Questions and answers on periodic table

 Here’s a list of chemistry questions related to the periodic table that can help test understanding of its concepts, properties, and trends. These questions range from basic to more advanced levels.

1. What is the periodic law?

   • The periodic law states that the properties of elements are a periodic function of their atomic numbers, meaning elements with similar properties occur at regular intervals when arranged by increasing atomic number.

2. How are elements arranged in the periodic table?

   • Elements are arranged in order of increasing atomic number (number of protons). They are organized into rows (periods) and columns (groups or families) based on their electronic configuration and recurring chemical properties.

3. What is the significance of an element’s atomic number?

   • The atomic number indicates the number of protons in the nucleus of an atom, which defines the element and determines its position on the periodic table.

4. What distinguishes metals from nonmetals on the periodic table?

   • Metals are typically good conductors of heat and electricity, malleable, ductile, and have a shiny appearance. Nonmetals, on the other hand, are poor conductors, brittle in solid form, and can be gases or dull solids.

5. What are the two main categories of elements found in the periodic table?

   • The two main categories are metals and nonmetals. A third category, metalloids, contains elements that exhibit properties of both metals and nonmetals.

6. What is a group (or family) in the periodic table?

   • A group is a vertical column in the periodic table where elements share similar chemical properties due to having the same number of valence electrons.

7. How many periods are in the modern periodic table?

   • There are seven periods in the modern periodic table.

8. What are the noble gases, and where are they located in the periodic table?

   • Noble gases are a group of inert gases, including helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn), located in Group 18 (or Group 0) of the periodic table.

9. Which group of elements is known for being highly reactive and why?

   • The alkali metals (Group 1) are known for being highly reactive due to their single valence electron, which they readily lose to form positive ions.

10. Explain the trend in atomic radius as you move down a group and across a period.

    • Down a group: Atomic radius increases because additional electron shells are added, increasing the distance between the outermost electrons and the nucleus.
    • Across a period: Atomic radius decreases because the increased nuclear charge pulls the electrons closer to the nucleus, resulting in a smaller radius.

11. What is ionization energy, and how does it change across periods and down groups?

    • Ionization energy is the energy required to remove an electron from an atom in its gaseous state.
      • Across a period: Ionization energy increases due to higher nuclear charge.
      • Down a group: Ionization energy decreases because the outermost electrons are further from the nucleus and more shielded by inner electrons.

12. Define electronegativity and discuss how it varies in the periodic table.

    • Electronegativity is the tendency of an atom to attract electrons in a chemical bond.
      • Across a period: Electronegativity increases due to increasing nuclear charge.
      • Down a group: Electronegativity decreases because of increased atomic radius and shielding effect.

13. What is the electron configuration of oxygen, and how does it relate to its position on the periodic table?

    • The electron configuration of oxygen (atomic number 8) is $1s^2 2s^2 2p^4$. This configuration indicates it has six valence electrons (2 in the s orbital and 4 in the p orbitals), placing it in Group 16 (or Group VI A) of the periodic table.

14. Identify the elements that belong to the transition metals and discuss their general properties.

    • Transition metals include elements in Groups 3-12 (e.g., iron (Fe), copper (Cu), nickel (Ni), etc.). They are characterized by the presence of d electrons, exhibit variable oxidation states, are often good conductors of heat and electricity, and can form colored compounds.

15. How do metalloids differ from metals and nonmetals? Provide examples.

    • Metalloids have properties intermediate between metals and nonmetals. They are semiconductors of electricity, which means they can conduct electricity better than nonmetals but not as well as metals. Examples include silicon (Si), germanium (Ge), and arsenic (As).

16. Describe how periodic trends can be explained by atomic structure and electron configurations.

    • Periodic trends are explained by the arrangement of electrons around the nucleus. As you move across a period, the increase in protons leads to a stronger nuclear charge, which affects properties like atomic radius, ionization energy, and electronegativity. Moving down a group adds more electron shells, increasing atomic size and decreasing ionization energy and electronegativity due to shielding.

17. What role do valence electrons play in determining an element's chemical reactivity?

    • Valence electrons are the outermost electrons in an atom and determine how an element interacts with others. Elements with fewer valence electrons (like alkali metals) tend to lose them easily and are more reactive, while those with nearly filled shells (like noble gases) are less reactive.

18. How does the concept of shielding affect ionization energy and electronegativity?

    • Shielding refers to the effect where inner electrons block the attraction between the nucleus and the outermost electrons. This decreases ionization energy and electronegativity, as the outer electrons feel less nuclear pull, making them easier to remove or less attracted to bonding.

