Chemistry Chapter 3: Full 50 JEE/NEET MCQ

Chemistry Chapter 3: Classification of Elements (50 Full MCQs)

1. The IUPAC name of element with atomic number 119 is:

A. Ununennium

B. Unbinilium

C. Ununnilium

D. Ununseptium

Answer: A

Solution: For 119: 1=un, 1=un, 9=enn. Name: Ununennium. Symbol: Uue.

2. Maximum number of elements in the 6th period is:

A. 18

B. 32

C. 8

D. 50

Answer: B

Solution: 6th period includes 6s, 4f, 5d, and 6p orbitals (2+14+10+6 = 32 elements).

3. Which block contains metals, non-metals, and metalloids?

A. s-block

B. p-block

C. d-block

D. f-block

Answer: B

Solution: The p-block is the only block containing all three types of elements.

4. The element with Z=114 belongs to:

A. Group 14, Period 7

B. Group 16, Period 7

C. Group 14, Period 6

D. Group 13, Period 7

Answer: A

Solution: It follows Lead (Z=82) in Group 14 (Carbon family).

5. General configuration of f-block elements is:

A. (n-2)f^1-14(n-1)d^0-1ns²

B. (n-1)f^1-14nd^0-1ns²

C. (n-2)f^0-14(n-1)d¹⁰ns²

D. nf^1-14(n-1)d¹ns²

Answer: A

Solution: Electrons fill the (n-2)f subshell.

6. Total number of elements in s-block is:

A. 12

B. 14

C. 10

D. 13

Answer: B

Solution: 2 from Group 1, 2 from Group 2 across 7 periods, but only 14 total (including H and He).

7. Screening effect order for the same shell is:

A. s > p > d > f

B. f > d > p > s

C. s = p = d = f

D. p > s > d > f

Answer: A

Solution: s-orbitals are closer to the nucleus and provide better shielding.

8. Element with Z=33 belongs to:

A. Period 3

B. Period 4

C. Group 13

D. Group 17

Answer: B

Solution: Configuration: [Ar] 3d¹⁰ 4s² 4p³. Highest n=4, so Period 4.

9. Which group contains only metals?

A. Group 1

B. Group 2

C. Group 17

D. Group 18

Answer: B

Solution: Group 1 contains Hydrogen (non-metal), while Group 2 contains only metals.

10. First noble gas is:

A. Neon

B. Helium

C. Argon

D. Radon

Answer: B

Solution: Helium (Z=2) is the first noble gas in Period 1.

11. Largest ionic radius among isoelectronic species O^2-, F^-, Na⁺, Mg²⁺ is:

A. Mg²⁺

B. Na⁺

C. F^-

D. O^2-

Answer: D

Solution: Size increases as nuclear charge (Z) decreases in isoelectronic species. O^2- has Z=8.

12. Atomic radius of Gallium (Ga) is less than Aluminum (Al) due to:

A. Lanthanide contraction

B. Poor shielding by 3d electrons

C. Inert pair effect

D. Large size

Answer: B

Solution: 3d electrons shield poorly, increasing Z_eff and shrinking the atom.

13. Across a period, atomic size decreases due to:

A. Decrease in Z_eff

B. Increase in Z_eff

C. Increase in shielding

D. Addition of shells

Answer: B

Solution: Effective nuclear charge increases while electrons are added to the same shell.

14. Which is the correct size order?

A. I^- > I > I⁺

B. I⁺ > I > I^-

C. I > I^- > I⁺

D. I^- > I⁺ > I

Answer: A

Solution: Anion > Parent Atom > Cation.

15. Lanthanide contraction is caused by:

A. 4f electrons

B. Poor shielding of 4f electrons

C. 5d electrons

D. 6s electrons

Answer: B

Solution: Poor shielding of 4f electrons causes the nucleus to pull 6s electrons more strongly.

16. van der Waals radius is ______ than covalent radius.

A. Smaller

B. Larger

C. Equal

D. Half

Answer: B

Solution: van der Waals radius accounts for non-bonding distance between atoms.

17. Which gas has the largest atomic radius in Period 2?

A. Lithium

B. Fluorine

C. Neon

D. Oxygen

Answer: C

Solution: Noble gases are measured by van der Waals radius, which is much larger than covalent radii.

