How would you justify the presence of 18 elements in the 5th period of the periodic table
Answers
Answer:
The order in which the enery of the available orbitals 4d,5s and 5p increases in 5s<4d<5p. The total number of orbitals available is 9. The maximum number of electrons that can be accommodated is 18, and therefore 18 elements are there in the 5th period.
Explanation:
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introduction ´ˎ˗
In first period there’s H and He with electronic configuration 1s¹ & 1s² respectively.
in 2nd period there’s 8 elements, from Li ( 1s² 2s¹ ) to Ne ( 1s² 2s² 2p⁶ )
do you see a pattern ? in first period electrons are filled in 1s subshells. In the second period the electrons after filling 1s, starts filling in 2s and 2p.
so, yeah the n or principal quantum number of the s and p subshells getting filled in a period is equal to the period number.
obviously for d subshell, the value of principal quantum number (n) doesn’t show the period number. But this n is equal to period number - 1
- (The reason behind this is that after 3p⁶ is filled, 4s subshell is filled instead of 3d sub shell. As n+l value of 4s is lower than the n+l value of 3d )
answer ´ˎ˗
In 5th period, the electrons start filling from 5s subshell and the filling ends at 5p subshell. So, now what are subshells that come between 5s and 5p?
Refer to the image to find out (based on aufbau principle of filling of electrons)
the subshell between 5s and 5p is 4d.
so, in 5th period, 5s, 4d and 5p subshells are filled with electrons.
now how many electrons are filled totally?
That will be = twice the number of orbitals in each subshell
= (2× 2l₁+1) + (2× 2l₂+1) + (2× 2l₃+1)
= (2 × 1) + (2×3) + (2×5) = 2+6+10 = 18
18 electrons are filled along the 5th period.
each element along a period is filled with 1 electron in its valence shell right? yeah so by this we can that there’s a total of 18 elements in 5th period !
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