Consider the Chromium (III) complexes [Cr(NH 3 ) 5 L] +n where L = Cl _ , H 2 O and NH 3 The complex compound absorbs visible energy is given in table for d-d transition. Match the table to solve give

Question

Consider the Chromium (III) complexes [Cr(NH₃)₅L]⁺ⁿ

where L = Cl^_ , H₂O and NH₃

The complex compound absorbs visible energy is given in table for d-d transition.

Match the table to solve given two questions

^-34

Given that : h = 6.6 × 10^-34 (J-sec) , velocity of light = 3 × 10 ⁸ m/s

R = 1 × 10 ⁷ m ^-1 , N_A = 6 × 10²³ ; Atomic number of Cr = 24

Calculate crystal field stabilization energy (in $kJ /$ mole) for the homoleptic complex.

TestHub · Accepted Answer

Solution: The homoleptic complex is [ C r ( N H 3 ​ ) 6 ​ ] 3 + . From the table, for L = NH₃, λ ma x ​ = 465 nm. Energy of one photon: E = λ h c ​ E = 465 × 1 0 − 9 m ( 6.6 × 1 0 − 34 J s ) × ( 3 × 1 0 8 m / s ) ​ E = 4.258 × 1 0 − 19 J Energy per mole: E m o l ​ = E × N A ​ E m o l ​ = ( 4.258 × 1 0 − 19 J / p h o t o n ) × ( 6 × 1 0 23 p h o t o n s / m o l ) E m o l ​ = 255.48 × 1 0 3 J / m o l = 255.48 k J / m o l For Cr³⁺ (d³) in an octahedral field, the electron configuration is t 2 g 3 ​ e g 0 ​ . CFSE = 3 × ( − 0.4 Δ o ​ ) + 0 × ( 0.6 Δ o ​ ) = − 1.2 Δ o ​ CFSE = − 1.2 × 255.48 k J / m o l = − 306.576 k J / m o l

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