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Crystal Field Split in Octahedral Complexes

[ T i H 2 O 6...

Question

$[T i (H_{2} O)_{6}]^{3 -}$ absorbs energy equal to wavelength 498 nm. On this basis crystal field splitting energy comes out to be $a \times 10^{- 3} J$ . What is the value of a?___

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Solution

$E = \frac{h c}{λ} = \frac{6.627 \times 10^{- 34} \times 3 \times 10^{8}}{498 \times 10^{- 12}} = 4 \times 10^{- 19} J$

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[T i (H_{2} O)_{6}]^{3 +}

absorbs light of wavelength 498 nm during a d–d transition. The octahedral splitting energy for the above complex is

\times 10^{- 19} J

. (Round off to the nearest integer).

h = 6.626 \times 10^{- 34} J s

; c = 3 \times 10^8~\text{ m s}^{-1}\).

Q. What is meant by crystal field splitting energy ? On the basis of crystal field theory, write the electronic configuration of

d^{4}

t_{2 g}

e_{g}

in an octahedral field when
(i)

Δ_{0} > P

Δ_{0} < P

Q. What is crystal field splitting energy? Draw a figure to show spitting of degenerate orbitals in an octahedral crystal field?

[F e (H_{2} O)_{6}]^{2 +}

has a crystal field splitting energy value of

10, 400 c m^{- 1}

and a pairing energy value of

17, 600 c m^{- 1}

. Then, the compound is :

Q. The crystal field splitting energy CFSE

(I n Δ_{0} u n i t s)

[C o F_{6}]^{3 -}

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Crystal Field Split in Octahedral Complexes

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