18. Radius of Fe²⁺ is ______ than Fe³⁺.

A. Larger

B. Smaller

C. Same

D. Triple

Answer: A

Solution: Higher positive charge leads to greater contraction of the electron cloud.

19. Atomic radius increases down a group due to:

A. Increase in Z_eff

B. Addition of new shell

C. Decrease in Z

D. Shielding effect

Answer: B

Solution: Adding a principal energy level outweighs the increase in nuclear charge.

20. Zr and Hf have similar radii due to:

A. Being in same group

B. Lanthanide contraction

C. Same Z

D. Shielding

Answer: B

Solution: Contraction in the 5d series (Hf) cancels out the expected size increase from Zr.

21. 1st Ionization Enthalpy is highest for:

A. B

B. C

C. N

D. O

Answer: C

Solution: Nitrogen has a stable half-filled 2p³ configuration.

22. Element with lowest 1st I.E. is:

A. Li

B. Cs

C. Fr

D. He

Answer: B

Solution: Cesium has the largest size and lowest effective nuclear pull on valence electrons.

23. Successive I.E. values are always:

A. I.E.1 > I.E.2

B. I.E.2 > I.E.1

C. I.E.1 = I.E.2

D. Random

Answer: B

Solution: Removing an electron from a positive ion is always harder due to increased Z/e ratio.

24. Be has higher I.E. than B due to:

A. Penetration effect of 2s

B. Size

C. Nuclear charge

D. Shielding

Answer: A

Solution: 2s electrons of Be are more penetrated towards the nucleus than 2p of B.

25. Noble gases have very high I.E. because:

A. Stable octet

B. Small size

C. High Z_eff

D. All of these

Answer: D

Solution: Full shells and high Z_eff make electron removal extremely difficult.

26. Order of 1st I.E. in Group 13:

A. B > Al > Ga > In > Tl

B. B > Tl > Ga > Al > In

C. B > Ga > Al > Tl > In

D. Al > B > Ga

Answer: B

Solution: Irregular trend due to d-block and f-block contractions.

27. Which ion is isoelectronic with Neon?

A. O^-

B. Mg²⁺

C. Cl^-

D. K⁺

Answer: B

Solution: Mg²⁺ has 10 electrons, same as Ne.

28. I.E. decreases down the group due to:

A. Size increase

B. Shielding increase

C. Both A and B

D. Z increase

Answer: C

Solution: Larger distance and more inner shells reduce the nuclear pull.

29. Highest second I.E. is of:

A. Na

B. Mg

C. Al

D. Si

Answer: A

Solution: Removing 2nd electron from Na⁺ involves breaking a stable noble gas core (2p⁶).

30. Metallic character is highest for:

A. s-block

B. p-block

C. d-block

D. f-block

Answer: A

Solution: Low I.E. allows s-block elements to lose electrons easily.

31. Most negative electron gain enthalpy is of:

A. F

B. Cl

C. Br

D. I

Answer: B

Solution: F is too small, causing interelectronic repulsion for the incoming electron.

32. Most electronegative element is:

A. O

B. F

C. Cl

D. N

Answer: B

Solution: Fluorine has the highest tendency to attract shared electrons (Pauling scale = 4.0).

33. Noble gases have ______ electron gain enthalpy.

A. Highly negative

B. Positive

C. Zero

D. Negative

Answer: B

Solution: Energy must be supplied to force an electron into a higher shell of a stable atom.

34. Electronegativity increases across a period because:

A. Size increases

B. Z_eff increases

C. Shielding increases

D. I.E. decreases

Answer: B

Solution: Stronger nuclear pull attracts bonding electrons more effectively.

35. Order of electronegativity:

A. F > O > N

B. N > O > F

C. O > F > N

D. F > N > O

Answer: A

Solution: Values are roughly F(4.0) > O(3.5) > N(3.0).

36. Pauling scale is based on:

A. Bond energies

B. I.E. and E.A.

C. Atomic size

D. Metallic character

Answer: A

Solution: It relates electronegativity to the extra stability of polar bonds.

