
What are the total number of orbitals and electrons for m=2 , if there are 30 protons in an atom?
A. 6 orbitals, 12 electrons
B. 5 orbitals, 10 electrons
C. 7 orbitals, 14 electrons
D. 4 orbitals, 8 electrons
Answer
220.2k+ views
Hint: Each electron has magnetic quantum number zero and number of protons is equal to the atomic number. From that, orbitals in it can be determined by electronic configuration, where the number of electrons is just double the number of orbitals.
Complete step-by-step answer:
In an atom, an atomic number is the same as the number of protons present inside the nucleus of that atom. We are given that number of protons are 30 in an atom such that its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}\] because number of electrons are equal to number of protons.
Here, the magnetic quantum number determines the orientation of different orbitals in spatial arrangement. For every corresponding n and l values of an electron, magnetic quantum numbers always have a zero. This means they are spherical shape orbitals. M stands for magnetic quantum number which goes from -1 to+1 , where l is angular quantum number.
We know now that, m value for s(l=0) orbital is only zero, for p(l=1) orbital it can be -1, 0 or +1 and for d orbitals(l=2) it can be -2, -1, 0, +1, +2.
The total number of possible orbitals with the same value of l are 2l +1. So, we can say that there is one s-orbital for l=0, three p-orbitals for l=1 and five d-orbitals for l=2.
As per the electronic configuration, every orbital is fully filled and each has an electron having magnetic quantum number zero. Thus, the total number of orbitals with m=0 are 7. Since every orbital occupies two electrons as per Pauli’s exclusion principle, the number of electrons are 14 in all the 7 orbitals.
Hence, the correct option is (C).
Note: The magnetic quantum number is used to distinguish the orbitals available within a subshell and to calculate the azimuthal component of the orientation of the orbital in space i.e. corresponding l values.
Complete step-by-step answer:
In an atom, an atomic number is the same as the number of protons present inside the nucleus of that atom. We are given that number of protons are 30 in an atom such that its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}\] because number of electrons are equal to number of protons.
Here, the magnetic quantum number determines the orientation of different orbitals in spatial arrangement. For every corresponding n and l values of an electron, magnetic quantum numbers always have a zero. This means they are spherical shape orbitals. M stands for magnetic quantum number which goes from -1 to+1 , where l is angular quantum number.
We know now that, m value for s(l=0) orbital is only zero, for p(l=1) orbital it can be -1, 0 or +1 and for d orbitals(l=2) it can be -2, -1, 0, +1, +2.
The total number of possible orbitals with the same value of l are 2l +1. So, we can say that there is one s-orbital for l=0, three p-orbitals for l=1 and five d-orbitals for l=2.
As per the electronic configuration, every orbital is fully filled and each has an electron having magnetic quantum number zero. Thus, the total number of orbitals with m=0 are 7. Since every orbital occupies two electrons as per Pauli’s exclusion principle, the number of electrons are 14 in all the 7 orbitals.
Hence, the correct option is (C).
Note: The magnetic quantum number is used to distinguish the orbitals available within a subshell and to calculate the azimuthal component of the orientation of the orbital in space i.e. corresponding l values.
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

How to Convert a Galvanometer into an Ammeter or Voltmeter

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

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

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

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

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

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

