P Block Elements

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New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Correct order of reducing character is

NH3 < PH3 < AsH3 < SbH3 < BiH3

New answer posted

a year ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Calamine -> ZnCO3    

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Hybridisation = sp3d2

Shape – square planar

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

X e F 2 + P F 5 → [ X e F ] + [ P F 6 ] −

New answer posted

a year ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

XeF2 : Hybrid orbital = sigma bond + L.P

= 2 + 3

= 5 = sp3d

Shape will be linear

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

HBrO2 has least existence

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

M n O 4 − + S 2 O 3 2 − → O H − M n O 2 + S O 4 2 −

O.S of S in    S O 4 2 − = + 6

New answer posted

a year ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

Both (A) and (B) are correct but R is not correct explanation of A. In both oxidation state of metal is +1, also both have similar lattice structure.  

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

2 C H 3 C H 2 C l → e t h e r 2 N a C 2 H 5 − C 2 H 5 + 2 N a C l ( W u r t z       R e a c t i o n )
2 C 6 H 5 C l → e t h e r 2 N a C 6 H 5 − C 6 H 5 + 2 N a C l ( W i t t i n g     R e a c t i o n )

 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

P 4 + 3 N a O H + 3 H 2 O → P H 3 + 3 N a H 2 P O 2

R e d P + a l k a l i → H 4 P 2 O 6 ( N o     P − H     b o n d )

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