37. Element with highest electron affinity:

A. Cl

B. F

C. O

D. S

Answer: A

Solution: Chlorine releases the most energy when accepting an electron.

38. Across a period, non-metallic character:

A. Increases

B. Decreases

C. Constant

D. First increases

Answer: A

Solution: Higher I.E. and EN favor non-metallic behavior.

39. Which is most electropositive?

A. Li

B. Na

C. K

D. Cs

Answer: D

Solution: Cesium loses its valence electron most easily.

40. Nature of oxides changes from left to right as:

A. Acidic to Basic

B. Basic to Acidic

C. Basic to Neutral

D. Neutral to Basic

Answer: B

Solution: Metals form basic oxides; non-metals form acidic oxides.

41. Diagonal relationship is between:

A. Li-Mg

B. Be-Al

C. B-Si

D. All of these

Answer: D

Solution: Diagonal elements share similar charge/radius ratios.

42. Amphoteric oxide is:

A. Na₂O

B. Al₂O₃

C. CO₂

D. SO₂

Answer: B

Solution: Aluminum oxide reacts with both acids and bases.

43. Strongest reducing agent in aqueous solution is:

A. Li

B. Na

C. K

D. Cs

Answer: A

Solution: Lithium has the highest hydration enthalpy due to small size.

44. Strongest oxidizing agent is:

A. F₂

B. Cl₂

C. I₂

D. O₂

Answer: A

Solution: Fluorine has the highest reduction potential.

45. Liquid non-metal is:

A. Hg

B. Br₂

C. Ga

D. S

Answer: B

Solution: Bromine is the only non-metal that is liquid at STP.

46. Most basic hydroxide:

A. NaOH

B. KOH

C. CsOH

D. LiOH

Answer: C

Solution: Basic strength of Group 1 hydroxides increases down the group.

47. Inert pair effect is most prominent in:

A. C

B. Si

C. Ge

D. Pb

Answer: D

Solution: s-electrons of the valence shell become reluctant to bond in heavier p-block elements.

48. Typical elements are:

A. Period 2

B. Period 3

C. Period 4

D. Period 1

Answer: B

Solution: Period 3 elements represent the properties of their respective groups.

49. Nature of Cl₂O₇:

A. Strongly acidic

B. Strongly basic

C. Amphoteric

D. Neutral

Answer: A

Solution: It reacts with water to form Perchloric acid (HClO₄), the strongest mineral acid.

50. Triad system was given by:

A. Mendeleev

B. Dobereiner

C. Newlands

D. Moseley

Answer: B

Solution: Johann Dobereiner arranged elements in groups of three with similar properties.

Screening constant (Sigma) calculation using Slater's rule for 4s electron.

1. Screening constant (σ) calculation using Slater's rule for 4s electron +

Correct Answer — General Method

For a 4s electron, calculate σ by Slater's rules according to the electron configuration of the atom.

Slater's Rules for a 4s Electron

For an electron in the ns or np orbital, such as 4s, Slater's rules are:

σ = 0.35 × (other electrons in the same 4s,4p group) + 0.85 × (electrons in the n−1 shell) + 1.00 × (electrons in n−2 or lower shells)

The electron being considered is not counted in calculating σ.

For a 4s electron:

• Other electrons in 4s, 4p → 0.35 each
• Electrons in 3s, 3p, 3d → 0.85 each
• Electrons in 1s, 2s, 2p → 1.00 each

Example: Calculate σ for a 4s Electron of Potassium (K, Z = 19)

Electronic configuration of K:

K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

We want the shielding constant for the 4s¹ electron.

Step 1: Same group

There are no other electrons in the 4s/4p group.

0 × 0.35 = 0

Step 2: n−1 shell

The n−1 shell is the third shell: 3s² + 3p⁶ = 8 electrons

8 × 0.85 = 6.80

Step 3: n−2 and lower shells

1s² + 2s² + 2p⁶ = 10 electrons

10 × 1.00 = 10

Step 4: Total screening constant

σ = 0 + 6.80 + 10
σ = 16.80

Therefore:

For K, σ(4s) = 16.80

Effective Nuclear Charge

After calculating σ, effective nuclear charge is:

Z_eff = Z − σ

For K:

Z_eff = 19 − 16.80
Z_eff = 2.20

Thus, the 4s electron of potassium experiences an effective nuclear charge of approximately +2.20.

