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Question

# A particular atom has the 4th shell as its valence shell. If the difference between the number of electrons between K and N shell and L and M shell is zero, find the electronic configuration of its stable ion.

A
2, 8, 8, 2
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B
2, 8, 18, 8
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C
2, 8, 8
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D
2, 8
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Solution

## The correct option is C 2, 8, 8Number of electrons in shell is given by 2n2where n= energy levelMaximum number of electrons in K shell⟶2n2=2×(1)2=2 [n=1]Maximum number of electrons in L shell⟶2n2=2×(2)2=2×4=8 [n=2]Maximum number of electrons in M shell⟶2n2=2×(3)2=2×9=18 [n=3]Maximum number of electrons in N shell⟶2n2=2×(4)2=2×16=32 [n=4]Difference between K and N shell is zero. K can accommodate maximum 2 electron. Therefore, for difference to be zero N should also have 2 electrons.∴K=2 electrons N=2 electronsDifference between L and M is also zero. Maximum electron in L=8. So, for difference to be zero M should also have 8 electron in it.∴L=8e− M=8e−Electrons per shell inK L M N [N is the 4th valence shell] 2 8 8 2Stable ion is formed by loss of 2 electrons from N shell.∴ Electronic configuration will be 2,8,8 .

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