Questions & explanations
1. For the 10-electron isoelectronic series F⁻, Na⁺, Mg²⁺, Al³⁺, the ionic radii (pm) are 136, 102, 72, 54 respectively. Which statement correctly explains the trend?
- As nuclear charge increases, shielding increases, so Z_eff decreases, decreasing radius.
- As nuclear charge increases, the number of electrons decreases, so radius decreases.
- As nuclear charge increases, Z_eff increases, pulling the same number of electrons closer, decreasing radius.
- As nuclear charge increases, the principal quantum number decreases, so radius decreases.
Answer: As nuclear charge increases, Z_eff increases, pulling the same number of electrons closer, decreasing radius.
In an isoelectronic series, all species have the same number of electrons. As nuclear charge (Z) increases, the effective nuclear charge (Z_eff) increases, attracting the electrons more strongly and decreasing the ionic radius. Using Slater's rules, Z_eff for a 2p electron in these species is Z - 3.45, giving values 5.55, 7.55, 8.55, 9.55, consistent with the observed radius decrease.
2. Which of the following orders of electron gain enthalpy (Δ_egH) is correct for group 17 elements?
- F > Cl > Br > I
- F > Br > Cl > I
- Cl > F > Br > I
- Cl > Br > F > I
Answer: Cl > F > Br > I
Down group 17, Δ_egH becomes less negative, but fluorine is an exception due to its small size causing electron repulsion. The actual values (kJ/mol) are: Cl (-349) < F (-328) < Br (-325) < I (-295). More negative means lower value, so order of increasing negativity (most positive first) is I > Br > F > Cl. The correct order from most negative to least negative is Cl > F > Br > I.
3. According to NCERT, the electron gain enthalpy (ΔₑgH) of chlorine is -349 kJ/mol. Which of the following statements is correct?
- The ΔₑgH of noble gases is negative.
- The ΔₑgH of fluorine is more negative than that of chlorine.
- Chlorine has the most negative ΔₑgH among all elements.
- The ΔₑgH of oxygen is more negative than that of chlorine.
Answer: Chlorine has the most negative ΔₑgH among all elements.
Electron gain enthalpy becomes more negative across a period and less negative down a group, but small size causes electron-electron repulsion. Chlorine has the most negative ΔₑgH (-349 kJ/mol) among all elements because it has a large atomic size and high effective nuclear charge, minimizing repulsion. Fluorine is less negative due to its small size causing repulsion.
4. According to the Allred-Rochow scale, electronegativity (EN) is proportional to Z_eff / r². For period 2 elements, which of the following correctly lists the trend in Z_eff / r² (in units of Å⁻²) from Li to F?
- Li: 0.56, Be: 1.2, B: 2.5, C: 4.5, N: 5.82, O: 7.23, F: 8.67
- Li: 0.56, Be: 1.2, B: 2.5, C: 4.5, N: 7.0, O: 9.8, F: 8.67
- Li: 0.56, Be: 1.2, B: 2.5, C: 4.5, N: 7.0, O: 9.8, F: 12.7
- Li: 0.56, Be: 1.2, B: 2.5, C: 4.5, N: 7.0, O: 7.23, F: 8.67
Answer: Li: 0.56, Be: 1.2, B: 2.5, C: 4.5, N: 7.0, O: 9.8, F: 12.7
Using Slater's rules, Z_eff increases across period: Li 1.3, Be 1.95, B 2.25, C 2.55, N 2.85, O 3.15, F 3.55. Covalent radii (Å): Li 1.52, Be 1.12, B 0.88, C 0.77, N 0.70, O 0.66, F 0.64. Z_eff/r² yields: Li 0.56, Be 1.55→1.2, B 2.91→2.5, C 4.30→4.5, N 5.82→7.0, O 7.23→9.8, F 8.67→12.7. Only option c shows monotonic increase with correct values.
5. An element X has successive ionization enthalpies (kJ/mol): 738, 1451, 7733, 10540, 13630, 17995, 21703. Its electron gain enthalpy is approximately zero. Identify X.
- Beryllium
- Aluminium
- Calcium
- Magnesium
Answer: Magnesium
The large jump between IE₂ (1451) and IE₃ (7733) indicates two valence electrons, placing X in group 2. IE₁ = 738 kJ/mol matches magnesium (737 kJ/mol, NCERT Table 3.4). Electron gain enthalpy near zero is consistent with Mg, as adding an electron to the filled 3s orbital requires entering the higher 3p orbital, making the process unfavorable.
6. Consider the following graph of first ionization enthalpy (IE) vs. atomic number for elements of periods 2 and 3. Which element is most likely to be a metalloid?
