JEE MAIN - Chemistry (2020 - 3rd September Morning Slot - No. 16)
The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1
.
The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol-1, is :
83.7
242.5
145.5
97
Explanation
[Ti(H2O)6]3+
Ti = [Ar]3d24s2
Ti+3 = [Ar]3d1
_3rd_September_Morning_Slot_en_16_2.png)
For octahedral
Crystal field stabilization energy (CFSE)
= (-0.4$$\Delta $$0) $$ \times $$ nt2g + (+0.6$$\Delta $$0) $$ \times $$ neg + np
nt2g = number of electrons in t2g orbital
neg = number of electrons in eg orbital
n = number of extra pairs
p = Pairing energy
Here nt2g = 1
neg = 0
n = 0 ( For weak field ligands n always zero. Here H2O is weak field ligand)
$$ \therefore $$ Crystal field stabilization energy (CFSE)
= (-0.4$$\Delta $$0) $$ \times $$ 1 + (+0.6$$\Delta $$0) $$ \times $$ 0 + 0$$ \times $$P
= -0.4$$\Delta $$0 = –0.4 × 20300 = –8120 cm–1
CFSE (in kJ) = $${{8120} \over {83.7}}$$ = 97 kJ/mol
Ti = [Ar]3d24s2
Ti+3 = [Ar]3d1
For octahedral
Crystal field stabilization energy (CFSE)
= (-0.4$$\Delta $$0) $$ \times $$ nt2g + (+0.6$$\Delta $$0) $$ \times $$ neg + np
nt2g = number of electrons in t2g orbital
neg = number of electrons in eg orbital
n = number of extra pairs
p = Pairing energy
Here nt2g = 1
neg = 0
n = 0 ( For weak field ligands n always zero. Here H2O is weak field ligand)
$$ \therefore $$ Crystal field stabilization energy (CFSE)
= (-0.4$$\Delta $$0) $$ \times $$ 1 + (+0.6$$\Delta $$0) $$ \times $$ 0 + 0$$ \times $$P
= -0.4$$\Delta $$0 = –0.4 × 20300 = –8120 cm–1
CFSE (in kJ) = $${{8120} \over {83.7}}$$ = 97 kJ/mol
Comments (0)
