(C) The valence shell electronic configuration is given as $5s^2\,5p^4$. The principal quantum number $n = 5$ indicates that the element belongs to pe

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(C) The valence shell electronic configuration is given as $5s^2\,5p^4$. The principal quantum number $n = 5$ indicates that the element belongs to period $5$. The total number of valence electrons is $2 + 4 = 6$. For $p$-block elements,the group number is calculated as $10 + \text{valence electrons} = 10 + 6 = 16$. Therefore,the element belongs to group $16$ and period $5$.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Extended or long form of periodic table

Medium

In the long form of the periodic table,the valence shell electronic configuration of $5s^2\,5p^4$ corresponds to the element present in

JEE MAIN 2015 All JEE MAIN

A
Group $16$ and period $6$
B
Group $17$ and period $6$
C
Group $16$ and period $5$
D
Group $17$ and period $5$

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Classification of Elements and Periodicity in Properties

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Extended or long form of periodic table

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The element with atomic number $36$ belongs to $\dots$ block in the periodic table.

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Assign the position of the element having outer electronic configuration in the periodic table:
$(i)$ $ns^{2} np^{4}$ for $n=3$
$(ii)$ $(n-1)d^{2} ns^{2}$ for $n=4$
$(iii)$ $(n-2)f^{7} (n-1)d^{1} ns^{2}$ for $n=6$

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The cause of periodicity of properties is

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Determine the position of the element in the periodic table for the following outer electronic configurations:
$(i)$ $ns^2 np^4$ where $n = 3$
$(ii)$ $(n-1)d^2 ns^2$ where $n = 4$
$(iii)$ $(n-2)f^7 (n-1)d^1 ns^2$ where $n = 6$

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In the modern periodic table,the period indicates the value of:

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In the long form of the periodic table,the valence shell electronic configuration of $5s^2\,5p^4$ corresponds to the element present in

Similar Questions

The element with atomic number $36$ belongs to $\dots$ block in the periodic table.

Assign the position of the element having outer electronic configuration in the periodic table:$(i)$ $ns^{2} np^{4}$ for $n=3$$(ii)$ $(n-1)d^{2} ns^{2}$ for $n=4$$(iii)$ $(n-2)f^{7} (n-1)d^{1} ns^{2}$ for $n=6$

The cause of periodicity of properties is

Determine the position of the element in the periodic table for the following outer electronic configurations:$(i)$ $ns^2 np^4$ where $n = 3$$(ii)$ $(n-1)d^2 ns^2$ where $n = 4$$(iii)$ $(n-2)f^7 (n-1)d^1 ns^2$ where $n = 6$

In the modern periodic table,the period indicates the value of:

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Assign the position of the element having outer electronic configuration in the periodic table:
$(i)$ $ns^{2} np^{4}$ for $n=3$
$(ii)$ $(n-1)d^{2} ns^{2}$ for $n=4$
$(iii)$ $(n-2)f^{7} (n-1)d^{1} ns^{2}$ for $n=6$

Determine the position of the element in the periodic table for the following outer electronic configurations:
$(i)$ $ns^2 np^4$ where $n = 3$
$(ii)$ $(n-1)d^2 ns^2$ where $n = 4$
$(iii)$ $(n-2)f^7 (n-1)d^1 ns^2$ where $n = 6$