Atomic & Ionic Radii, Screening & Isoelectronic — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Atomic & Ionic Radii, Screening & Isoelectronic MCQs with step-by-step solutions (40 questions). Part of Classification of Elements and Periodicity. Practise online on Prepizo — no login needed.
▶ Practise Atomic & Ionic Radii, Screening & Isoelectronic online (free)
Questions with solutions
Q1 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
For the same element, the correct order of atomic radii is:
A. r(covalent) > r(metallic) > r(vdW)
B. r(metallic) > r(vdW) > r(covalent)
C. r(van der Waals) > r(metallic) > r(covalent) ✓ Correct
D. All are equal
Solution: van der Waals radius (non-bonded contact) is largest; covalent radius (shared overlap) is smallest.
Q2 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
Across a period (left to right), atomic radius generally:
A. Increases
B. Remains constant
C. First decreases then increases
D. Decreases ✓ Correct
Solution: Nuclear charge (and Z_eff) rises while the shell stays the same, pulling electrons in — radius decreases.
Q3 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
Down a group, atomic radius generally:
A. Increases ✓ Correct
B. First increases then decreases
C. Stays constant
D. Decreases
Solution: A new shell is added each period, so radius increases down a group.
Q4 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
A cation is always ______ than its parent atom:
A. Larger
B. Equal in size to
C. Smaller ✓ Correct
D. Twice as big as
Solution: Losing electron(s) reduces electron–electron repulsion and often removes a shell, so the cation is smaller.
Q5 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
An anion is always ______ than its parent atom:
A. Equal in size to
B. Larger ✓ Correct
C. Half the size of
D. Smaller
Solution: Adding electron(s) increases repulsion and the electron cloud expands, so the anion is larger.
Q6 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
The screening (shielding) effect refers to:
A. The increase in nuclear charge
B. The pairing of electrons
C. The removal of an electron
D. The reduction of nuclear attraction on valence electrons by inner electrons ✓ Correct
Solution: Inner electrons "screen" the outer ones from the full nuclear charge.
Q7 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
The effective nuclear charge is given by:
A. Z_eff = Z + σ
B. Z_eff = Z − σ ✓ Correct
C. Z_eff = σ − Z
D. Z_eff = Z × σ
Solution: Z_eff = actual nuclear charge minus the screening constant σ.
Q8 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
Isoelectronic species are those having:
A. The same size
B. The same mass number
C. The same number of electrons ✓ Correct
D. The same number of protons
Solution: Isoelectronic species have identical electron counts but different nuclear charges.
Q9 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
Among isoelectronic species, the one with the highest nuclear charge (Z) is the:
A. Most reactive metal
B. Largest
C. Same size as the others
D. Smallest ✓ Correct
Solution: More protons pulling the same number of electrons ⇒ smaller size.
Q10 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
The order of screening power of orbitals in the same shell is:
A. p > s > d > f
B. f > d > p > s
C. s = p = d = f
D. s > p > d > f ✓ Correct
Solution: Penetration order s > p > d > f gives the same order of shielding ability.
Q11 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
The abnormally small covalent radius of the noble gases is NOT usually compared because they are measured as:
A. Metallic radii
B. Ionic radii
C. van der Waals radii (non-bonded), which are larger ✓ Correct
D. Covalent radii
Solution: Noble gases don't form normal bonds, so their radii are van der Waals radii — hence appear larger than the preceding halogen.
Q12 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
The metallic radius is defined as:
A. The distance to the nucleus of a bonded non-metal
B. Half the bond length in a covalent molecule
C. The radius of the isolated atom
D. Half the internuclear distance between two adjacent metal atoms in a metallic crystal ✓ Correct
Solution: It is half the closest internuclear separation in the solid metal.
Q13 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
The gradual small decrease in size across the lanthanides is called the:
A. Isoelectronic contraction
B. Diagonal contraction
C. Lanthanide contraction ✓ Correct
D. Screening effect
Solution: Poor shielding by 4f electrons causes the steady size decrease known as lanthanide contraction.
Q14 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · theory
Poor shielding is offered by which type of electrons (leading to lanthanide contraction)?
A. inner-shell s electrons
B. s electrons
C. f electrons ✓ Correct
D. p electrons
Solution: f orbitals are diffuse and shield poorly, so Z_eff on outer electrons increases across the lanthanides.
Q15 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · theory
For the same element, the anion is larger than the cation mainly because:
A. The cation gains a shell
B. The anion has more protons
C. The anion has more electrons and less nuclear pull per electron ✓ Correct
D. They have equal Z_eff
Solution: The anion has excess electrons (more repulsion, lower Z_eff per electron), the cation has fewer.
Q16 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · theory
Covalent radius is measured as:
A. The ionic separation
B. The van der Waals distance
C. Half the internuclear distance between two identical covalently-bonded atoms ✓ Correct
D. The full bond length
Solution: For X–X, covalent radius = (bond length)/2.
Q17 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · numerical
The correct increasing order of atomic radius for period-2 elements Li, Be, B, C is:
A. B < C < Be < Li
B. C < B < Be < Li ✓ Correct
C. Be < Li < C < B
D. Li < Be < B < C
Solution: Radius decreases across a period, so Li is largest and C smallest: C < B < Be < Li.
