Chemistry MCQs for Class 11 Chapter – 3 Classification of Elements and Periodicity in properties

Q.1. Periodic classification of elements is used to examine the
(a) periodic trends in physical properties of elements
(b) periodic trends in chemical properties of elements
(c) Both (a) and (b)
(d) None of the above

Answer

(c) Periodic classification of elements follow a logical consequence of electronic configuration of atoms which is used to examine the physical and chemical properties of the elements.

Q.2. Increasing order of electro negativity is
(a) Bi < P < S < Cl
(b) P < Bi < S < Cl
(c) S < Bi < P < Cl
(d) Cl < S < Bi < P

Answer

(a) Bi < P < S < Cl
Explanation:
Increasing order of electro negativity is Bi < P < S < Cl.

Q.3. How many elements are there in 6th period of periodic table?
(a) 18
(b) 8
(c) 30
(d) 32

Answer

(d) 6th period consists of 32 elements.

Q.4. Which of the following pairs has both members from the same period of the periodic table.
(a) Na – Ca
(b) Na – Cl
(c) Ca – Cl
(d) Cl – Br

Answer

(b) Na and Cl both belongs to III period.

Q.5. The chemistry of lithium is very similar to that of magnesium even though they are placed in different groups. Its reason is:
(a) Both are found together in nature
(b) Both have nearly the same size
(c) Both have similar electronic configuration
(d) The ratio of their charge and size (i.e. charge density) is nearly the sameAnswer

Answer

(d) The ratio of their charge and size (i.e. charge density) is nearly the same<br. The chemistry of lithium is very similar to that of magnesium even though they are placed in different groups because of diagonal relationship.

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Q.6. The electronic configuration of halogen is
(a) ns² np⁶
(b) ns² np³
(c) ns² np⁵
(d) ns²

Answer

Answer: (c) ns² np⁵
Explanation:
Halogens has 7 electrons in his valance shell (ns²np⁵).

Q.7. Which of the following is non-metallic ?
(a) B
(b) Be
(c) Mg
(d) Al

Answer

(a) Metallic character decreases down group and increases along a period.

Q.8. The order of increasing sizes of atomic radii among the elements O, S, Se and As is :
(a) As < S < O < Se
(b) Se < S < As < O
(c) O < S < As < Se
(d) O < S < Se < As

Answer

(c) On moving down in a group atomic radii increases due to successive addition of extra shell hence
O < S < Se Further As is in group 15 having one less electron in its p orbital hence have higher atomic radii than group 16 elements.
i.e., O < S < Se < As

Q.9. The element with atomic number 35 belongs to
(a) d – Block
(b) f – Block
(c) p – Block
(d) s – Block

Answer

(c) p – Block
The electronic configuration of element with atomic number 35 is [Ar]3d¹⁰4s² 4p⁵. The valence electron belongs to p block. Therefore, it is a p-block element.

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Q.10. Why the size of an anion is larger than the parent atom?
(a) Due to increased repulsion among the electrons.
(b) Due to decrease in effective nuclear charge.
(c) Due to increased in effective nuclear charge.
(d) Both (a) and (b)

Answer

(d) The size of an anion will be larger than that of the parent atom because the addition of one or more electron(s) would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Q.11. The correct order of first ionization potential among following elements, Be, B, C, N and O is
(a) B < Be < C < O < N
(b) B < Be < C < N < O
(c) Be < B < C < N < O
(d) Be < B < C < O < N

Answer

(a) B < Be < C < O < N
The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionisation energy. The ionisation potential decreases as the size of the atom decreases. Atoms with fully or partly filled orbitals have high ionisation potential.

Q.12. Which of the following properties generally decreases along a period?
(a) Ionization Energy
(b) Metallic Character
(c) Electron Affinity
(d) Valency.

Answer

(b) Metallic Character
The IE, EA increases along the period. The valency initially increases then decreases. The metallic character decreases along the period.

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Q.13. The electron affinity for the inert gases is –
(a) zero
(b) high
(c) negative
(d) positive

Answer

(b) O < S < F < Cl
Electron gain enthalpy –141, –200, – 333, – 349 kJ mol^–1

Q.14. On the Pauling’s electronegativity scale the element next to F is
(a) N
(b) Cl
(c) O
(d) Ne.

Answer

(c) O
Explanation:
Pauling explained electronegativity as the power of an atom in a molecule to attract electrons towards it. When we analyse the trend in periodic table, we can see that the degree of electronegativity decreases while going down the groups, while it increases across the periods. In the case of a covalent bond, based on the bond energies, Pauling calculated the differences in electronegativity between atoms in the bond and assigned a value of 4 to fluorine, which is the most electronegative element, and others were calculated with respect to that value. Hence, on paulings scale, the element next to fluorine is Oxygen.

Q.15. Representative elements are those which belong to
(a) p and d – Block
(b) s and d – Block
(c) s and p – Block
(d) s and f – BlockAnswer

Answer

(c) s and p – Block
Explanation: Elements in which all the inner shells are complete but outer shell is incomplete is known as representative elements i.e. Those elements which have less than 8 electrons in outermost shell are representative. s and p block elements except inert gas is known as representative elements.

Q.16. The group number, number of valence electrons, and valency of an element with the atomic number 15, respectively, are:
(a) 16, 5 and 2
(b) 15, 5 and 3
(c) 16, 6 and 3
(d) 15, 6 and 2Answer

Answer

(b) 15, 5 and 3
Explanation:
Atomic number (Z) =15 =P → [Ne] 3s² 3p³ Phosphorus belongs to 15th group Number of valence electrons 3s²3p³ = 5 and valency = 3 in ground state.

