PROPERTIES OF S BLOCK
Name of block = (A) s - block
(B) p - block
(C) d - block
(D) f - block
Properties s - block
Configuration Formula = ns1–2
On the basis of nature = Soft metal
Tendency = To lost of electron but easily.
Properties = (1) Formation → It forms - (a) Metallic bond & (b) Ionic bond.
It forms strong metallic bonding.
Reason →
Causes of strong metallic bond - Due to easily lost of electron.
Variation of Metallic Character
Depend upon = (a) Group (size)
(b) Period (V.E)
(1) Metallic Character in a Group
Based on - Size of an atom.
Relation - Relation of metallic character with the size
⇨ Directly proportional
Conclusion to above -
metallic character is ∝ Size of an atom in a group.
Example
iA
Li
Na
k
Rb
Cs
Arrange in order on the basis metallic character -
Cs > Rb > K > Na > Li
ⅱA
Be
Mg
Ca
Sr
Ba
Arrange in order on the basis metallic character -
Ba > Sr > Ca > Mg > Be
(2) Metallic Character in a period
Based on = Valence Electron (V.E)
Relation - Relation of metallic character with valance electron (v.e)
⇨ Inversely Proportional.
Conclusion to above -
metallic character is ∝ 1/V.E in a Period.
Example -
(a) Li & Be
Solution → Li = 1s2 , s1 Valence Electron (V.E) = 1
Be = 1s2 , 2s2 V.E = 2
Arrange in order on the basis metallic character in a period -
Li > Be
(b) Na & Mg
Solution → Na = 3s1 V.E = 1
Mg = 3s2 V.E = 2
Arrange in order on the basis metallic character in a period -
Na > Mg
(3) On the basis of ionization potential (I.P)
It having I.P comparison to any block
Causes of less I.P
Due to easily lost of electron the requirement of energy will be less
Variation of ionization potential = ( a) In a Group Decrease
(b) In a period increase
Example
(a)
iA
Li
Na
k
Rb
Cs
Arrange in order on the basis ionization potential -
Li > Na > K > Rb > Cs
(b) Li & Be - Arrange in order
Be > Li
(c) on the basis of melting point (m.p) & boiling point ( b.p)
⇨ It having less m.p & b.p in comparison to any block.
Causes - It having less strength
Variation of m.p & b.p = ( a) In a Group Decrease
(b) In a period increase
(4) Acidic Character
⇨ It Show acidic character but best
⇨ Causes of acidic character -
Ans ⇨ Due to lost of electron
Example - s - block & d - block
Above the blocks arrange on the basis of acidic character
s - block > d - block
Na, Mg, Sc, Mn
arrange in order on the basis of acidic character
Solution - Na = 3s1 V.E = 1
Mg = 3s2 V.E = 2
s - block
Sc = 3d14s2 V.E = 3
Mn = 3d54s2 V.E = 7
Na > Mg > Sc > Mn
Final conclusion = (a) In a Group increases
(b) In a period Decreases.
(5) Basic Character / Alkaline properties
It do not show basic character
Oxidizing character
It do not show Oxidizing properties
Causes - s - block having tendency to lost of electron but oxidizing properties having tendency to gain of electron to other electron so it causes s - block do not so Oxidizing properties.
Reducing properties
It show Reducing character = (a) In a Group increases
(b) In a period Decreases.
Example - Na, Mg, K, Ca
Solution → Na = 3s1 V.E = 1
Mg = 3s2 V.E = 2 (g.p 2)
K = 4s1 V.E = 1
Ca = 4s2 V.E = 2 (g.p 2)
K > Na > Ca > Mg
(8) Flame test
It show characteristic feature of s - block
It show flame test because s - block element having tendency to lost of electron to non metal but the s - block element are unable to return in a won state ( that is ground State ) then it causes called flame test or in case of salt ( That is combustion of metal and non metal ) in which the soft metal ( except Li & Be) which are losses electron to non metal but soft metal are unable to return own electron in ground state then it causes to show flame test.
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