
Total number of orbitals associated with the 3rd shell will be ____.
A) 2
B) 4
C) 9
D) 3
Answer
219.9k+ views
Hints: The Number of orbitals can be easily calculated according to the orbit concerned. First, we need to calculate the number of subshells, then the number of orbitals, and the number of elements.
Formula used:
Number of orbitals = ${n^2}$, where $n$ is the number of orbit(s).
Complete step by step solution:
Firstly, Rutherford stated that the electrons revolve around the nucleus. Later, Bohr came to the conclusion that the electrons revolve around the nucleus but in a stationary orbit and never jump into other orbits until they are forced to do so, like supplying energy to it or absorbing energy from it. After that Bohr stated that there are orbitals associated with respective orbits namely s (sharp), p (principle), d (diffused), f (fundamental), and so on. These orbitals have their respective limits of holding electrons such as s=2, p=6, d=10, f=14, and so on.
Each orbital will contain half shells to their limited electron numbers. Each shell of orbitals contains 2 electrons of opposite spin. The first orbit contains only s orbital; 2nd orbit contains s, p orbitals; 3rd orbit contains s, p, d orbitals; 4th orbit contains s, p, d, f orbitals and it continues further. Now the number of orbitals concerned for a particular can be determined by the square of the number of that orbit.
Since, for the first orbit, $n = 1$
Number of subshell $ = {n^2} = {1^2} = 1$
Hence, the 9 orbitals are associated with the 3rd orbit i.e. one s, three p, and five d orbitals.
For 2nd orbit, $n = 2$
Number of subshell $ = {2^2} = 4$
For 3rd orbit, $n = 3$
Number of subshell $ = {3^2} = 9$
Note: There is only one orbital in the 2s subshell. But as there are three directions in which a p-orbital can point, there are three orbitals in the 2p subshell. One component of these orbitals is oriented along the X-axis, another along the Y-axis, and the last one is along the Z-axis of a coordinate system and these orbitals are known as the 2px, 2py, and 2pz orbitals.
Formula used:
Number of orbitals = ${n^2}$, where $n$ is the number of orbit(s).
Complete step by step solution:
Firstly, Rutherford stated that the electrons revolve around the nucleus. Later, Bohr came to the conclusion that the electrons revolve around the nucleus but in a stationary orbit and never jump into other orbits until they are forced to do so, like supplying energy to it or absorbing energy from it. After that Bohr stated that there are orbitals associated with respective orbits namely s (sharp), p (principle), d (diffused), f (fundamental), and so on. These orbitals have their respective limits of holding electrons such as s=2, p=6, d=10, f=14, and so on.
Each orbital will contain half shells to their limited electron numbers. Each shell of orbitals contains 2 electrons of opposite spin. The first orbit contains only s orbital; 2nd orbit contains s, p orbitals; 3rd orbit contains s, p, d orbitals; 4th orbit contains s, p, d, f orbitals and it continues further. Now the number of orbitals concerned for a particular can be determined by the square of the number of that orbit.
Since, for the first orbit, $n = 1$
Number of subshell $ = {n^2} = {1^2} = 1$
Hence, the 9 orbitals are associated with the 3rd orbit i.e. one s, three p, and five d orbitals.
For 2nd orbit, $n = 2$
Number of subshell $ = {2^2} = 4$
For 3rd orbit, $n = 3$
Number of subshell $ = {3^2} = 9$
Note: There is only one orbital in the 2s subshell. But as there are three directions in which a p-orbital can point, there are three orbitals in the 2p subshell. One component of these orbitals is oriented along the X-axis, another along the Y-axis, and the last one is along the Z-axis of a coordinate system and these orbitals are known as the 2px, 2py, and 2pz orbitals.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

