The element with atomic number 31 belongs to
which group and period respectively?
(1) 13, 4
(2) 14,3
(3) 4, 13
(4) 4,8
Answers
Answer:
It belongs to group 13 and period 4 ( no.1).
Answer:
(((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 .