NEET Previous Year Questions (2014-2024): Classification of Elements & Periodicity in Properties | Chemistry Class 11 PDF Download

2024

Q2: Arrange the following elements in increasing order of first ionization enthalpy:
Li, Be, B, C, N
Choose the correct answer from the options given below:
(a) Li < Be < B < C < N
(b) Li < B < Be < C < N
(c) Li < Be -< C < B < N
(d) Li < Be < N < B < C              [2024]
Ans: (b)
The first ionization enthalpy, also known as ionization energy, is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous phase to form a cation. The trend of first ionization energies generally increases across a period from left to right in the periodic table. This is due to the increasing nuclear charge and the decreasing atomic radius, which cause the valence electrons to be attracted more strongly to the nucleus.
However, there are notable exceptions based on the electron configuration stability and electron pairing in orbitals. Let's analyze the given elements:
Lithium (Li): Being the first element in the period, it has the smallest nuclear charge and only one electron in its outer shell, which makes it easy to remove an electron.
Beryllium (Be): This element has two electrons in the 2s orbital. The removal of one electron slightly disturbs the fully filled 2s sub-shell, creating more stability than having an unpaired electron. Therefore, Be has a higher ionization energy than Li.
Boron (B): This element has a half-filled 2p orbital configuration (one electron in the 2p orbital), which is relatively less stable compared to a full or empty p orbital, leading to a slightly lower ionization energy than Be.
Carbon (C): With two electrons in separate 2p orbitals (following Hund's rule), C experiences more effective nuclear shielding and electron-electron repulsion compared to a single electron in Boron's 2p orbital. This makes it relatively easier to remove an electron from B than from C, but harder than removing one from Be.
Nitrogen (N): It has exactly half-filled 2p orbitals, which provides extra stability and hence has a higher ionization energy than Carbon. Contrarily, the configuration of three p electrons is stable owing to the exchange energy and symmetric distribution in space.
Given these points, we can order the elements by increasing first ionization enthalpy as follows:
Li < B < Be < C < N
This matches with Option B. Thus, the correct answer is:
Option B
Li < B < Be < C < N

2022

Q1: The IUPAC name of an element with atomic number 119 is      (NEET 2022 Phase 1)
(a) ununoctium 
(b) ununennium 
(c) unnilennium 
(d) unununnium
Ans: (b)

IUPAC name of element : 119 : ununennium

Q2: Gadolinium has a low value of third ionisation enthalpy because of     (NEET 2022 Phase 1)
(a) small size
(b) high exchange enthalpy
(c) high electronegativity
(d) high basic character
Ans: (b)
Electronic configuration of Gadolinium
Gd :- [Xe] 4f⁷ 5d¹ 6s²
In case of 3^rd ionisation enthalpy electron will be removed from 5d and resultant configuration will be [Xe]4f⁷ that is stable electronic configuration as it will have high exchange energy, hence less energy will be required to remove 3^rd electron.

Q3: The correct order of first ionization enthalpy for the given four elements is :
(a) C < F < N < O
(b) C < N < F < O
(c) C < N < O < F
(d) C < O < N < F    (NEET 2022 Phase 2)
Ans: (d)

• Generally, on moving left to right in a period. First ionization enthalpy of elements increases due to increase in effective nuclear charge.
• Due to more stable half-filled outer electronic configuration (2s²2p³) of N, its first ionization enthalpy is more than O. 

So, correct order of IP is : C < O < N < F

Q4: Decrease in size from left to right in actinoid series is greater and gradual than that in lanthanoid series due to
(a) 5f orbitals have greater shielding effect
(b) 4f orbitals are penultimate
(c) 4f orbitals have greater shielding effect
(d) 5f orbitals have poor shielding effect    (NEET 2022 Phase 2)
Ans: (d)
Due to more diffused nature of 5f orbitals as compared to 4f orbitals the shielding effect of 5f is poor, resulting in the decrease in size from left to right in actinoid series which is greater and gradual than that in lanthanoid series

