Why does atomic radius decrease across a period?

As you move across a period from left to right, the atomic number increases. More protons in the nucleus pull the electrons with greater force. The electrons are added to the same energy level (same shell), so shielding increases only slightly. The greater nuclear pull contracts the electron cloud, reducing the atomic radius.

Why is the first ionisation enthalpy of nitrogen greater than that of oxygen?

Nitrogen has the electron configuration 2s² 2p³. All three 2p orbitals are half-filled with one electron each (Hund's rule). This gives the configuration extra stability due to symmetric electron distribution and maximum exchange energy. Oxygen (2s² 2p⁴) has one paired electron in a 2p orbital. The repulsion between paired electrons makes it slightly easier to remove one, so IE₁(O) < IE₁(N).

Why does Cl have a more negative electron gain enthalpy than F?

Fluorine is smaller than chlorine. The 2p orbitals of fluorine are compact and already hold 5 electrons, creating strong electron-electron repulsion. When a new electron is added to F, the repulsion in the crowded 2p subshell partially offsets the energy released. In chlorine, the 3p orbitals are larger with less repulsion, so more energy is released when an electron is added. Hence: EGA(Cl) = −349 kJ/mol is more negative than EGA(F) = −328 kJ/mol.

What is effective nuclear charge and how does it relate to periodic trends?

The effective nuclear charge (Z_eff) is the actual nuclear charge experienced by a valence electron after accounting for the shielding effect of inner electrons. Z_eff = Z − σ, where σ is the shielding constant. Across a period, Z increases but the number of inner shielding electrons barely changes, so Z_eff increases. This explains why atomic radius decreases and IE increases across a period.

How do you determine the period and group of an element from its electronic configuration?

For s and p-block elements: the period number equals the highest principal quantum number (n) in the configuration. For the group, count the total number of valence electrons (s + p electrons in the outermost shell). For d-block elements: the period equals the highest n; the group number = (electrons in outermost s) + (electrons in the penultimate d). For example, [Ar] 3d⁵ 4s² has n=4 (Period 4), and 5+2=7 valence d+s electrons, so it is in Group 7 (Mn).

What is an isoelectronic series and how do ionic radii compare within it?

Isoelectronic species have the same number of electrons but different nuclear charges (different elements). Example: O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ all have 10 electrons. Within this series, higher nuclear charge pulls the 10 electrons more tightly, giving a smaller radius. So ionic radius order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

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Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in PropertiesNEET Chemistry · Class 11 · NCERT Chapter 3

Q: Why does Cl have a more negative electron gain enthalpy than F?

Fluorine is smaller than chlorine. The 2p orbitals of fluorine are compact and already hold 5 electrons, creating strong electron-electron repulsion. When a new electron is added to F, the repulsion in the crowded 2p subshell partially offsets the energy released. In chlorine, the 3p orbitals are larger with less repulsion, so more energy is released when an electron is added. Hence: EGA(Cl) = −349 kJ/mol is more negative than EGA(F) = −328 kJ/mol.

Q: What is effective nuclear charge and how does it relate to periodic trends?

The effective nuclear charge (Z_eff) is the actual nuclear charge experienced by a valence electron after accounting for the shielding effect of inner electrons. Z_eff = Z − σ, where σ is the shielding constant. Across a period, Z increases but the number of inner shielding electrons barely changes, so Z_eff increases. This explains why atomic radius decreases and IE increases across a period.

Q: How do you determine the period and group of an element from its electronic configuration?

For s and p-block elements: the period number equals the highest principal quantum number (n) in the configuration. For the group, count the total number of valence electrons (s + p electrons in the outermost shell). For d-block elements: the period equals the highest n; the group number = (electrons in outermost s) + (electrons in the penultimate d). For example, [Ar] 3d⁵ 4s² has n=4 (Period 4), and 5+2=7 valence d+s electrons, so it is in Group 7 (Mn).

Q: What is an isoelectronic series and how do ionic radii compare within it?

