A complex compound in which the oxidation num | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is calculated as: $x + 4(0) = 0 \Rightarrow x = 0$ (since $CO$ is a neutral ligand). For other options

Vedclass · Accepted Answer

$[Ni(CO)_4]$

Vedclass · Answer

(C) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is calculated as: $x + 4(0) = 0 \Rightarrow x = 0$ (since $CO$ is a neutral ligand). For other options: $(a)$ $K_4[Fe(CN)_6]: 4(+1) + x + 6(-1) = 0 \Rightarrow x = +2$ $(b)$ $K_3[Fe(CN)_6]: 3(+1) + x + 6(-1) = 0 \Rightarrow x = +3$ $(d)$ $[Pt(NH_3)_4]Cl_2: x + 4(0) + 2(-1) = 0 \Rightarrow x = +2$

$A$ complex compound in which the oxidation number of a metal is zero is

Similar Questions

The $IUPAC$ name of $[Cr(NH_3)_4 Cl_2] NO_3$ is .......

The $IUPAC$ name of $[Pt(NH_3)_4 NO_2Cl] SO_4$ is .............

Which of the following formulae represents the trioxalatocobaltate$(III)$ ion?

State the $IUPAC$ name of the following complex: $Cis-platin$.

The name of the complex ion,$[Fe(CN)_6]^{3-}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module