Q5: Fluorine is a stronger oxidising agent than chlorine because :   (NEET 2022 Phase 2)
(a) F-F bond has a low enthalpy of dissociation.
(b) Fluoride ion (F^$-$) has high hydration enthalpy.
(c) Electron gain enthalpy of fluorine is less negative than chlorine.
(d) Fluorine has a very small size.
Choose the most appropriate answer from the options given :
(a) (b) and (c) only
(b) (a) and (b) only
(c) (a) and (c) only
(d) (a) and (d) only
Ans: (b)
Fluorine is a stronger oxidising agent than chlorine due to
(i) Low dissociation enthalpy of F-F bond
(ii) High hydration enthalpy of F− ion

2020

Q1: Identify the incorrect match.     (NEET 2020)

(a) (3), (iii)
(b) (4), (iv)
(c) (1), (i)
(d) (2), (ii)
Ans: b
101 - Unnilunium - Mendelevium
103 - Unniltrium - Lawrencium
106 - Unnilhexium - Seaborgium
111 - Unununium - Roentgenium
110 - Ununnilium - Darmstadtium

Q2: Match the following :    (NEET 2020)

Which of the following is correct option?

(a) 

(b)

(c)

(d)

Ans: (a)
CO : Neutral oxide
BaO : Basic oxide
Al₂O₃ : Amphoteric oxide
Cl₂O₇ : Acidic oxide

2019

Q1: For the second period elements the correct increasing order of first ionisation enthalpy is:    (NEET 2019)
(a) Li < Be < B < C < N < O < F < Ne
(b) Li < B < Be < C < O < N < F < Ne
(c) Li < B < Be < C < N < O < F < Ne
(d) Li < Be < B < C < O < N < F < Ne
Ans: (b)

Ionisation enthalpy increases in a period from left to right, because of increased nuclear charge and decrease in atomic radii. But because of half filled orbitals of Be, N and fully filled orbitals of Ne, stability of those atoms inceases.
So, the correct increasing order of first ionisation enthalpy will be :
Li < B < Be < C < O < N < F < Ne

2018

Q1: The correct order of atomic radii in group 13 elements is             (NEET 2018)
(a) B < Al < In < Ga < Tl
(b) B < Al < Ga < In < Tl
(c) B < Ga < Al < Tl < In
(d) B < Ga < Al < In < Tl
Ans: (d)
Ga is slightly smaller than Al due poor shielding of d e– so Zeff increasing.
So, Atomic size : B < Ga < Al < In < Tl

Q2: Which of the following oxides is most acidic in nature?            (NEET 2018)
(a) MgO
(b) BeO
(c) BaO
(d) CaO
Ans: (b)
In metals, on moving down the group, metallic character increases, so basic nature increases hence most acidic will be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic oxides.

2017

Q1: The element Z = 114 has been discovered recently. It will belong to which of the following family/groupand electronic configuration ?    (NEET 2017)
(a)
(b)
(c)
(d)
Ans: (a)

2015

Q1: The species Ar, K⁺ and Ca²⁺ contain the same number of electrons. In which order do their radii increase ?    
(a) K⁺< Ar < Ca²⁺
(b) Ar < K⁺< Ca²⁺
(c) Ca²⁺ < Ar < K⁺
(d) Ca²⁺ < K⁺ < Ar      (AIPMT 2015 Cancelled Paper)
Ans: (d)
⇒ Ca²⁺ < K⁺ < Ar
Ar, K⁺ and Ca²⁺ are isoelectronic i.e with the same number of electrons, 18. For isoelectronic species, ionic radii decrease with increases in effective (relative) positive charge. Also Ar, K and Ca belong to the same period.

2014

Q.9. Which of the following orders of ionic radii is correctly represented?    (NEET 2014)
(a) F^- > O^2- > Na⁺
(b) Al³⁺ > Mg²⁺ > N^3- 
(c) H^- > H^. > H
(d) Na⁺ > F > O^2- 
Ans: (c)
Cation loose electrons are smaller in size than the parent atom, where anions gain electrons are larger in size than the parent atom.
Hence the order is H^- > H^. > H⁺