Q18 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · numerical
The correct order of size for the isoelectronic species N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ is:
A. N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ ✓ Correct
B. Al³⁺ > Mg²⁺ > Na⁺ > F⁻ > O²⁻ > N³⁻
C. All equal
D. F⁻ > O²⁻ > N³⁻ > Na⁺ > Al³⁺ > Mg²⁺
Solution: All have 10 electrons; size decreases as Z increases (7→13), so N³⁻ (Z 7) is largest, Al³⁺ (Z 13) smallest.
Q19 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · numerical
Among the isoelectronic species O²⁻, F⁻ and Na⁺, the largest is:
A. O²⁻ ✓ Correct
B. F⁻
C. All equal
D. Na⁺
Solution: Same 10 electrons; O²⁻ has the fewest protons (8), so the weakest pull and largest size.
Q20 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · numerical
Which is the smallest among the isoelectronic species Na⁺, Mg²⁺, Al³⁺?
A. All equal
B. Na⁺
C. Al³⁺ ✓ Correct
D. Mg²⁺
Solution: Al³⁺ has the highest nuclear charge (13) for the same 10 electrons ⇒ smallest.
Q21 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · numerical
The screening constant σ (by Slater's rules) for the valence 3s electron of sodium (Na, Z = 11) is:
A. 2.2
B. 10.0
C. 6.8
D. 8.8 ✓ Correct
Solution: σ = (8 electrons in n=2)×0.85 + (2 in n=1)×1.00 = 6.8 + 2.0 = 8.8.
Q22 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · numerical
Using Slater's rules, the effective nuclear charge on the 3s electron of Na (Z = 11) is:
A. 2.2 ✓ Correct
B. 11.0
C. 1.0
D. 8.8
Solution: Z_eff = Z − σ = 11 − 8.8 = 2.2.
Q23 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · numerical
By Slater's rules, Z_eff for a 2p electron of fluorine (F, Z = 9, config 1s²2s²2p⁵) is:
A. 5.2 ✓ Correct
B. 3.8
C. 9.0
D. 4.55
Solution: σ = (6 others in n=2)×0.35 + (2 in n=1)×0.85 = 2.1 + 1.7 = 3.8; Z_eff = 9 − 3.8 = 5.2.
Q24 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · numerical
The increasing order of size for Na, Mg, Al (period 3) is:
A. Al < Mg < Na ✓ Correct
B. Mg < Al < Na
C. Al < Na < Mg
D. Na < Mg < Al
Solution: Radius decreases across the period ⇒ Al smallest, Na largest.
Q25 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · numerical
The order of atomic radius down group 1 (Li, Na, K, Rb) is:
A. K < Na < Li < Rb
B. Li < Na < K < Rb ✓ Correct
C. Rb < K < Na < Li
D. Na < Li < K < Rb
Solution: Radius increases down a group as shells are added.
Q26 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · numerical
By Slater's rules, Z_eff on a 2p electron of carbon (C, Z = 6) is:
A. 3.25 ✓ Correct
B. 4.15
C. 2.75
D. 6.0
Solution: σ = (3 others in n=2)×0.35 + (2 in n=1)×0.85 = 1.05 + 1.70 = 2.75; Z_eff = 6 − 2.75 = 3.25.
Q27 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · numerical
The correct order of ionic radii Cl⁻, K⁺, Ca²⁺ (all isoelectronic with Ar) is:
A. Cl⁻ > K⁺ > Ca²⁺ ✓ Correct
B. K⁺ > Cl⁻ > Ca²⁺
C. Ca²⁺ > K⁺ > Cl⁻
D. All equal
Solution: All have 18 electrons; size decreases with Z (17→20): Cl⁻ (17) > K⁺ (19) > Ca²⁺ (20).
Q28 — Atomic & Ionic Radii, Screening & Isoelectronic · medium · numerical
Which has the largest radius: Fe, Fe²⁺, Fe³⁺?
A. Fe²⁺
B. Fe³⁺
C. Fe ✓ Correct
D. All equal
Solution: The neutral atom has the most electrons and least effective pull per electron ⇒ Fe > Fe²⁺ > Fe³⁺.
Q29 — Atomic & Ionic Radii, Screening & Isoelectronic · easy · numerical
The Z_eff on the outer electron generally ______ across a period, causing the radius to shrink:
A. Stays constant
B. Decreases
C. Increases ✓ Correct
D. Becomes zero
Solution: Each added proton is poorly shielded by same-shell electrons, so Z_eff rises across the period.
Q30 — Atomic & Ionic Radii, Screening & Isoelectronic · hard · numerical
By Slater's rules, Z_eff for the 3s electron of magnesium (Mg, Z = 12) is:
A. 12.0
B. 9.15
C. 3.85
D. 2.85 ✓ Correct
Solution: σ = (1 other in n=3)×0.35 + (8 in n=2)×0.85 + (2 in n=1)×1.0 = 0.35 + 6.8 + 2.0 = 9.15; Z_eff = 2.85.