In an atom the number of electrons in N shell is equal to the number of electrons in K,L,M,N shells.Answer the following questions.
a)Which is the outermost shell?
b)How many electrons are there in the outermost shell?
c)What is the atomic number
d)Write the electronic configuration between elements.
TPS:
you can't have as much electron in N shell as in K,L,M,N shells
sorry its only K,L,M shells
plz answer
what is electronic configuration between elements?
sorry.its electronic configuration of elements
sorry your answer is wrong .according to aufbau principle the orbital with least electronic configuration should be filled first.so the outer most shell is P.2 electrons are present in outermost shell .the atomic number is 66.the electronic configuration is 2,8,18,26,8,2
sorry your answer is wrong .according to aufbau principle the orbital with least electronic configuration should be filled first.so the outer most shell is P.2 electrons are present in outermost shell .the atomic number is 66.t
sorry your answer is wrong .according to aufbau principle the orbital with least electronic configuration should be filled first.so the outer most shell is P.2 electrons are present in outermost shell .the atomic number is 66.t
18+8+2 = 28, not 26
And when did start teaching aufbau principle to 14 year olds??
i am class 10.in physics structure of atom ,aufbau principle is there
2,8,18,28,8,2
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The number of electrons in N shell is equal to the number of electrons in K,L,M shells. So if we start filling electrons in an element, it will be
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ ........
Number of electrons in K, L and M( shells = 2+2+6+2+6+10 = 28
So in N shell (4s, 4p, 4d and 4f), number of electrons should be 28.
Number of electrons in 4s = 2
Number of electrons in 4p = 6
Number of electrons in 4d = 10
So in 4f, we need = 28 - (2+6+10) = 10 electrons.
So the element will have configuration till 4f¹⁰.
So configuration is = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁰.
a) outermost shell is P (6s)
b) 2 electrons are there in the outermost shell. (in 6s)
c) atomic number = 2+2+6+2+6+2+10+6+2+10+6+2+10 = 66
d) configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁰
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ ........
Number of electrons in K, L and M( shells = 2+2+6+2+6+10 = 28
So in N shell (4s, 4p, 4d and 4f), number of electrons should be 28.
Number of electrons in 4s = 2
Number of electrons in 4p = 6
Number of electrons in 4d = 10
So in 4f, we need = 28 - (2+6+10) = 10 electrons.
So the element will have configuration till 4f¹⁰.
So configuration is = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁰.
a) outermost shell is P (6s)
b) 2 electrons are there in the outermost shell. (in 6s)
c) atomic number = 2+2+6+2+6+2+10+6+2+10+6+2+10 = 66
d) configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁰
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