19. Discuss the properties and applications of alkali metals and alkaline earth metals.

    • Alkali Metals (Group 1): Highly reactive, soft, low density, and low melting points. They are used in batteries (like lithium) and in various chemical reactions.
    • Alkaline Earth Metals (Group 2): Less reactive than alkali metals, but still reactive, harder than alkali metals, and have higher melting points. They are used in fireworks (like magnesium) and in construction (like calcium).

20. What are lanthanides and actinides, and why are they placed separately in the periodic table?

    • Lanthanides (elements 57-71) and actinides (elements 89-103) are placed separately to keep the periodic table compact. They have similar properties and involve f orbitals in their electron configurations, which can complicate the overall structure of the table.

21. How can the periodic table be used to predict the properties of undiscovered elements?

    • By understanding periodic trends and properties of known elements, scientists can make educated predictions about the properties of undiscovered elements based on their expected position in the periodic table, particularly their reactivity, electron configuration, and physical properties.

22. Using the periodic table, predict which element would have a higher ionization energy: sodium (Na) or magnesium (Mg), and explain your reasoning.

    • Magnesium (Mg) has a higher ionization energy than sodium (Na) because Mg is further to the right in the same period and has a higher nuclear charge, making it harder to remove an electron.

23. How does the reactivity of halogens change as you move down the group? Provide examples.

    • The reactivity of halogens decreases as you move down the group. For example, fluorine (F) is more reactive than iodine (I) because its smaller atomic size allows it to attract electrons more effectively.

24. If an element has an atomic number of 12, identify its group, period, and whether it is a metal, nonmetal, or metalloid.

    • An element with atomic number 12 is magnesium (Mg). It is located in Group 2 (alkaline earth metals) and Period 3. Magnesium is a metal.

25. Given the following elements: Carbon (C), Silicon (Si), and Germanium (Ge), explain the trend in their properties.

    • Carbon (C) is a nonmetal, silicon (Si) is a metalloid, and germanium (Ge) is a metalloid. As you move down this group (Group 14), the elements transition from nonmetals to metalloids, and their metallic character increases. For example, conductivity increases from carbon to germanium.

26. Which element would you expect to be more metallic: aluminum (Al) or phosphorus (P)? Justify your answer using periodic trends.

• Aluminum (Al) is expected to be more metallic than phosphorus (P) because it is located to the left of phosphorus on the periodic table, which is consistent with the trend of increasing metallic character as you move from right to left across a period.
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WAEC Chemistry Questions on Periodic Table

Question 1: 

Arrange the following elements in order of increasing size: K (Potassium), Li (Lithium), and Na (Sodium). Provide reasons for your order.

Answer:

To arrange K, Li, and Na in order of increasing size, we need to consider their atomic radii. Atomic size generally increases down a group in the periodic table. Here's how we can determine their relative sizes:

1. Position in the Periodic Table:
   • Li is in Group 1 (alkali metals) and Period 2.
   • Na is in Group 1 and Period 3.
   • K is in Group 1 and Period 4.

Since all three elements belong to the same group (Group 1), their atomic size will depend on their period number. As we move down the group, the atomic size increases because:

• Each successive element has an additional electron shell, which increases the distance between the nucleus and the outermost electron.
• The additional inner shells also cause electron shielding, reducing the effective nuclear charge felt by the outermost electron, allowing the atom to expand.

2. Order of Atomic Size:
   • Li (Lithium) has the smallest atomic radius because it has the fewest electron shells (only 2).
   • Na (Sodium) is larger than Li because it has an additional electron shell (3 electron shells).
   • K (Potassium) is the largest because it has even more electron shells (4 shells).

Increasing Order of Atomic Size:

$$\text{Li}<\text{Na}<\text{K}$$

Reasons:

• Li is the smallest because it has the fewest electron shells (2).
• Na is larger than Li due to the addition of a third electron shell.
• K is the largest because it has the most electron shells (4), with more significant electron shielding and less effective nuclear charge felt by the outermost electron.

Question 2:

Arrange the following species in order of increasing size and explain your order. Na⁺, Mg²⁺, and Ne [Atomic number: Ne = 10, Na = 11, Mg = 12]

Answer:

Species Information:

• Ne (Neon): Atomic number 10. It is a neutral atom with 10 electrons.
• Na⁺ (Sodium ion): Atomic number 11. After losing 1 electron, it has 10 electrons.
• Mg²⁺ (Magnesium ion): Atomic number 12. After losing 2 electrons, it also has 10 electrons.

