²(aq) + 2Cl⁻(aq) (pH = 6.5)
NOTE: Group 1 and group 2 chlorides are also neutral with few exceptions.

Q.31: What are acidic chlorides? / Prove that chloride of aluminium is acidic in nature?
Ans: Acidic Chlorides:
If we move in period 3, from aluminium to sulphur all chlorides react with water to make acidic solution with pH less than 7 this process is called hydrolysis.
Explanation:
When AlCl₃ is added to water, it ionize to give aluminium and chloride ions in solution. Al³⁺ ion is hydrated and causes a water molecule to lose an H⁺ ion, this process is hydrolysis.
The following reaction occurs:
AlCl₃(s) + 3H₂O(l) ———> Al(OH)₃(s) + 3HCl(aq)
(Aluminium hydroxide) (Hydrochloric acid)