
Electronic configuration $\left[ {Kr} \right]4{d^{10}}$ belongs to:
A) Period (4) Group (10) Block (d)
B) Period (5) Group (12) Block (d)
C) Period (5) Group (10) Block (d)
D) Period (4) Group (12) Block (d)
Answer
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Hint:In this question, we have to locate the position of the element in the periodic table. It can be done by keeping the following points in mind:
The principal quantum number of the valence shell represents the period of the element.
The sub-shell in which the last electron is filled corresponds to the block of the element.
Group of the element is predicted from the electrons present in the outermost (n) or penultimate (n-1) shell as mentioned below:
-For s-block elements: group number is equal to number of s electrons
-For p- block elements: group number is equal to $10 + $ number of ns and np electrons
-For d- block elements: group number is equal to sum of the number of (n-1) d and ns electrons
-For f- block elements: group number is three.
Complete solution:
We are given with electronic configuration $\left[ {Kr} \right]4{d^{10}}$:
Firstly, predicting the period of the element,
We know that period is predicted by the principal quantum number of the valence shell. Since the outermost electron enters the fourth shell, the element belongs to the fourth period.
Now, predicting the block of the element,
Since, the last electron is filled in 4d-subshell, the element belongs to the d block.
Now, predicting the group of the element,
We know that for d- block elements: group number is equal to sum of the number of (n-1) d and ns electrons.
For the element given in question,
ns electrons $ = \,0$
(n-1) d electrons $ = \,10$
Hence, group number $ = \,0 + \,10 = \,10$
So, the element belongs to the 4th period, d block and group 10.
Hence, A is the correct option.
Note:We must be careful while doing these types of questions because certain times the electronic configuration is provided in a way where the orbital in which the last electron goes is present in between the configuration and in the right most end which we normally check. For example, the electronic configuration of iron is sometimes given as $\left[ {Ar} \right]$$4{s^2}\,3{d^6}$.
The principal quantum number of the valence shell represents the period of the element.
The sub-shell in which the last electron is filled corresponds to the block of the element.
Group of the element is predicted from the electrons present in the outermost (n) or penultimate (n-1) shell as mentioned below:
-For s-block elements: group number is equal to number of s electrons
-For p- block elements: group number is equal to $10 + $ number of ns and np electrons
-For d- block elements: group number is equal to sum of the number of (n-1) d and ns electrons
-For f- block elements: group number is three.
Complete solution:
We are given with electronic configuration $\left[ {Kr} \right]4{d^{10}}$:
Firstly, predicting the period of the element,
We know that period is predicted by the principal quantum number of the valence shell. Since the outermost electron enters the fourth shell, the element belongs to the fourth period.
Now, predicting the block of the element,
Since, the last electron is filled in 4d-subshell, the element belongs to the d block.
Now, predicting the group of the element,
We know that for d- block elements: group number is equal to sum of the number of (n-1) d and ns electrons.
For the element given in question,
ns electrons $ = \,0$
(n-1) d electrons $ = \,10$
Hence, group number $ = \,0 + \,10 = \,10$
So, the element belongs to the 4th period, d block and group 10.
Hence, A is the correct option.
Note:We must be careful while doing these types of questions because certain times the electronic configuration is provided in a way where the orbital in which the last electron goes is present in between the configuration and in the right most end which we normally check. For example, the electronic configuration of iron is sometimes given as $\left[ {Ar} \right]$$4{s^2}\,3{d^6}$.
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