Question

plz answer the above question​

Answer 1

Answer:

outer orbit's last number

Answer 2

# $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}$

1. Number of electrons = 2+2+6+2+6+2 = 20

2. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}$-- Calcium(Metal)

    So, it loses 2electrons........

    The electronic configuration of nearest noble gas element is      

    $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$

3. $1s^{2}$ -- K shell

   $2s^{2} 2p^{6}$ -- L shell

   $3s^{2} 3p^{6}$ -- M shell

   $4s^{2}$ -- N shell

   Ca loses 1 electron from $4s^{2}$...

   So, the orbital is -- s-orbital

4. Group number is 2

   Period number is 4

5. 2 valence electrons are present in it.

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