(a) The total number of
electrons in an atom with
atomic number 110 having
(n+l) = 8
is equal to 'p
(b) The atomic number of
first atom in which the last
electron enters 5g
electronic configuration
according to
(n+l)
rule is
'Q'. Find P+Q?
Answers
Explanation:
The electronic configuration of element with at.no. 105 is
1s
2
,2s
2
p
6
,3s
2
p
6
d
10
,4s
2
p
6
d
10
4f
14
,5s
2
p
6
d
10
f
14
,6s
2
p
6
d
3
7s
2
Also we know than angular quantum no. of sub-orbits are
for s, l=0
p, l=1
d, l=2
f, l=3
in this configuration orbitals with n+l=8 is
5f,(n+l)=5+3=8, and no of electron present in 5f is 14
6d,(n+l)=6+2=8 and no of electron present in 6d is 3
so total no of electron present in n+l=8 is 14+3=17
Given : a) n+l = 8 , it have the atomic number 110
b) electronic configuration of last electron given is
5g.
Find : P+Q
Solution :
- In this we have to find the value of P and Q where P= the total number of electrons in an atom with atomic number 110 having (n+l) = 8.
- And Q = the atomic number of first atom in which the last electron enters 5g electronic configuration according to (n+l) rule.
- a) So first we have to find the value of P .
- In this, atomic no. 110 is given whose electronic configuration is [Rn] 5f¹⁴6d⁹7s¹.
- So, in this n+l = 8 js for the following orbitals -
5f¹⁴ , here l = 3 ( for f the azimuthal quantum no. is 3)
and n = 5 (principal quantum number).
n+l = 5+3 = 8. No. of electrons present in it is 14.
- Also for ,6d¹⁰ , l= 2 , n= 6 . n+l = 6+2 = 8. No. of electrons present in it is 10
- So, the total number of electrons in an atom with atomic number 110 having (n+l) = 8 , P is 14+ 10 = 24.
- b) The atomic number of first atom in which the last electron enters 5g electronic configuration according to (n+l) rule is :
- So, first we should know what is n+l rule , according to n+l rule the one with lower n+l rule value filled first.
- So, here for 5g , n+l = 5+4 = 9 (as for g , l =9 ) . Here all the orbitals will be filled whose n+l value is less than 9 and if there is equal n+l value then the orbital having lower n value filled first.
- So, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d¹⁰ 7p⁶ 8s² 5g¹ .
- When we count all the elections which is equal to 121 . And as we know no. of electrons = no. of protons and which in return equals to atomic no.
- Hence, The atomic number of first atom in which the last electron enters 5g electronic configuration according to (n+l) rule , Q is 121.
- Therefore, P+Q = 24+121 = 145.