The CFSE for - CoCl 6-4- complex is 18000 cm -1 the - for - CoCl 4-2- will beA- 18000 cm -1B- 16000 cm -1C- 8000 cm -1D- 2000 cm -1

The correct option is C 8000 cm^ 1Given CFSE for [CoCl_6]^4 nbsp;= 18000 cm^ 1 [CoCl_4]^2 → nbsp;Tetrahedral complex Hence, Δ _t = ? ∵Δ_t = 49Δ_0 ∴Δ_t ...

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Standard XII

Physics

Young's Modulus of Elasticity

The CFSE for ...

Question

The CFSE for $[C o C l_{6}]^{4 -}$ complex is 18000 $c m^{- 1}$ the $Δ$ for $[C o C l_{4}]^{2 -}$ will be

18000 c m^{- 1}

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16000 c m^{- 1}

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8000 c m^{- 1}

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2000 c m^{- 1}

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Solution

The correct option is C $8000 c m^{- 1}$
Given CFSE for $[C o C l_{6}]^{4 -} = 18000 c m^{- 1}$
$[C o C l_{4}]^{2 -} \to$ Tetrahedral complex
Hence, $Δ_{t} = ?$
$∵ Δ_{t} = \frac{4}{9} Δ_{0} ∴ Δ_{t} for [C o C l_{4}]^{2 -} = \frac{4}{9} \times 18000 c m^{- 1} = 8000 c m^{- 1}$

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The CFSE for [CoCl6]4− complex is 18000 cm−1 the Δ for [CoCl4]2− will be

The correct option is C 8000 cm−1Given CFSE for [CoCl6]4−=18000 cm−1 [CoCl4]2−→ Tetrahedral complex Hence, Δt=? ∵Δt=49Δ0∴Δt for [CoCl4]2−=49×18000 cm−1=8000 cm−1

The CFSE for $[C o C l_{6}]^{4 -}$ complex is 18000 $c m^{- 1}$ the $Δ$ for $[C o C l_{4}]^{2 -}$ will be

The correct option is C $8000 c m^{- 1}$
Given CFSE for $[C o C l_{6}]^{4 -} = 18000 c m^{- 1}$
$[C o C l_{4}]^{2 -} \to$ Tetrahedral complex
Hence, $Δ_{t} = ?$
$∵ Δ_{t} = \frac{4}{9} Δ_{0} ∴ Δ_{t} for [C o C l_{4}]^{2 -} = \frac{4}{9} \times 18000 c m^{- 1} = 8000 c m^{- 1}$