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

the EAN  of nickel in k2[Ni(CN)4]

Grade:Upto college level

7 Answers

saisandeep Mattaparthi
84 Points
11 years ago

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

                 =28-2+2(4)

                =34

Theresh Babu Benguluri
39 Points
11 years ago

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

             = 25+8

             =33

KADAMBALA VAMSI
35 Points
11 years ago

34

mannam kiran kumar
40 Points
11 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
32 Points
11 years ago

ean= z-x+y 

here,

z=28

x=2

y=4*2

ean=28-2+8

    =34

Ojas
20 Points
5 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
15 Points
5 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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