If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/mol D) 2.72 x 10 5 kJ/mol E) 2.72 x 10 9 kJ/mol Group of answer choices

The energy of the complex is

Explanation:

To calculate the energy of the complex, we use the equation given by Planck which is:

where,

= Wavelength of the complex =    (Conversion factor:   )

h = Planck's constant =

c = speed of light =

= Avogadro's number =

= energy of the complex

Putting values in above equation, we get:

Conversion factor used:  1 kJ = 1000 J

Hence, the energy of the complex is

The complex energy is option(C) .

Explanation:

To calculate the energy of the complex, by using Planck's equation.

Where,

h = Planck's constant = .

c = speed of light = .

= Avogadro's number = .

E = energy of the complex.

Substitute all the required values in the equation:

.

The energy of the complex is .

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