I did my work and i felt like I\'m missing something. I want to compare my answe

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11, The number of orbitals in a shell with n = 3 is 12. Which one of the atoms in the set: Y. Cr, Mg. N, Ba, Se, Sn has the largest first ionization energy? 13. Which one of the atoms in the set: Mg. Cr, N, Sn. Ba, Sn, Se would you expect to have the smallest first ionization energy?

Explanation / Answer

Ans 11 : 9

The number of orbitals is given by the formula : n2

where n is the number of orbit

So here : 33

= 9

Ans 12 : N

The ionization energy refers to the amount of energy require to remove the electron from the outermost shell.

Since N is smallest here , it will require maximum energy.

Ans 13 : Ba

Ba has the largest atomic radius here , amongst all others , due to which lowest amount of energy will be required to remove the electron.

Ans 19 : D

Technetium has 43 electrons.

and Kr has 36 , so 7 electrons after Kr according to Aufbau principle will be 5s2 4d5

Ans 21 : B) radium

This is the group of alkaline earth metals , which is group II . The general electronic configuration of these elements is ns2.

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