Similarly,second and third periods can accommodate 8 and 18 electrons respectively. Since their outermost shells can contain only 8 electrons, there are only 8 elements in boththe periods. -pllz explain this again.
i shall be thankful.

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Solution

Using the formula...
2n^2.
For second period...
n=2
Hence the maximum possible number of electrons in 2nd period = 2 x (2)² =2 x4 = 8.
For 3rd period..
n=3
Hence the maximum possible number of electrons in 3rd period = 2 x (3)² = 18

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Similarly, second and third periods can accommodate 8 and 18 electrons respectively. Since their outermost shells can contain only 8 electrons, there are only 8 elements in both the periods. -pllz explain this again.i shall be thankful.

Using the formula...2n2.For second period...n=2Hence the maximum possible number of electrons in 2nd period = 2 x (2)2 =2 x4 = 8.For 3rd period..n=3Hence the maximum possible number of electrons in 3rd period = 2 x (3)2 = 18

Similarly,second and third periods can accommodate 8 and 18 electrons respectively. Since their outermost shells can contain only 8 electrons, there are only 8 elements in boththe periods. -pllz explain this again.
i shall be thankful.

Using the formula...
2n^2.
For second period...
n=2
Hence the maximum possible number of electrons in 2nd period = 2 x (2)² =2 x4 = 8.
For 3rd period..
n=3
Hence the maximum possible number of electrons in 3rd period = 2 x (3)² = 18