Explanation:

1. All three species have the same number of electrons (10 electrons), making them isoelectronic. Isoelectronic species have the same electron configuration, but their sizes differ due to the nuclear charge (the number of protons in the nucleus).
2. Nuclear charge increases as you move from Ne (10 protons) to Na⁺ (11 protons) and Mg²⁺ (12 protons).
3. Stronger nuclear charge pulls electrons closer to the nucleus, reducing the size of the ion or atom. As the number of protons increases, the attraction between the nucleus and the electrons increases, making the species smaller.

Order of Increasing Size:

• Mg²⁺ < Na⁺ < Ne

Explanation of Order:

• Mg²⁺ has the greatest nuclear charge (12 protons) with 10 electrons, so the electrons are pulled in most strongly, making Mg²⁺ the smallest.
• Na⁺ has 11 protons and 10 electrons, so it is larger than Mg²⁺ but smaller than Ne.
• Ne has the smallest nuclear charge (10 protons) with 10 electrons, so its electron cloud is not pulled in as tightly, making it the largest of the three.

Final Order:

• Mg²⁺ (smallest) < Na⁺ < Ne (largest)

Question 3:

 From the following atoms: ₇N, ₈O, ₉F, ₁₂Mg, ₁₃Al, ₁₆S, ₁₇Cl, ₁₈Ar, ₁₉K, and ₂₀Ca, identify two anions and two cations that are isoelectronic with argon (Ar).

Answer:

To find ions that are isoelectronic with argon (Ar, atomic number 18), we need to identify ions that have the same number of electrons as argon, which has 18 electrons.

Isoelectronic ions:

• Anions are formed when atoms gain electrons.
• Cations are formed when atoms lose electrons.

Argon (Ar) has 18 electrons.

Anions:

To form anions that are isoelectronic with argon, we look for elements with fewer than 18 electrons that can gain electrons to reach 18.

1. Chlorine (Cl) has 17 electrons. When it gains 1 electron, it becomes Cl⁻:

   $$\text{Cl}(17{\text{e}}^{-})+1{\text{e}}^{-}\to {\text{Cl}}^{-}(18{\text{e}}^{-})$$

   So, Cl⁻ is isoelectronic with argon.

2. Sulfur (S) has 16 electrons. When it gains 2 electrons, it becomes S²⁻:

   $$\text{S}(16{\text{e}}^{-})+2{\text{e}}^{-}\to {\text{S}}^{2-}(18{\text{e}}^{-})$$

   So, S²⁻ is also isoelectronic with argon.

Cations:

To form cations that are isoelectronic with argon, we look for elements with more than 18 electrons that can lose electrons to reach 18.

1. Potassium (K) has 19 electrons. When it loses 1 electron, it becomes K⁺:

   $$\text{K}(19{\text{e}}^{-})-1{\text{e}}^{-}\to {\text{K}}^{+}(18{\text{e}}^{-})$$

   So, K⁺ is isoelectronic with argon.

2. Calcium (Ca) has 20 electrons. When it loses 2 electrons, it becomes Ca²⁺:

   $$\text{Ca}(20{\text{e}}^{-})-2{\text{e}}^{-}\to {\text{Ca}}^{2+}(18{\text{e}}^{-})$$

   So, Ca²⁺ is isoelectronic with argon.

Final Answer:

• Anions isoelectronic with argon: Cl⁻ and S²⁻
• Cations isoelectronic with argon: K⁺ and Ca²⁺

Question 4:

Arrange the following ions in order of increasing size, providing your reasons: ₁₉K⁺, ₁₇Cl⁻, ₁₆S²⁻, ₂₀Ca²⁺.

Solution:

Step 1: Understanding the ions involved

1. K⁺ (Potassium ion): Atomic number 19, loses 1 electron to form K⁺, resulting in 18 electrons.
2. Cl⁻ (Chloride ion): Atomic number 17, gains 1 electron to form Cl⁻, resulting in 18 electrons.
3. S²⁻ (Sulfide ion): Atomic number 16, gains 2 electrons to form S²⁻, resulting in 18 electrons.
4. Ca²⁺ (Calcium ion): Atomic number 20, loses 2 electrons to form Ca²⁺, resulting in 18 electrons.

