What is EAN of Co in ≤ft[{Co}≤ft({NH}₃)₆]³+} ?

32 ≤ft[{Co}≤ft({NH}₃)₆]³+} : Oxidation state of cobalt is +3 and six ligands donate 12 electrons. EAN = Atomic no. of metal (Z) - Number of electrons lost by metal to form the ion (X)+ Number of electrons donated by ligands (Y) {Z}=27 ; {X}=3 ; {Y}=12 EAN of {aligned} {Co}³+} & ={Z}-{X}+{Y} \\ & =27-3+12=36 {aligned}

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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 56 of 150

Marks: +1, -0

What is EAN of Co in

\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}

36
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32
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ :
Oxidation state of cobalt is +3 and six ligands donate 12 electrons.
EAN = Atomic no. of metal (Z) - Number of electrons lost by metal to form the ion (X)+ Number of electrons donated by ligands (Y)
$\mathrm{Z}=27 ; \mathrm{X}=3 ; \mathrm{Y}=12$
EAN of
$\begin{aligned} \mathrm{Co}^{3+} & =\mathrm{Z}-\mathrm{X}+\mathrm{Y} \\ & =27-3+12=36 \end{aligned}$

Validate

Solution:

Concept:

EAN = (atomic number of metal) − (electrons lost to form the ion) + (electrons donated by ligands).

Explanation:

Atomic number of Co,

\mathrm{Z} = 27

Oxidation state of Co is

+3

, so electrons lost,

\mathrm{X} = 3

Six

\mathrm{NH}_3

ligands donate

6 \times 2 = 12

electrons, so

\mathrm{Y} = 12

Therefore,

\mathrm{EAN} = \mathrm{Z} - \mathrm{X} + \mathrm{Y} = 27 - 3 + 12 = 36

Answer:

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