Question

(a) How is group, period and block of elements predicted. (b) Give the electronic configuration and in terms of period group and block where would you locate the elements with Z= 17, 19. 24, 26, 29, 31,34,38, 40, 51,55,114 ,117 and 120 ?

Answer 1

(((a))) PREDICTION of PERIOD
=======================
In the electronic configuration of an atom , the principal quantum number of valance electron represents the PERIOD of the element .
example :- Ca ( atomic no = 20 )
configuration of Ca : 1s², 2s² , 2p^6 , 3s² , 3p^6, 4s² .
valance electrons = 2 { in 4S }
Principal Quantum no = 4
hence, period of Ca is 4th

PREDICTION of GROUP
======================
The orbital { S, P , d , or f } which receives the last electton represents the block of the element.
➡ The group of the element is predicted from the block of the elements as : -

(1) if the element belongs to S - block , it's group number is equal to the number of valance electrons .

(2) if the element belongs to P -block ,
Group number = 10 + number of valance electrons
= 10 +nS electrons + nP electrons

(3) if the elements belongs to d -block ,
Group number = number of electrons in (n -1)d subshell + number of electrons in nS subshell .

(4) if the element belong to f - subshell , group number is always 3 .



(((b)))
Z = 17
configuration : 1s² , 2s² , 2p^6, 3s², 3P^5
hence,
Period : 3rd
group : 10 + 2 + 5 = 17th

Z = 19
configuration : 1s² , 2s² , 2p^6, 3s² , 3p^6 , 4s¹
period : 4th
group : 1st

Z = 24
configuration: 1s², 2s² , 2p^6, 3s² , 3P^6, 3d^5 , 4s¹
period : 4th
group : 6th

Z = 26
configuration: 1s² , 2s² , 2P^6, 3s² ,3P^6, 3d^6, 4s²
period : 4th
group : 8th

Z = 29
configuration : [Ar] 3d^10 , 4s¹
period : 4th
group : 11th

Z = 31
configuration: [Ar]3d^10, 4s², 4P¹
period : 4th
group : 13th

Z = 34
configuration : [Ar] 3d^10, 4s², 4P^4
period : 4th
group: 16th

Z = 38
Configuration: [Kr] 4d^2
period : 4th
group: 2nd

Z = 40
configuration: [Kr]4d^4
period :4th
group : 4th

Z = 51
configuration: [Kr]4d^10, 5s², 5P^ 3
period : 5th
group: 15th

Z = 55
configuration: [Xe] 6s¹
period : 6th
group : 1st

Z =114
configuration : [Rn]7s², 5f^14, 6d^10,7p²
period : 7th
group : ( 2 + 10 + 2) th = 14th

Z = 117
configuration : [Rn]7s², 5f¹⁴, 6d^10, 7P^5
period : 7th
group : 17th

Z = 120
configuration: [Rn]7s², 5f¹⁴, 6d^10, 7P^6, ... Z = 120 not exist in Periodic table
only 118 elements exist .

Answer 2

Answer:

Explanation:

For p block,just add 12.Like in case-2p⁴.Here just add 12 and we will get group.