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  4. Number of ions present in K4[Fe(CN)6] is

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Q. Number of ions present in $K_{4}[Fe(CN)_{6}]$ is

Coordination Compounds

Solution:

$K_{4}[Fe(CN)_{6}]$isacomplexsalt.Onionisationitwilldissociate in $4K^{+}$ and $[Fe(CN)_{6}]^{4-}$ ion. Hence, in $K_{4}[Fe(CN)_{6}]$ five ions are present.