Question

A.A main group element with the valence electron configuration 5s2 is in periodic group It forms a monatomic ion with a charge of B. A main group element with the valence electron configuration 4s24p periodic group is in It forms a monatomic ion with a charge of

Answer 1

A. It will be easy to loose 2 e

So it will form 2+

B. It will be easy to gain 2e and complete it's octet

So, 2- is correct

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Answer 2

A. Ans. An element with the valence electron configuration 5s² is in 2^nd periodic group.

It forms a monoatomic ion with +2 charge (i.e.cation).

B. Ans. An element with valence shell configuration 4s²4p⁴ is in 16 periodic group.

It will form a monoatomic ion with -2 charge (i.e.anion).

Explanation:-

Rules to find group no.-

1. If elements have 1 or 2 e- in their valence shell then,

   Group no. of the element = No. of valence shell e-

   ex. Na has valence shell configuration 3s¹. Since it has i1 e- in its valence shell so it is in 1st group of the periodic table.

   Mg has valence shell configuration 3s². Since, it has 2 e- in its valence shell so it is in 2nd group of the periodic table.

2. If elements have more than 2 (3,4,5,6,7,8) e- in their valence shell then,

   Group no. of the element = 10 + no. of valence shell electrons

   ex. O has valence shell configuration 2s²2p⁴. Since it has 6 e- in its valence shell so it is in (10 + 6) 16 group of the periodic table.

   Al has valence shell configuration 3s²3p¹. Since it has 3 e- in its valence shell so it is in (10+3) 13 group of the periodic table.

Rule to find period no.-

Period no. of an element = the principal quantum no.(n) of the valence shell

Rule to find block (s,p,d and f) of an element-

Block of an element = type of orbital which receive last e-

Rules to find that given element will form cation or anion:-

i. If elements have 1 or 2 or 3 e- in their valence shell, they are metals and they have the tendency to lose one or more e- due to which they will form positive charge ion (cation)

ex. Na has 1 e- in its valence shell, so it will lose 1 e- to form Na⁺ ion.

Mg has 2 e- in its valence shell, so it will lose 2 e- to form Mg⁺² ion.

ii. if elements have more than 3 e- (4,5,6,7,8) in their valence shell, they are nometals and have tendency to gain one or more e- to form negative charged ion (anion).

ex. O has 6 e- in its valence shell so it will gain two e-.

      N has 5 e- in its valence shell so it will gain 3 e-.

Reason to lose or gain e- :

Except nobel gases, other elements in the periodic table are very reactive because they do not have stable electronic configuration and they can get stable configuration only when they have noble gas configuration i.e. 8 e- in their valence shell. They can attain 8 e- in their valence shell by losing or gaining one or more e-.

When they gain e-, get +ve charge form (cation) and when they lose e-, get -ve charge ion (form anion).

No. of e- lose or gain by the element = no. of +ve or -ve charge on that element

A. According to Question:

the valence shell configuration of an element is 5s²

Group no. = 2  because it has 2 e- in its valence shell.

Period no. = 5  because the value of n=5

Block  = p  because last e- enter into p orbital (subshell)

 given element with configuration 5s² has 2 e- in its valence shell is a metal and it has tendency to lose of 2 e- from its valence shell and will form +2 charge ion to get nobel gas configuration.

B. element with valence shell configuration 4s²4p⁴ has 6 e- in its valence shell so it is in group 16 (10+6).

Period no. = 4  because it has n=4

Block = p  bacause last e- enter into p orbital (subshell).

since, it has 6 e- in its valence shell so it is nonmetal and has tendency to gain 2 e- to complete its octet. ( i.e. to gain 8 e- ) or they can get noble gas configuration and after gaining 2 e- it will get -2 charge (form anion).