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Classification of Elements and Periodicity in Properties

RE-NEET (UG) 2024MatchingEasy📈 Topic in 25% papers+4−1
Ques. 1RE-NEET (UG) 2024

Match List I with List II: Choose the correct answer from the options given below:

  1. A-I, B-II, C-IV, D-III
  2. A-I, B-IV, C-III, D-II
  3. A-I, B-IV, C-II, D-III
  4. A-IV, B-I, C-II, D-III

Answer

(3) A-I, B-IV, C-II, D-III

Solution

A. Lanthanoid: Ce (Z = 58) (I).

B. d-block element: Mn (Z = 25) (IV).

C. p-block element: As (group 15) (II).

D. s-block element: Cs (group 1) (III).

Thus A-I, B-IV, C-II, D-III.

Answer: (3)

Keep practising — Classification of Elements and Periodicity in Properties

Ques. 2NEET (UG) 2023

The element expected to form largest ion to achieve the nearest noble gas configuration is

  1. F
  2. N
  3. Na
  4. O
Ques. 3NEET (UG) 2022

The IUPAC name of an element with atomic number 119 is

  1. ununennium
  2. unnilennium
  3. unununnium
  4. ununoctium
Ques. 4NEET (UG) 2022 (Phase 2)

The correct order of first ionization enthalpy for the given four elements is :

  1. C < F < N < O
  2. C < N < F < O
  3. C < N < O < F
  4. C < O < N < F
Ques. 5NEET (UG) 2022 (Phase 2)

If first ionization enthalpies of elements X and Y are 419 kJ mol-1 and 590 kJ mol-1, respectively and second ionization enthalpies of X and Y are 3069 kJ mol-1 and 1145 kJ mol-1, respectively. Then correct statement is :

  1. Both X and Y are alkaline earth metals
  2. X is an alkali metal and Y is an alkaline earth metal
  3. X is an alkaline earth metal and Y is an alkali metal
  4. Both X and Y are alkali metals
Ques. 6NEET (UG) 2020

Identify the incorrect match.

  1. (b)-(ii)
  2. (c)-(iii)
  3. (d)-(iv)
  4. (a)-(i)
Ques. 7NEET (UG) 2020

Match the following : Which of the following is correct option?

  1. (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
  2. (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)
  3. (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
  4. (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
Ques. 8NEET (UG) 2019

For the second period elements the correct increasing order of first ionization enthalpy is :

  1. Li < B < Be < C < N < O < F < Ne
  2. Li < Be < B < C < O < N < F < Ne
  3. Li < B < Be < C < O < N < F < Ne
  4. Li < Be < B < C < N < O < F < Ne
Ques. 9NEET 2017

The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration ?

  1. Carbon family, [Rn] 5f14 6d10 7s2 7p2
  2. Oxygen family, [Rn] 5f14 6d10 7s2 7p4
  3. Nitrogen family, [Rn] 5f14 6d10 7s2 7p6
  4. Halogen family, [Rn] 5f14 6d10 7s2 7p5
Ques. 10NEET 2016 (Phase 1)

In which of the following options the order of arrangement does not agree with the variation of property indicated against it ?

  1. I < Br < Cl < F (increasing electron gain enthalpy)
  2. Li < Na < K < Rb (increasing metallic radius)
  3. Al3+ < Mg2+ < Na+ < F- (increasing ionic size)
  4. B < C < N < O (increasing first ionisation enthalpy)
Ques. 11AIPMT 2015

The formation of the oxide ion, O2-(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g) + e- O-(g); H = -141 kJ mol-1 O-(g) + e- O2-(g); H = +780 kJ mol-1 Thus, process of formation of O2- in gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that,

  1. O- ion has comparatively smaller size than oxygen atom
  2. oxygen is more electronegative
  3. addition of electron in oxygen results in larger size of the ion
  4. electron repulsion outweighs the stability gained by achieving noble gas configuration.

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