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Q.17. Diagonal relationship is shown by
(a) All elements with their diagonally downward elements towards right
(b) Most of the elements of second period
(c) All 3d series elements
(d) None of the above

Answer

(d)

Q.18. Which of the following oxides is amphoteric in character?
(a) SnO₂
(b) CO₂
(c) SiO₂
(d) CaO

Answer

Answer: (a) SnO₂
Explanation:
CaO is basic; CO₂ is acidic; SiO₂ is weakly acidic. SnO₂ is amphoteric.

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Q.19. In the modern periodic table, the period indicates the value of:
(a) Atomic Number
(b) Atomic Mass
(c) Principal Quantum Number
(d) Azimuthal Quantum Number

Answer

(c) Principal Quantum Number

The periodic table is a tabular arrangement of the chemical elements, organized on the basis of their atomic number (number of protons in the nucleus), electron configurations, and recurring chemical properties. The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Q.20. Which of the following sequence correctly represents the decreasing acidic nature of oxides ?
(a) Li2O > BeO > B2O3 > CO2 > N2O3
(b) N2O3 > CO2 > B2O3 > BeO > Li2O
(c) CO2 > N2O3 > B2O3 > BeO > Li2O
(d) B2O3 > CO2 > N2O3 > Li2O > BeO

Answer

(b) On passing from left to right in a period acidic character of the normal oxides of the elements increases with increase in electronegativity.

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Q.21. Arrange S, O and Se in ascending order of electron affinity
(a) Se < S < O
(b) Se < O < S
(c) S < O < Se
(d) S < Se < O

Answer

(a) Se < S < O
Explanation:
Correct order of electron affinity is Se < S < O. In a group electron affinity decreases with increase in atomic number.

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Q.22. which of the following elements are found in pitch blende?
(a) Actinium and protoactinium
(b) Neptunium and plutonium
(c) Actinium only
(d) Both (a) and (b)

Answer

(d) Neptunium and plutonium like actinium and protoactinium are also found in pitch.

Q.23. In the modern periodic table , the period indicates the value of:
(a) Atomic Number
(b) Atomic Mass
(c) Principal Quantum Number
(d) Azimuthal Quantum Number

Answer

(c) Principal Quantum Number
Explanation:
The periodic table is a tabular arrangement of the chemical elements, organized on the basis of their atomic number (number of protons in the nucleus), electron configurations, and recurring chemical properties.
The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

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Q.24. The electronic configuration of an element is 1s², 2s² 2p⁶, 3s² 3p³. What is the atomic number of the element which is just below the above element in the periodic table
(a) 31
(b) 34
(c) 33
(d) 49

Answer

Answer: (c) 33
Explanation:
33−1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p³

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Q.25. The reduction in atomic size with increase in atomic number is a characteristic of elements of-
(a) d−block
(b) f−block
(c) Radioactive series
(d) High atomic masses

Answer

(b) f−block
Explanation:
The reduction in atomic size with increase in atomic number is a characteristic of elements of f- block. It is known as lanthanoid contraction and actinoid contraction. This is due to poor shielding of electrons present in f subshell.

Q.26. The number of elements in the 5th period of the periodic table is
(a) 3
(b) 9
(c) 8
(d) 18

Answer

(d) 18
Explanation:
While filling 5th shell according to Aufbau principle 5s, 5p, 4d filled so 2 + 6 + 10 = 18 electrons or elements are present in 5th shell. Further we start filling the 4d orbital which can take 10 electrons. So in the 4th principal quantum energy states, we can fill 18 electrons. Thus 5th period has 18 elements.

Q.27. Which of the following forms the most stable gaseous negative ion?
(a) F
(b) Cl
(c) Br
(d) I

Answer

(b) Cl
Explanation:
The element which forms the most stable gaseous negative ion is fluorine.

Q.28. On the Pauling’s electro negativity scale the element next to F is
(a) N
(b) Cl
(c) O
(d) Ne

Answer

(c) O
Explanation:
Pauling explained electro negativity as the power of an atom in a molecule to attract electrons towards it. When we analyse the trend in periodic table, we can see that the degree of electro negativity decreases while going down the groups, while it increases across the periods. In the case of a covalent bond, based on the bond energies, Pauling calculated the differences in electro negativity between atoms in the bond and assigned a value of 4 to fluorine, which is the most electro negative element, and others were calculated with respect to that value. Hence, on paulings scale, the element next to fluorine is Oxygen.

Q.29. The element californium belongs to a family of :
(a) Alkali metal family
(b) Actinide series
(c) Alkaline earth family
(d) Lanthanide series

Answer

Answer: (b) Actinide series
Explanation:
Atomic number of californium is 98 and its electronic configuration is
Rn⁸⁶ 7s² 5f¹⁰
so it is a f-block element and as it is in 7th period, it is a part of actinide series.

Q.30. In the long form of the periodic table, the valence shell electronic configuration of 5s²5p⁴ corresponds to the element present in:
(a) Group 16 and period 6
(b) Group 17 and period 6
(c) Group 16 and period 5
(d) Group 17 and period 5

Answer

Answer: (c) Group 16 and period 5
Explanation:
Tellurium (Te) has 5s²5p⁴ valence shell configuration. It belongs to group 16 and present in period 5 of the periodic table.