In the long form of the periodic table, the valence shell electronic configuration of \[5{s^2}5{p^4}\] corresponds to the elements present in
A. Group 16 and period 6
B. Group 17 and period 5
C. Group 16 and period 5
D. Group 17 and period 6
Answer
650.7k+ views
Hint: We must know that the electronic configuration of an element describes two things. First, the number of electron shells and another is the number of valence electrons in one atom of the element. And the electronic configuration of an atom directly relates to its position in the Modern Periodic Table.
Complete answer:
The electronic configuration provided in the question is \[5{s^2}5{p^4}\]. So, from the electronic configuration, we can say that the principal quantum number of elements is 5 and the valence electrons are 6.
The principal quantum number of 5 corresponds to the fifth period of the modern periodic table and 6 valence electrons correspond to group 16 of the periodic table.
So, we can conclude that the electronic configuration provided in the question is the same as the electronic configuration of tellurium.
Atomic number of tellurium is 52 and its electronic configuration is as below:
\[\left[ {Kr} \right]{\text{ }}4{d^{10}}5{s^2}5{p^4}\]
We know the Tellurium element belongs to group 16 and lies in period 5.
So, the element is Tellurium (\[Te\]).
Thus, we can conclude that in the long form of the periodic table, the valence shell electronic configuration of \[5{s^2}5{p^4}\] corresponds to the elements present in group 16 and period 5.
Hence, the correct option is option C.
Note:
We need to have knowledge about determining principal quantum number and valence electrons as it will help to obtain the information of groups and periods in the modern periodic table. In the electronic configuration, n denotes the principal quantum number, and the value of n ranges starts from 1 to the shell containing the outermost electron of that atom. For example, cesium (Cs) have the outermost valence electron in the shell with energy level 6, so an electron in cesium can have an n value from 1 to 6.
Complete answer:
The electronic configuration provided in the question is \[5{s^2}5{p^4}\]. So, from the electronic configuration, we can say that the principal quantum number of elements is 5 and the valence electrons are 6.
The principal quantum number of 5 corresponds to the fifth period of the modern periodic table and 6 valence electrons correspond to group 16 of the periodic table.
So, we can conclude that the electronic configuration provided in the question is the same as the electronic configuration of tellurium.
Atomic number of tellurium is 52 and its electronic configuration is as below:
\[\left[ {Kr} \right]{\text{ }}4{d^{10}}5{s^2}5{p^4}\]
We know the Tellurium element belongs to group 16 and lies in period 5.
So, the element is Tellurium (\[Te\]).
Thus, we can conclude that in the long form of the periodic table, the valence shell electronic configuration of \[5{s^2}5{p^4}\] corresponds to the elements present in group 16 and period 5.
Hence, the correct option is option C.
Note:
We need to have knowledge about determining principal quantum number and valence electrons as it will help to obtain the information of groups and periods in the modern periodic table. In the electronic configuration, n denotes the principal quantum number, and the value of n ranges starts from 1 to the shell containing the outermost electron of that atom. For example, cesium (Cs) have the outermost valence electron in the shell with energy level 6, so an electron in cesium can have an n value from 1 to 6.
Recently Updated Pages
Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Father s age is three times the sum of the ages of-class-11-maths-CBSE

Give a comparative account of the classes of kingdom class 11 biology CBSE

The ceiling of a long hall is 25m high What is the class 11 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Draw a welllabelled diagram of a plant cell class 11 biology CBSE

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

