
Which of the following can be represented by the configuration \[\left[ {{\rm{Kr}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\]?
A. Ca
B. Sr
C. Ba
D. Ra
Answer
163.5k+ views
Hint: Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium have similar properties and hence are placed in group 2 of the periodic table. These metals are known as alkaline earth metals. Their common outer electronic configuration is \[{\rm{n}}{{\rm{s}}^{\rm{2}}}\].
Complete Step by Step Answer:
The elements in which the final electron joins the s-orbital are organised in an s-block.
In this block, there are two groups, namely group-1 and group-2.
Group 1 metals involve Lithium, Sodium, Potassium, Rubidium, Cesium and Francium.
Group-2 holds Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.
These elements are the alkaline earth metals as they show alkalinity and are found in the earth.
All the given elements are alkaline earth metals belonging to group-2.
Their common outer electronic configuration is \[{\rm{n}}{{\rm{s}}^{\rm{2}}}\].
We have to find out which of the following options alkaline earth metals have configuration \[\left[ {{\rm{Kr}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
A. Ca
Its atomic number is 20.
The electronic configuration is \[\left[ {{\rm{Ar}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, A is incorrect.
B. Sr
Its atomic number is 38.
The electronic configuration is \[\left[ {{\rm{Kr}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, B is correct.
C. Ba
Its atomic number is 56.
The electronic configuration is \[\left[ {{\rm{Xe}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, C is incorrect.
D. Ra
Its atomic number is 88.
The electronic configuration is \[\left[ {{\rm{Rn}}} \right]{\rm{7}}{{\rm{s}}^{\rm{2}}}\].
So, D is incorrect.
So, option B is correct.
Note: The electronic configuration of alkaline earth metals is represented in terms of noble gas configuration. So, to find out the metal which has the given configuration we have to remember the atomic number of alkaline earth metals and noble gases. The atomic number of Kr is 36. The atomic number of Sr is 38. These two atomic numbers are close. The addition of two electrons to the 5s subshell of Kr will give the configuration of Sr.
Complete Step by Step Answer:
The elements in which the final electron joins the s-orbital are organised in an s-block.
In this block, there are two groups, namely group-1 and group-2.
Group 1 metals involve Lithium, Sodium, Potassium, Rubidium, Cesium and Francium.
Group-2 holds Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.
These elements are the alkaline earth metals as they show alkalinity and are found in the earth.
All the given elements are alkaline earth metals belonging to group-2.
Their common outer electronic configuration is \[{\rm{n}}{{\rm{s}}^{\rm{2}}}\].
We have to find out which of the following options alkaline earth metals have configuration \[\left[ {{\rm{Kr}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
A. Ca
Its atomic number is 20.
The electronic configuration is \[\left[ {{\rm{Ar}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, A is incorrect.
B. Sr
Its atomic number is 38.
The electronic configuration is \[\left[ {{\rm{Kr}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, B is correct.
C. Ba
Its atomic number is 56.
The electronic configuration is \[\left[ {{\rm{Xe}}} \right]{\rm{5}}{{\rm{s}}^{\rm{2}}}\].
So, C is incorrect.
D. Ra
Its atomic number is 88.
The electronic configuration is \[\left[ {{\rm{Rn}}} \right]{\rm{7}}{{\rm{s}}^{\rm{2}}}\].
So, D is incorrect.
So, option B is correct.
Note: The electronic configuration of alkaline earth metals is represented in terms of noble gas configuration. So, to find out the metal which has the given configuration we have to remember the atomic number of alkaline earth metals and noble gases. The atomic number of Kr is 36. The atomic number of Sr is 38. These two atomic numbers are close. The addition of two electrons to the 5s subshell of Kr will give the configuration of Sr.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

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

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
