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Periodic Table and Properties Practice Questions & Answers

Analysis of the Periodic Table's organization and the predictive power of periodic trends.

Trends Covered:

  • Atomic & Ionic Radii: Size variations across periods and groups.
  • Ionization Energy: Energy required to remove electrons.
  • Electron Affinity & Electronegativity: Attraction for electrons.
  • Classifications: Metals, Non-metals, Metalloids, and s/p/d/f blocks.

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Dmitri Mendeleev is credited with the development of the first widely accepted periodic table. What was the primary property he used to arrange the elements?

  • Atomic number

  • Atomic mass

  • Electron configuration

  • Number of neutrons

View Answer & Explanation
Correct Answer: Option B -

Atomic mass

Explanation:

Mendeleev arranged elements in order of increasing atomic mass. It was later, through the work of Henry Moseley, that the periodic table was reorganized by atomic number.

Which scientist modified the Periodic Law to state that the properties of elements are periodic functions of their atomic numbers rather than atomic masses?

  • Dmitri Mendeleev

  • J.J. Thomson

  • Henry Moseley

  • Niels Bohr

View Answer & Explanation
Correct Answer: Option C -

Henry Moseley

Explanation:

Henry Moseley's experiments with X-ray spectra showed that the atomic number (number of protons) is the fundamental property determining an element's identity, correcting Mendeleev's mass-based arrangement.

Elements in the s-block and p-block (excluding the noble gases) are collectively known as:

  • Transition elements

  • Inner transition elements

  • Representative elements

  • Transuranium elements

View Answer & Explanation
Correct Answer: Option C -

Representative elements

Explanation:

The elements in Groups 1, 2, and 13–18 (s and p blocks) are often called representative or main-group elements because their outer shells are being filled.

Which general electronic configuration represents the d-block elements (Transition Metals)?

  • (n1)d110ns02(n-1)d^{1-10} ns^{0-2}(n1)d110ns02

  • ns2np16ns^2 np^{1-6}ns2np16

  • (n2)f114(n1)d01ns2(n-2)f^{1-14} (n-1)d^{0-1} ns^2(n2)f114(n1)d01ns2

  • ns1ns^1ns1 to ns2ns^2ns2

View Answer & Explanation
Correct Answer: Option A -

(n1)d110ns02(n-1)d^{1-10} ns^{0-2}(n1)d110ns02

Explanation:

Transition metals are characterized by the filling of inner d-orbitals. The general configuration is (n1)d110ns02(n-1)d^{1-10} ns^{0-2}(n1)d110ns02.

An element has the atomic number 35. To which block and group does it belong?

  • s-block, Group 2

  • p-block, Group 17

  • d-block, Group 10

  • p-block, Group 15

View Answer & Explanation
Correct Answer: Option B -

p-block, Group 17

Explanation:

Atomic number 35 corresponds to Bromine (BrBrBr). Its configuration ends in 4s23d104p54s^2 3d^{10} 4p^54s23d104p5. Since the last electron enters a p-orbital, it is in the p-block. With 7 valence electrons (s2p5s^2 p^5s2p5), it belongs to Group 17 (Halogens).

Which of the following sets of atomic numbers belongs to the Alkali Metals (Group 1)?

  • 3, 11, 19, 37

  • 4, 12, 20, 38

  • 9, 17, 35, 53

  • 2, 10, 18, 36

View Answer & Explanation
Correct Answer: Option A -

3, 11, 19, 37

Explanation:

Group 1 elements are Li (3), Na (11), K (19), Rb (37), etc. Option (b) is Alkaline Earth Metals, (c) is Halogens, and (d) is Noble Gases.

The metalloids (semi-metals) are located along a 'staircase' line on the periodic table. Which of the following elements is a metalloid?

  • Aluminum (AlAlAl)

  • Silicon (SiSiSi)

  • Phosphorus (PP​P)

  • Sulfur (SS​S)

View Answer & Explanation
Correct Answer: Option B -

Silicon (SiSiSi)

Explanation:

Silicon (SiSiSi) is a classic metalloid, exhibiting properties intermediate between metals and nonmetals. Aluminum is a metal; Phosphorus and Sulfur are nonmetals.

Which of the following correctly describes the trend of atomic radius across a period (left to right) and its cause?

  • Increases due to additional electron shells

  • Decreases due to increasing effective nuclear charge (ZeffZ_{eff}Zeff)

  • Increases due to increased electron-electron repulsion

  • Decreases due to the lanthanoid contraction

View Answer & Explanation
Correct Answer: Option B -

Decreases due to increasing effective nuclear charge (ZeffZ_{eff}Zeff)

Explanation:

Across a period, electrons are added to the same shell while protons are added to the nucleus. This increases the effective nuclear charge (ZeffZ_{eff}Zeff), pulling electrons closer and shrinking the radius.

Arrange the following species in order of increasing radius: Mg2+Mg^{2+}Mg2+, Na+Na^+Na+, NeNeNe, FF^-F.

  • Mg2+<Na+<Ne<FMg^{2+} < Na^+ < Ne < F^-Mg2+<Na+<Ne<F

  • F<Ne<Na+<Mg2+F^- < Ne < Na^+ < Mg^{2+}F<Ne<Na+<Mg2+

  • Mg2+<Na+<F<NeMg^{2+} < Na^+ < F^- < NeMg2+<Na+<F<Ne

  • Ne<F<Na+<Mg2+Ne < F^- < Na^+ < Mg^{2+}Ne<F<Na+<Mg2+

View Answer & Explanation
Correct Answer: Option A -

Mg2+<Na+<Ne<FMg^{2+} < Na^+ < Ne < F^-Mg2+<Na+<Ne<F

Explanation:

These species are isoelectronic (all have 10 electrons). The size is determined by the nuclear charge. MgMgMg has 12 protons, NaNaNa 11, NeNeNe 10, FF​F 9. Higher nuclear charge pulls the same number of electrons tighter. Thus, Mg2+Mg^{2+}Mg2+ is smallest and FF^-F (with the fewest protons) is largest. (Note: FF^-F is typically larger than NeNeNe due to lower Z).

Why is the radius of a cation always smaller than its neutral parent atom (e.g., Na+<NaNa^+ < NaNa+<Na)?

  • The cation gains neutrons

  • The nuclear charge decreases

  • Loss of valence electrons increases effective nuclear charge per electron and may remove a shell

  • Electron-electron repulsion increases

View Answer & Explanation
Correct Answer: Option C -

Loss of valence electrons increases effective nuclear charge per electron and may remove a shell

Explanation:

Removing an electron reduces electron-electron repulsion and, in many cases (like NaNa+Na \to Na^+NaNa+), removes the outermost shell entirely. The remaining electrons are held more tightly by the nucleus.

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