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Roni posted an Question

August 01, 2020 • 02:49 am

30 points

IIT JAM
Chemistry (CY)

Please solve both question 3&4 ..

please solve both question 3&4 .. periodic table as soon as possible

2 Answer(s)

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Suman Kumar

Use this simple formula If n=period number x=number of electrons in one orbit Number of elements in nth period= x*{(n+2)/2}^2 if n is even
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x*{(n+1)/2}^2. if n is odd
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Now Here n= 9.
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3*((9+1)/2)^2 = 75
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Priyanshu kumar Best Answer

check this👇
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you can also directly calculate using this formula
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hope it will helps roni😊
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yes sir thank s
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please accept the answer roni if you got it😊
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Priyanshu kumar

The maximum number of elements in a period is equal to the maximum number of electrons in its outer most shell. As we move from one period to the other the number of electrons it can accommodate in its orbital increases (in the order of s, p, d and f). But the maximum number of electrons a period can accommodate is = 2 + 6 + 10 + 14 = 32 So, we would have 32 elements in the 10th period.
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3 option C is correct
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got it roni??
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any doubt in this??
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hmm sir .. I think if g , h orbital present than .. 2+6+10+14+18+22=72 electron present
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no roni if we go from one period to another in a period..then orbitals increases...so s,p,d,f
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