Important Exam Point

Do not use one fixed σ value for every 4s electron. The screening constant depends on the atomic number and electronic configuration of the atom/ion being considered. For example, for Ca:

Ca: [Ar]4s²

For one 4s electron, the other 4s electron contributes:

1 × 0.35

while the eight 3s/3p electrons contribute:

8 × 0.85

and the ten electrons in n−2 and lower shells contribute:

10 × 1.00

Hence:

σ = 0.35 + 6.80 + 10
σ = 17.15

JEE/NEET Shortcut

For a 4s/4p electron, remember:

Same shell/group → 0.35
One shell below → 0.85
Two or more shells below → 1.00

Then use:

Z_eff = Z − σ

30 Related MCQs with Solutions

1. According to Slater's rule, the contribution of another electron in the same ns/np group is:

A. 0.10

B. 0.35

C. 0.85

D. 1.00

Answer: B. 0.35

2. For a 4s electron, electrons in the 3s and 3p orbitals contribute:

A. 0.35 each

B. 0.50 each

C. 0.85 each

D. 1.00 each

Answer: C. 0.85 each

3. For a 4s electron, electrons in 1s, 2s and 2p contribute:

A. 0.35 each

B. 0.50 each

C. 0.85 each

D. 1.00 each

Answer: D. 1.00 each

4. Which electron is excluded while calculating its own screening?

A. All inner electrons

B. The electron under consideration

C. All 1s electrons

D. All d electrons

Answer: B

5. The electronic configuration of K is:

A. [Ar]4s¹

B. [Ne]3s²

C. [Ar]3d¹

D. [Kr]5s¹

Answer: A. [Ar]4s¹

6. The screening constant for the 4s electron of K is:

A. 10.00

B. 16.80

C. 18.15

D. 19.00

Answer: B. 16.80

7. For K, the 3s²3p⁶ electrons contribute:

A. 6.80

B. 8.00

C. 10.00

D. 2.80

Answer: A. 6.80

8. For K, the electrons in 1s²2s²2p⁶ contribute:

A. 3.50

B. 6.80

C. 8.50

D. 10.00

Answer: D. 10.00

9. Effective nuclear charge is calculated using:

A. Z + σ

B. Z − σ

C. σ − Z

D. Z × σ

Answer: B. Z − σ

10. The Z_eff of the 4s electron in K is:

A. 1.20

B. 2.20

C. 3.20

D. 4.20

Answer: B. 2.20

11. Slater's rules are used to estimate:

A. Atomic mass

B. Effective nuclear charge

C. Neutron number

D. Melting point

Answer: B

12. The shielding effect causes the outer electron to experience:

A. Full nuclear charge

B. Reduced effective nuclear attraction

C. No nuclear attraction

D. Negative nuclear charge

Answer: B

13. For a 4s electron, which orbital belongs to the same principal shell?

A. 3p

B. 3d

C. 4p

D. 2p

Answer: C. 4p

14. For an ns/np electron, the n−1 shell contributes:

A. 0.35 per electron

B. 0.50 per electron

C. 0.85 per electron

D. 1.50 per electron

Answer: C

15. For a 4s electron, the n−2 shell corresponds to:

A. n=3

B. n=2

C. n=1

D. n=5

Answer: B. n=2

16. For K, the number of electrons contributing 1.00 each is:

A. 2

B. 8

C. 10

D. 18

Answer: C. 10

17. For Ca ([Ar]4s²), another 4s electron contributes:

A. 0.35

B. 0.85

C. 1.00

D. 2.00

Answer: A. 0.35

18. The screening constant for one 4s electron of Ca is:

A. 16.80

B. 17.15

C. 18.00

D. 20.00

Answer: B. 17.15

19. For Ca, the approximate Z_eff of a 4s electron is:

A. 1.85

B. 2.85

C. 3.85

D. 4.85

Answer: B. 2.85

20. Which quantity increases when σ decreases for a fixed Z?

A. Z_eff