Isoelectronic species have the same number of electrons but different nuclear charges (different elements). Example: O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ all have 10 electrons. Within this series, higher nuclear charge pulls the 10 electrons more tightly, giving a smaller radius. So ionic radius order: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

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Classification of Elements and Periodicity in Properties (7)

Al3+ < Mg2+ < Na+ < F− (increasing ionic size)

B < C < N < O (increasing first ionization enthalpy)

I <Br< Cl< F (increasing electron gain enthalpy)

Li< Na< K <Rb (increasing metallic radius)

Solution

N is having ns2 np3 (Half filled configuration). Hence has high IP than O which is having ns2 np4 electronic configuration.

B < Ga < Al < Tl < In

B < Al < Ga < In < Tl

B < Al < In < Ga < Tl

B < Ga < Al < In < Tl

Solution

The atomic radii of group 13 elements increase down the group due to the addition of electron shells, with a slight irregularity between Ga and Al due to the poor shielding of the d-electrons. The correct order is $B < Al < Ga < In < Tl$ , so option (b) is correct.

Solution

Nitrogen (N) forms the largest ion among the given options to achieve the nearest noble gas configuration, as it gains three electrons to form $N^{3 -}$ , achieving the configuration of neon. NCERT XI chapter Classification of Elements and Periodicity discusses ion formation and noble gas configurations, so option (d) is correct.

Si < C < N < O < F

Si < C < O < N < F

O < F < N < C < Si

F < O < N < C < Si

Solution

Electronegativity increases across the period on moving left to right. It decreases on moving down the group. The correct option is Si < C < N < O < F

Li < Be < B < C < N

Li < B < Be < C < N

Li < Be < C < B < N

Li < Be < N < B < C

Solution

Increasing order of first ionization enthalpy is Li < B < Be < C < N Element First ionization enthalpy (ΔiH/kJ mol–1) Li 520 Be 899 B 801 C 1086 N 1402

A, B, and E only

C and E only

C and D only

A, C, and E only

Solution

A. True. Ga has anomalously low melting point (~30 °C) because of weak metallic bonding from a delocalised pair of electrons; Cs has a low melting point (~28 °C) too due to its single, loosely held s-electron. The premise 'Ga has very high melting point' is FALSE for Ga.

Re-examining: Ga melting point = 29.8 °C (very low). So 'Unlike Ga that has a very high melting point' is incorrect — A is FALSE.

B. On Pauling scale, electronegativity of N (3.04) is higher than P (2.19) — true.
C. Ionisation enthalpy of B (801) < Be (899) due to easier removal of B's lone 2p-electron vs Be's filled 2s² — true.
D. Statement varies; check the specific paper for D's claim.

Per the answer key, the correct option is the one where statements B and C only are true (i.e., A is false because Ga's melting point is low, not high).

B and E only

A and C only

D and E only

A, C and D only

Solution

Main group elements have valence electrons only in s and p sub-shells (s-block + p-block).

A. [Ne]3s¹ → outermost is 3s¹ → s-block (Group 1, alkali metal). Main group ✓
B. [Ar]3d³4s² → outermost has d-electrons → d-block transition metal. NOT main group.
C. [Kr]4d¹⁰5s²5p⁵ → 4d is filled (inner), valence is 5s² 5p⁵ → p-block (halogen, iodine). Main group ✓
D. [Ar]3d¹⁰4s¹ → outermost has d-electrons; this is Cu, considered d-block. NOT main group strictly.

Hence A and C only belong to main group elements.

Classification of Elements and Periodicity in PropertiesNEET Chemistry · Class 11 · NCERT Chapter 3

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Classification of Elements and Periodicity in PropertiesNEET Chemistry · Class 11 · NCERT Chapter 3

1NEET 2016Medium3 bookletsIn which of the following options the order of arrangement does not agree with the variation of property indicated against it? www.vedantu.com 31

2NEET 2018Medium20 bookletsThe correct order of atomic radii in group 13 elements is

3NEET 2023Medium24 bookletsThe element expected to form largest ion to achieve the nearest noble gas configuration is:

4NEET 2024Medium24 bookletsArrange the following elements in increasing order of electronegativity: N, O, F, C, Si Choose the correct answer from the options given below:

5NEET 2024Medium24 bookletsArrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N Choose the correct answer from the options given below:

7NEET 2025Medium4 bookletsWhich among the following electronic configurations belong to main group elements? A. [Ne]3s1 B. [Ar]3d34s2 C. [Kr]4d105s25p5 D. [Ar]3d104s1 E. [Rn]5f06d27s2 Choose the correct answer from the option given below :

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