Questions & explanations
1. Which of the following statements about electron gain enthalpy is correct?
- Electron gain enthalpy of nitrogen is more negative than that of oxygen.
- Electron gain enthalpy of beryllium is more negative than that of boron.
- Electron gain enthalpy of chlorine is the most negative among halogens.
- Electron gain enthalpy of noble gases is highly negative.
Answer: Electron gain enthalpy of chlorine is the most negative among halogens.
Among halogens, chlorine has the highest negative electron gain enthalpy (greater than fluorine) because fluorine's very small size leads to high electron-electron repulsion, reducing the energy released. Nitrogen (group 15) has less negative electron gain enthalpy than oxygen due to extra stability of half-filled p orbitals. Beryllium (group 2) has less negative electron gain enthalpy than boron because of stable s² configuration. Noble gases have positive (endothermic) electron gain enthalpies.
2. The first ionization enthalpy of nitrogen is higher than that of oxygen because:
- Nitrogen has a half-filled 2p subshell
- Nitrogen has a smaller atomic radius
- Oxygen has a higher nuclear charge
- Oxygen has a half-filled 2p subshell
Answer: Nitrogen has a half-filled 2p subshell
Nitrogen has electronic configuration 1s² 2s² 2p³, a half-filled p-subshell which is extra stable due to exchange energy and symmetrical distribution. Removing an electron from this stable configuration requires more energy than removing an electron from oxygen (2p⁴), where electron-electron repulsion makes removal easier.
3. Which of the following correctly represents the order of first ionization enthalpy for the elements Na, Mg, Al, and Si?
- Na < Mg < Al < Si
- Na > Mg > Al > Si
- Si < Al < Mg < Na
- Na < Al < Mg < Si
Answer: Na < Al < Mg < Si
First ionization enthalpy generally increases across a period but with anomalies. Mg has higher IE than Al due to the stable 3s² configuration. Na has the lowest, Si the highest. The correct increasing order is Na (496 kJ/mol) < Al (577 kJ/mol) < Mg (738 kJ/mol) < Si (787 kJ/mol).
4. An element has the electronic configuration [Ar] 3d¹⁰ 4s² 4p³. To which group and period does it belong?
- Group 15, Period 4
- Group 15, Period 3
- Group 13, Period 4
- Group 16, Period 4
Answer: Group 15, Period 4
The highest principal quantum number is n = 4, so the element is in period 4. The valence electrons are 4s² 4p³, giving total 5 valence electrons; in the p-block, group number = 10 + number of valence electrons = 10 + 5 = 15. Hence it belongs to group 15.
5. Electronegativity is defined as the measure of the tendency of an atom to:
- A) Release energy when an electron is added
- B) Attract shared electrons towards itself in a chemical bond
- C) Remove an electron from its valence shell
- D) Lose electrons to form a positive ion
Answer: B) Attract shared electrons towards itself in a chemical bond
Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself. Option A describes electron affinity, C describes ionization energy, and D describes electropositive or metallic character.
6. Which of the following statements correctly describes the trend of atomic radius in the periodic table?
- Atomic radius increases across a period and decreases down a group.
- Atomic radius decreases across a period and increases down a group.
- Atomic radius increases across a period and increases down a group.
- Atomic radius decreases across a period and decreases down a group.
Answer: Atomic radius decreases across a period and increases down a group.
Across a period, effective nuclear charge increases, pulling the electron cloud inward, so atomic radius decreases. Down a group, new electron shells are added, increasing the distance from the nucleus, so atomic radius increases.
7. Among the following elements, which one exhibits the highest metallic character?
- A) Beryllium (Be)
- B) Boron (B)
- C) Carbon (C)
- D) Nitrogen (N)
Answer: A) Beryllium (Be)
Metallic character decreases across a period from left to right. In period 2, beryllium is the leftmost element among the given options and therefore has the highest metallic character. Boron, carbon, and nitrogen are nonmetals.
8. Which of the following is the correct order of ionic radii for the given isoelectronic species?
- O²⁻ > F⁻ > Na⁺ > Mg²⁺
- O²⁻ > F⁻ > Mg²⁺ > Na⁺
- Na⁺ > Mg²⁺ > F⁻ > O²⁻
- F⁻ > O²⁻ > Na⁺ > Mg²⁺
Answer: O²⁻ > F⁻ > Na⁺ > Mg²⁺
All species (O²⁻, F⁻, Na⁺, Mg²⁺) have 10 electrons (isoelectronic). Ionic radius increases as nuclear charge decreases. Nuclear charges: O²⁻ (8), F⁻ (9), Na⁺ (11), Mg²⁺ (12). Thus O²⁻ is largest and Mg²⁺ is smallest.
9. Which of the following statements correctly represents the modern periodic law?
- Properties of elements are periodic functions of their atomic masses.
- Properties of elements are periodic functions of their atomic numbers.
- Properties of elements are periodic functions of their valencies.
- Properties of elements are periodic functions of their neutron numbers.
Answer: Properties of elements are periodic functions of their atomic numbers.
The modern periodic law, given by Moseley, states that physical and chemical properties of elements are periodic functions of their atomic numbers, not atomic masses as in Mendeleev's original law.
10. Which element of the third period can exhibit the highest range of oxidation states?
- A) Sodium (Na)
- B) Magnesium (Mg)
- C) Aluminium (Al)
- D) Chlorine (Cl)
Answer: D) Chlorine (Cl)
Chlorine shows a variety of oxidation states from -1 (e.g., HCl) to +7 (e.g., HClO₄). Sodium, magnesium, and aluminium predominantly show only +1, +2, and +3 oxidation states respectively.
11. An element has the electronic configuration [Xe] 4f⁶ 6s². To which block of the periodic table does it belong?
- s-block
- p-block
- d-block
- f-block
Answer: f-block
The last electron enters the 4f subshell, which is an inner (n-2)f orbital. Elements with the filling of f-orbitals are classified as f-block elements (lanthanoids and actinoids).
12. According to the IUPAC systematic nomenclature, what is the name for the element with atomic number 114?
- A) Ununquadium
- B) Ununhexium
- C) Ununpentium
- D) Ununoctium
Answer: A) Ununquadium
For atomic number 114, the roots are: 1 = un, 1 = un, 4 = quad, resulting in 'ununquadium'. Ununhexium is element 116, ununpentium is 115, and ununoctium is 118.