The $[T i (N C S)_{6}]^{3 -}$ ion exhibits a single absorption band at 544 nm. What will be the crystal field splitting energy $(i n k J m o l^{- 1})$ of the complex? $(h = 6.626 \times 10^{- 34} J . s; C = 3.0 \times 10^{8} m / s; N_{A} = 6.02 \times 10^{23} i o n s / m o l e$ .

240

No worries! We‘ve got your back. Try BYJU‘S free classes today!

220

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

270

No worries! We‘ve got your back. Try BYJU‘S free classes today!

250

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 220
The expression for the crystal field splitting energy is $= h v = h c$ . Here, h is the Planck's constant, v is the frequency of light, c and are the speed and the wavelengths of light respectively.
Substitute values in the above expression.
$= \frac{(6.626 \times 10^{34} J . s) \times (3.0 \times 108 m / s)}{544 \times 10^{9} m} = 3.65 \times 10^{19} J / i o n$
Convert the crystal field splitting energy from J/ion to kJ/mol.
$= (3.65 \times 10^{19} J / i o n) \times (6.02 \times 10^{23} i o n s / m o l e) = \frac{21.973 \times 10^{4}}{1000} k J m o l^{1} = 219.73 k J m o l^{1}$

Suggest Corrections

Similar questions

[T i (N C S)_{6}]^{3 -}

(h = 6.626 \times 10^{- 34} J s; c = 3 \times 10^{8} m / s; N_{A} = 6.02 \times 10^{23} i o n s / m o l e)

{[T i (N C S)_{6}]}_{6}^{3 -}

({in kJmol}^{- 1})

(h = 6.626 \times 10^{- 34} J s; c = 3.0 \times 10^{8} m / s; N_{A} = 6.02 \times 10^{23} ions/mole)

[T i (N C S)_{6}]^{3 -}

544 n m

(Planck constant = 6.626 \times 10^{- 34} J s)

[T i (N C S)_{6}]^{3 -}

(h = 6.626 \times 10^{- 34} J s; c = 3 \times 10^{8} m / s; N_{A} = 6.02 \times 10^{23} i o n s / m o l e)

[T i (N C S)_{6}]^{3 -}

544 n m

21979 \times 10^{- x} k J . m o l^{- 1}

Join BYJU'S Learning Program