Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0, \[1,{\text{ }}2,{\text{ }}......{\text{ }}n{\text{ }} + {\text{ }}1\], where \[n\] is the principal quantum number. Then, the element with atomic number:
(1) \[13\] has a half-filled valence subshell
(2) \[9\] is the first alkali metal
(3) \[8\] is the first noble gas
(4) \[6\] has a \[2p\] -valence subshell
Answer
301.5k+ views
Hint: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
Complete Step by Step Solution:
The electron configuration of an element is determined by the value of\[n\]. The first orbital, or energy level, can hold a maximum of \[2\] electrons. The second level can hold a maximum of \[8\] electrons, the third level can hold a maximum of \[18\] electrons, and so on. The element with atomic number n will have an electron configuration that corresponds to the level that can hold \[n\] electrons. The ionisation energy of an element is the amount of energy required to remove an electron from the atom. The higher the ionisation energy, the more difficult it is to remove an electron. The ionisation energy of an element with atomic number \[n\] is determined by the value of 2, 8, 18. In general, the higher the value of 2, 8, 18, the higher the ionisation energy. The atomic radius of an element is the distance from the centre of the nucleus to the outermost electron. The atomic radius of an element with atomic number \[n\] is determined by the value of \[n\]. In general, the higher the value of\[n\], the larger the atomic radius.
Option 1) \[13{\text{ }}:{\text{ }}1s21p62s21d3\] is not half filled.
Option 2) \[9{\text{ }}:{\text{ }}1s21p62s1\] is the first alkali metal because after losing one electron, it will achieve first noble gas configuration.
Option 3) \[8{\text{ }}:{\text{ }}1s21p6\]is the first noble gas because after \[1p6{\text{ }}e-\] will enter \[2s\] hence new period.
Option 4) \[6{\text{ }}:{\text{ }}1{\text{ }}2{\text{ }}1p4\]has \[1p\] valence subshell.
(1) \[_{13}X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^5}\;-\] Half filled.
(2) \[_9X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^1}\;-\] Not alkali metal
(3) \[_8X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;-\] Second Noble gas
Option 1 is the right answer.
Note: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
Complete Step by Step Solution:
The electron configuration of an element is determined by the value of\[n\]. The first orbital, or energy level, can hold a maximum of \[2\] electrons. The second level can hold a maximum of \[8\] electrons, the third level can hold a maximum of \[18\] electrons, and so on. The element with atomic number n will have an electron configuration that corresponds to the level that can hold \[n\] electrons. The ionisation energy of an element is the amount of energy required to remove an electron from the atom. The higher the ionisation energy, the more difficult it is to remove an electron. The ionisation energy of an element with atomic number \[n\] is determined by the value of 2, 8, 18. In general, the higher the value of 2, 8, 18, the higher the ionisation energy. The atomic radius of an element is the distance from the centre of the nucleus to the outermost electron. The atomic radius of an element with atomic number \[n\] is determined by the value of \[n\]. In general, the higher the value of\[n\], the larger the atomic radius.
Option 1) \[13{\text{ }}:{\text{ }}1s21p62s21d3\] is not half filled.
Option 2) \[9{\text{ }}:{\text{ }}1s21p62s1\] is the first alkali metal because after losing one electron, it will achieve first noble gas configuration.
Option 3) \[8{\text{ }}:{\text{ }}1s21p6\]is the first noble gas because after \[1p6{\text{ }}e-\] will enter \[2s\] hence new period.
Option 4) \[6{\text{ }}:{\text{ }}1{\text{ }}2{\text{ }}1p4\]has \[1p\] valence subshell.
(1) \[_{13}X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^5}\;-\] Half filled.
(2) \[_9X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^1}\;-\] Not alkali metal
(3) \[_8X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;-\] Second Noble gas
Option 1 is the right answer.
Note: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
Recently Updated Pages
Normality of 03 M phosphorus acid H3PO3 is A 05 B 06 class 11 chemistry JEE_Main

A molecule with highest bond energy A Fluorine B Chlorine class 11 chemistry JEE_Main

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

Covalent compounds generally have low melting and boiling class 11 chemistry JEE_Main

When an acid reacts with a metal carbonate or metal class 11 chemistry JEE_Main

The degeneracy of hydrogen atom that has equal energy class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2026-27 Free PDF Download (Sign-in Required)

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Different Types of Solutions in Chemistry

What Are Elastic Collisions in One Dimension?

Effective Nuclear Charge for JEE

