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the EAN of nickel in k2[Ni(CN)4]

KADAMBALA VAMSI , 12 Years ago
Grade Upto college level
anser 7 Answers
saisandeep Mattaparthi

Last Activity: 12 Years ago

the formula= atomic no-oxidation state of central metal atom+2(no of Ligands)

                 =28-2+2(4)

                =34

Theresh Babu Benguluri

Last Activity: 12 Years ago

EAN ===== 27-2+2(4)

             = 25+8

             =33

KADAMBALA VAMSI

Last Activity: 12 Years ago

34

mannam kiran kumar

Last Activity: 12 Years ago

atomic no of metal - oxidation state of central metal + 2* no.of ligands

Ni (z=28), oxidation state +2 , no of ligands = 4

28-2+2(4) = 18 it si argan.

dipesh agrawal

Last Activity: 12 Years ago

ean= z-x+y 

here,

z=28

x=2

y=4*2

ean=28-2+8

    =34

Ojas

Last Activity: 6 Years ago

 The sum of the number of electrons, donated by all ligands and those present on the central metal ion or atom in complex is called as effective atomic number (EAN).
 
•        Generally EAN of central metal ion will be equal to the number of electrons in the nearest noble gas.
•        If the EAN of the central metal is equal to the number of electrons in the nearest noble gas then the complex possess greater stability.
 
EAN = [(atomic number of central metal) – (the oxidation state of the metal) + (the number of electrons gained by the metal from the ligands through co-ordination)] = [Z metal – (ox.state of the metal) + 2(coordination number of the metal)]. Except in case of NO 3 is multiplied to coordination number of the metal.
Here:
Z metal (Ni) =  28, coordination number of the metal { n(CN) = 4},and ox.state of the metal = 2
EAN = 28 – (2) + 2(4) = 34
Pritesh Khairnar

Last Activity: 6 Years ago

By the formula we get its EAN 34 because atomistic number of nickel is 28 and subtracting two and adding 8 é of ligand we get 34
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