Step 2: All ions are isoelectronic

All these ions have 18 electrons, making them isoelectronic (same electron configuration). However, their sizes differ based on their nuclear charge.

Step 3: How nuclear charge affects size

• The number of protons in the nucleus affects the attraction between the nucleus and the electrons.
• Higher nuclear charge (more protons) results in greater attraction between the nucleus and the electrons, pulling them closer and making the ion smaller.
• Lower nuclear charge (fewer protons) results in weaker attraction, allowing the electrons to spread out more, making the ion larger.

Step 4: Compare nuclear charges

• K⁺ has 19 protons.
• Cl⁻ has 17 protons.
• S²⁻ has 16 protons.
• Ca²⁺ has 20 protons.

Step 5: Order of increasing size

• Ca²⁺ has the highest nuclear charge (20 protons), pulling electrons in more tightly, making it the smallest.
• K⁺ has 19 protons, slightly less nuclear charge than Ca²⁺, so it is larger than Ca²⁺.
• Cl⁻ has 17 protons, so it is larger than both K⁺ and Ca²⁺.
• S²⁻ has the lowest nuclear charge (16 protons), so it experiences the weakest pull on its electrons, making it the largest.

Final Order (increasing size):

Ca²⁺ < K⁺ < Cl⁻ < S²⁻

Explanation:

• Ca²⁺ is the smallest because it has the highest nuclear charge (20 protons).
• K⁺ is next, with 19 protons pulling the electrons more tightly than Cl⁻ and S²⁻.
• Cl⁻ is larger than K⁺ because it has fewer protons (17).
• S²⁻ is the largest, with the fewest protons (16) and the weakest attraction on its electrons.

Question 5:

Indicate which set of species are isoelectronic, and then write the electronic configurations of the isoelectronic sets from the species below:

Species	O²⁻	Ar	Be⁺	Ne	Cl⁻	He⁺	Li
Atomic number	8	18	4	10	17	2	3

b) Choose one of the isoelectronic sets in (a) above and deduce whether the species in the set have the same size or not.

Solution:

Step 1: Identifying isoelectronic species

• Isoelectronic species have the same number of electrons but different numbers of protons.
• Let's calculate the number of electrons in each species:

1. O²⁻:

   • Oxygen's atomic number = 8 (8 protons and normally 8 electrons).
   • After gaining 2 electrons to form O²⁻, it has 10 electrons.

2. Ar:

   • Argon's atomic number = 18.
   • Since it is a neutral atom, it has 18 electrons.

3. Be⁺:

   • Beryllium's atomic number = 4 (4 protons and normally 4 electrons).
   • After losing 1 electron to form Be⁺, it has 3 electrons.

4. Ne:

   • Neon’s atomic number = 10.
   • As a neutral atom, it has 10 electrons.

5. Cl⁻:

   • Chlorine’s atomic number = 17 (17 protons and normally 17 electrons).
   • After gaining 1 electron to form Cl⁻, it has 18 electrons.

6. He⁺:

   • Helium’s atomic number = 2 (2 protons and normally 2 electrons).
   • After losing 1 electron to form He⁺, it has 1 electron.

7. Li:

   • Lithium’s atomic number = 3.
   • As a neutral atom, it has 3 electrons.

Step 2: Determining the isoelectronic sets

From the electron counts:

• O²⁻ has 10 electrons.
• Ne has 10 electrons.

Thus, O²⁻ and Ne are isoelectronic.

• Cl⁻ has 18 electrons.
• Ar has 18 electrons.

Thus, Cl⁻ and Ar are also isoelectronic.

Step 3: Electronic configurations of the isoelectronic sets

• O²⁻ and Ne:
  Both have 10 electrons, so their electronic configuration is:

  $$1s^{2}2s^{2}2p^6$$

• Cl⁻ and Ar:
  Both have 18 electrons, so their electronic configuration is:

  $$1s^{2}2s^{2}2p^{6}3s^{2}3p^6$$

Step 4: Choosing an isoelectronic set and comparing sizes

Let's choose the Cl⁻ and Ar set for comparison.

• Cl⁻ and Ar have the same number of electrons, so they are isoelectronic. However, their sizes differ due to the number of protons:
  • Cl⁻ has 17 protons pulling on 18 electrons.
  • Ar has 18 protons pulling on 18 electrons.

Since Ar has more protons, the effective nuclear charge (attractive force of the nucleus on the electrons) is stronger in Ar than in Cl⁻. This results in Ar being smaller than Cl⁻ because the electrons are pulled closer to the nucleus in argon.