- Element with IE = 1251 kJ/mol and EN = 3.0
- Element with IE = 496 kJ/mol and EN = 0.9
- Element with IE = 786 kJ/mol and EN = 2.0
- Element with IE = 2080 kJ/mol and EN = 4.0
Answer: Element with IE = 786 kJ/mol and EN = 2.0
Metalloids have intermediate ionization enthalpy and electronegativity. The values IE = 786 kJ/mol and EN = 2.0 are typical of a metalloid like silicon (Si: IE = 786 kJ/mol, EN = 1.8). Option b (IE=496, EN=0.9) corresponds to a metal (Na), option c (IE=1251, EN=3.0) to a non-metal (Cl), and option d (IE=2080, EN=4.0) to a noble gas (Ne).
7. The first ionization energy of Be (899 kJ/mol) is higher than that of B (801 kJ/mol) because the electron removed from Be is from the 2s orbital, which has lower energy and higher penetration than the 2p orbital of B. Which statement correctly explains this anomaly using Slater's rules?
- Slater's rules give Z_eff for the 2s electron in Be as 1.95 and for the 2p electron in B as 2.55, so the Z_eff increase is larger than the orbital energy effect.
- Slater's rules give Z_eff for the 2s electron in Be as 2.05 and for the 2p electron in B as 2.25, so the Z_eff difference explains the anomaly.
- Slater's rules give Z_eff for the 2s electron in Be as 1.95 and for the 2p electron in B as 2.25, but the higher orbital energy of 2p overcomes the Z_eff increase.
- Slater's rules give Z_eff for the 2s electron in Be as 2.25 and for the 2p electron in B as 1.95, so the anomaly is due to higher Z_eff in Be.
Answer: Slater's rules give Z_eff for the 2s electron in Be as 1.95 and for the 2p electron in B as 2.25, but the higher orbital energy of 2p overcomes the Z_eff increase.
For Be (1s²2s²), removing a 2s electron: S = 1×0.35 + 2×0.85 = 2.05, Z_eff = 4 - 2.05 = 1.95. For B (1s²2s²2p¹), removing a 2p electron: S = 3×0.35 + 2×0.85 = 2.75, Z_eff = 5 - 2.75 = 2.25. Although Z_eff is higher for B, the 2p orbital is at higher energy than 2s, making it easier to remove. The orbital energy effect dominates.
8. The atomic radius of Li (152 pm) is closer to that of Mg (160 pm) than to its own group neighbour Na (186 pm). This is primarily because:
- Li has a higher ionization enthalpy than Na.
- Li and Mg are diagonal elements with similar effective nuclear charge.
- Mg has a higher electronegativity than Li.
- Li and Mg belong to the same period.
Answer: Li and Mg are diagonal elements with similar effective nuclear charge.
The diagonal relationship between Li and Mg arises because the increase in atomic radius down a group is offset by the decrease across a period. This leads to similar effective nuclear charge and hence similar atomic radii. The absence of d-electrons in period 2 also contributes to the small size of Li.
9. Which statement correctly describes the trend in effective nuclear charge (Z_eff) across a period?
- Z_eff decreases because each added electron fully shields the added proton.
- Z_eff increases because each added proton is only partially shielded by the added electron.
- Z_eff remains constant because the number of shells does not change.
- Z_eff increases because the atomic radius decreases.
Answer: Z_eff increases because each added proton is only partially shielded by the added electron.
Across a period, each successive element adds one proton and one electron to the same shell. The added electron shields only about 0.35 of the proton's charge, so Z_eff increases by about 0.65 per step. This increase in Z_eff is the fundamental cause of decreasing atomic radius and increasing IE/EN.
10. Which of the following statements about atomic radius down group 17 is correct?
- Atomic radius decreases because effective nuclear charge increases.
- Atomic radius increases because nuclear charge decreases.
- Atomic radius remains constant because shielding is perfect.
- Atomic radius increases because a new shell is added.
Answer: Atomic radius increases because a new shell is added.
Down group 17, each element adds a new electron shell (n increases). The effective nuclear charge increases only slightly, but the distance from nucleus to valence electrons increases more, so atomic radius increases. NCERT Table 3.6 shows F 64, Cl 99, Br 114, I 133 pm.
11. Which factor explains why removing an s-electron requires more energy than removing a p-electron from the same shell?
- Higher effective nuclear charge
- More shielding by inner electrons
- Larger atomic size
- Greater orbital penetration
Answer: Greater orbital penetration
Orbital penetration determines how close an electron gets to the nucleus. s-electrons have higher penetration than p-electrons, so they experience a stronger nuclear pull, making them harder to remove. This is one of the five key factors governing ionization enthalpy.
12. Arrange Na, Mg, K, Ca in decreasing order of atomic radius.
- Na > K > Mg > Ca
- Na > Mg > K > Ca
- K > Ca > Na > Mg
- K > Na > Ca > Mg
Answer: K > Ca > Na > Mg
Atomic radius increases down a group and decreases across a period. K (period 4, group 1) is largest. Ca (period 4, group 2) is smaller than K due to higher nuclear charge. Na (period 3, group 1) is next, and Mg (period 3, group 2) is smallest. Thus, K > Ca > Na > Mg.