B. Atomic mass

C. Neutron number

D. Principal quantum number

Answer: A. Z_eff

21. Greater shielding generally causes:

A. Greater effective nuclear attraction

B. Lower effective nuclear attraction

C. No change in attraction

D. Nuclear decay

Answer: B

22. Which electrons shield a 4s electron most effectively according to Slater's rules?

A. Same 4s/4p group: 0.35 each

B. 3s/3p: 0.85 each

C. 2s/2p: 1.00 each

D. All have equal contribution

Answer: C. 2s/2p electrons contribute 1.00 each

23. Which factor is NOT directly used in calculating σ by Slater's rule?

A. Number of electrons

B. Electron grouping

C. Principal shell position

D. Atomic mass

Answer: D. Atomic mass

24. If σ increases while Z remains constant, Z_eff:

A. Increases

B. Decreases

C. Becomes infinite

D. Becomes negative necessarily

Answer: B. Decreases

25. The 4s electron of K is shielded by how many inner-shell electrons?

A. 8

B. 10

C. 18

D. 19

Answer: C. 18 inner-shell electrons

26. In Slater's method, the shielding contribution of a 3d electron to a 4s electron is:

A. 0.35

B. 0.50

C. 0.85

D. 1.00

Answer: C. 0.85

27. For a 4s electron in a transition-metal atom, which group is especially important to check?

A. 1s only

B. 3d electrons

C. 5s electrons only

D. 6p electrons

Answer: B. 3d electrons

28. The main purpose of the shielding constant σ is to represent:

A. Nuclear charge reduction due to other electrons

B. Number of protons

C. Number of neutrons

D. Atomic volume

Answer: A

29. For a 4s electron, which contribution is strongest per electron?

A. Same-group electron

B. n−1 electron

C. n−2 or lower electron

D. All are equal

Answer: C. 1.00 per electron

30. The correct sequence for calculating Z_eff is:

A. Find Z → find σ → calculate Z−σ

B. Find σ → calculate Z+σ

C. Find atomic mass → calculate σ

D. Find neutrons → calculate σ

Answer: A

Advertisement

Nature of Cl 2 ​ O 7 ​ (Acidic/Basic).

1. Nature of Cl₂O₇ (Acidic/Basic) +

Correct Answer

Cl₂O₇ is an acidic oxide.

Detailed Explanation

Cl₂O₇ is dichlorine heptoxide. It is the highest oxide of chlorine, in which chlorine has an oxidation state of +7.

2Cl + 7O = 0
2x + 7(−2) = 0
x = +7

It is a non-metal oxide. Oxides of non-metals generally show acidic character, especially when the element is in a high oxidation state. Cl₂O₇ reacts with water to form perchloric acid:

Cl₂O₇ + H₂O → 2HClO₄

Since it produces HClO₄, a strong acid, the oxide is classified as an acidic oxide.

Periodic Trend Concept

For chlorine, the acidic character of its oxides increases with oxidation state:

Cl₂O < Cl₂O₃ < ClO₂ < Cl₂O₇

The higher oxidation state increases the electron-withdrawing character of chlorine and makes the oxide more acidic.

Exam Tip: Higher oxide of a non-metal generally has greater acidic character. Cl₂O₇ is therefore strongly acidic.

Quick Comparison

Na₂O → Basic
Al₂O₃ → Amphoteric
SiO₂ → Acidic
P₂O₅ → Acidic
Cl₂O₇ → Strongly Acidic

30 Related MCQs with Solutions

1. The nature of Cl₂O₇ is:

A. Basic

B. Acidic

C. Amphoteric

D. Neutral

Answer: B. Acidic
It reacts with water to form HClO₄.