Conclusion:

• Cl⁻ and Ar are isoelectronic, but Cl⁻ is larger than Ar due to its weaker nuclear charge (fewer protons attracting the same number of electrons).

Question 6:

Explain briefly the statement. the radius of the radius of Mg²⁺ is smaller than that of Na

Solution:

The statement "the radius of Mg²⁺ is smaller than that of Na⁺" can be explained by comparing the effective nuclear charge and the number of protons in each ion.

1. Number of electrons and protons:

• Mg²⁺ (Magnesium ion) has an atomic number of 12, which means it has 12 protons in the nucleus. After losing two electrons to form Mg²⁺, it has 10 electrons.
• Na⁺ (Sodium ion) has an atomic number of 11, which means it has 11 protons in the nucleus. After losing one electron to form Na⁺, it also has 10 electrons.

Both ions, Mg²⁺ and Na⁺, are isoelectronic (both have 10 electrons), but their nuclear charges are different.

2. Effective nuclear charge:

• Mg²⁺ has 12 protons pulling on 10 electrons, creating a stronger attractive force on the electrons because of its higher nuclear charge.
• Na⁺ has 11 protons pulling on 10 electrons, which means the nuclear attraction is slightly weaker compared to Mg²⁺.

3. Size and electron attraction:

Since Mg²⁺ has a greater positive charge (12 protons vs. 11 protons in Na⁺), it pulls the electrons closer to the nucleus, resulting in a smaller ionic radius. In contrast, Na⁺ has a slightly weaker attraction on its electrons, leading to a larger ionic radius compared to Mg²⁺.

Conclusion:

The radius of Mg²⁺ is smaller than that of Na⁺ because Mg²⁺ has a higher nuclear charge (12 protons) compared to Na⁺ (11 protons), and both ions have the same number of electrons. The stronger nuclear attraction in Mg²⁺ pulls the electrons closer, reducing the size of the ion.

Question 7:

Deduce the expected sequence of the radii of F-, Cl- and Br- showing which factors influence the observed sequence.

Solution:

To deduce the expected sequence of the radii of F⁻, Cl⁻, and Br⁻, we need to consider several factors that influence atomic and ionic radii, including nuclear charge, electron-electron repulsion, and principal quantum number.

Step 1: Identifying the species

• F⁻ (Fluoride ion): Atomic number 9; has gained 1 electron, so it has 9 protons and 10 electrons.
• Cl⁻ (Chloride ion): Atomic number 17; has gained 1 electron, so it has 17 protons and 18 electrons.
• Br⁻ (Bromide ion): Atomic number 35; has gained 1 electron, so it has 35 protons and 36 electrons.

Step 2: Analyzing the factors influencing ionic radii

1. Nuclear Charge:

   • As we move down the group from F to Cl to Br, the number of protons in the nucleus increases (9 in F, 17 in Cl, 35 in Br).
   • A higher nuclear charge means a stronger attractive force on the electrons, which tends to pull the electron cloud closer to the nucleus.

2. Principal Quantum Number:

   • The principal quantum number (n) indicates the energy level of the electrons. As we move from F to Cl to Br, we increase in principal quantum number:
     • F⁻ has electrons in the n=2 shell.
     • Cl⁻ has electrons in the n=3 shell.
     • Br⁻ has electrons in the n=4 shell.
   • A higher principal quantum number means that the outer electrons are in higher energy levels, which are farther from the nucleus, resulting in a larger ionic radius.

3. Electron-Electron Repulsion:

   • The addition of electrons increases electron-electron repulsion, which tends to push the electrons further apart, leading to a larger ionic radius.
   • However, this effect is often outweighed by the influence of the nuclear charge as we move down the group.

Step 3: Expected Sequence of Ionic Radii

Given these considerations, the expected sequence of ionic radii (from smallest to largest) is:

$$\text{F⁻}<\text{Cl⁻}<\text{Br⁻}$$

Conclusion

• F⁻ is the smallest because it has the least nuclear charge (9 protons) attracting its electrons.
• Cl⁻ is larger than F⁻ due to its greater nuclear charge (17 protons) and the higher principal quantum number (n=3).
• Br⁻ is the largest because it has the highest nuclear charge (35 protons) and its electrons are in the highest principal quantum number (n=4), leading to a larger radius.

Thus, while nuclear charge increases from F⁻ to Br⁻, the effect of the increased principal quantum number dominates, resulting in a clear sequence of increasing ionic radii.