2. Cl₂O₇ reacts with water to produce:

A. HCl

B. HClO

C. HClO₃

D. HClO₄

Answer: D. HClO₄

3. The oxidation state of chlorine in Cl₂O₇ is:

A. +1

B. +3

C. +5

D. +7

Answer: D. +7

4. Cl₂O₇ is classified as a:

A. Metallic oxide

B. Non-metal oxide

C. Basic salt

D. Amphoteric hydroxide

Answer: B. Non-metal oxide

5. Which acid is formed by hydration of Cl₂O₇?

A. HClO

B. HClO₂

C. HClO₃

D. HClO₄

Answer: D

6. The reaction Cl₂O₇ + H₂O gives:

A. 2HCl

B. 2HClO

C. 2HClO₃

D. 2HClO₄

Answer: D

7. Which is the highest oxide of chlorine?

A. Cl₂O

B. Cl₂O₃

C. ClO₂

D. Cl₂O₇

Answer: D. Cl₂O₇

8. Which oxide is strongly acidic?

A. Na₂O

B. MgO

C. Cl₂O₇

D. CaO

Answer: C

9. Which oxide is basic?

A. Cl₂O₇

B. SO₃

C. Na₂O

D. P₂O₅

Answer: C. Na₂O

10. Which oxide is amphoteric?

A. Na₂O

B. Al₂O₃

C. SO₃

D. Cl₂O₇

Answer: B. Al₂O₃

11. Acidic character of non-metal oxides generally increases with:

A. Decreasing oxidation state

B. Increasing oxidation state

C. Increasing metallic character

D. Increasing basicity

Answer: B

12. Which chlorine oxide corresponds to HClO₄?

A. Cl₂O

B. Cl₂O₃

C. ClO₂

D. Cl₂O₇

Answer: D

13. HClO₄ is known as:

A. Hypochlorous acid

B. Chlorous acid

C. Chloric acid

D. Perchloric acid

Answer: D. Perchloric acid

14. The central element in Cl₂O₇ is:

A. Oxygen

B. Chlorine

C. Hydrogen

D. None

Answer: B. Chlorine

15. Cl₂O₇ is the anhydride of:

A. HCl

B. HClO

C. HClO₃

D. HClO₄

Answer: D. HClO₄

16. The nature of SO₃ is:

A. Basic

B. Acidic

C. Amphoteric

D. Neutral

Answer: B. Acidic

17. Which oxide is a basic oxide?

A. K₂O

B. SO₃

C. Cl₂O₇

D. CO₂

Answer: A. K₂O

18. Which oxide is acidic in nature?

A. CaO

B. MgO

C. P₂O₅

D. Na₂O

Answer: C

19. Cl₂O₇ + H₂O → ?

A. 2HCl

B. HClO₃

C. 2HClO₄

D. Cl₂ + O₂

Answer: C. 2HClO₄

20. The oxidation state of oxygen in ordinary oxides is generally:

A. +2

B. −2

C. +1

D. −1

Answer: B. −2

21. Which has the highest oxidation state of chlorine?

A. HCl

B. HClO

C. HClO₃

D. HClO₄

Answer: D. HClO₄

22. The acidic character of chlorine oxides generally becomes stronger when chlorine oxidation state:

A. Decreases

B. Increases

C. Remains unchanged

D. Becomes zero

Answer: B

23. Which is a non-metal oxide?

A. Na₂O

B. CaO

C. Cl₂O₇

D. BaO

Answer: C

24. Which oxide is the anhydride of perchloric acid?

A. Cl₂O

B. Cl₂O₃

C. ClO₂

D. Cl₂O₇

Answer: D

25. Which pair contains only acidic oxides?

A. Na₂O, K₂O

B. SO₃, Cl₂O₇

C. CaO, MgO

D. Na₂O, Al₂O₃

Answer: B

26. Which oxide gives a strong acid on reaction with water?

A. Na₂O

B. CaO

C. Cl₂O₇

D. MgO

Answer: C

27. Cl₂O₇ is best described as:

A. Basic anhydride

B. Acidic anhydride

C. Amphoteric oxide

D. Neutral oxide

Answer: B. Acidic anhydride

28. Which oxide corresponds to chlorine in oxidation state +7?

A. Cl₂O

B. Cl₂O₃

C. ClO₂

D. Cl₂O₇

Answer: D

29. Which statement is correct for Cl₂O₇?

A. It forms a base with water

B. It forms HClO₄ with water

C. It is a metallic oxide

D. It is amphoteric

Answer: B

30. The best classification of Cl₂O₇ for competitive exams is:

A. Basic oxide

B. Amphoteric oxide

C. Acidic oxide

D. Neutral oxide

Answer: C. Acidic oxide

Advertisement