DEPARTMENT OF CHEMISTRY
CH 102 (INORGANIC): TUTORIAL NO.4
1. CrF2 :
Two long bonds : 2.43 Å
Four short bonds: ~ 2.00 Å
Cr 2+ is surrounded by 6 fluoride ions in an Oh environment.
Cr 2+ is d4 high spin; t2g3eg1
The unsymmetrical distribution of electrons in eg leads to Jahn – Teller Distortion.
This observation suggests that the eg electron is in dz2 orbital, and dx2- y2 is empty.
2. When a substance is subjected to an applied magnetic field , H, a magnetization,
I, is induced.
I/H = χv = Volume susceptibility.
The χv volume susceptibility, is related to ‘ gram susceptibility’ χ , and ‘ molar
Susceptibility Χm .
χ = χv/d and χm = χv.M/d;
d = density; M = molecular weight.
Also, χm = Nµ2/3kT (from Curie's Law)
Where
N = Avogadro Number
µ = magnetic moment of the molecule in B.M.
k = Boltzmann constant
T = absolute temperature.
By substituting numerical values for N and k, we obtain
χm = 0.125 µ2/ T (or) µ 2.83 χ
mT
3. (a) K2NiF6
NiF6 2 - ≡ Ni 4+ ≡ d 6
Very high charge; Hence high CFSE and pairing tendency ( c.f. Co3+)
(CFSE depends on the charge on the metal)
d 6 ≡ t2g6eg0 ≡ no unpaired electrons.
The complex is hence diamagnetic.
(b) NiCl2(NH3)2
NH3 is a strong field ligand. Hence, no tetrahedral structure.
For square planar µ = 0: So, no sq. planar.
For Oh
Ground state : t2g6eg2 no orbital contribution.
Exited state : t2g5eg3 Orbital contribution possible.
So, the structure is
NH3
Cl Cl
Ni
Cl Cl
NH3
(c) NiCl2(PEt3)2 ; phosphine is strong field ligand
µ = 0 and hence Square planar.
(d) NiCl2(Ph3AsO)2 ; both arsine-oxide (this a neutral oxygen donor) and chloride are
weak field ligands. Hence tetrahedral geometry is preferred. In addition, the presence of
bulky ligands also stabilize lower coordination numbers (i.e. Td). Since µ = 3.95 B.M.,
which is much higher than the spin-only value, this is a tetrahedral case with orbital
contribution in ground state.
(4). Orbital contribution is possible for configurations with asymmetrically filled t2g (of
Oh) and t2 (of Td), because of the possibility of rotation of the electronic charge around
the molecules.
For ‘l s or l’ due to the different symmetric of dz2 and d x2- y2 , such a rotation is not
possible.
The possible configuration for Oh are;
t2g1eg0 : t2g2eg0 : t2g4eg2 : t2g5eg2 .(high spin).
Similarly work-out low spin.
Do a similar exercise for Td
(Remember that all Td complexes are high spin).
(5) CoCl42-
Co2+ ; d 7
Td
Three unpaired spins; No orbital contribution. Symmetrically filled t2
CoF6 2 –
Co 2+
Oh
Same three unpaired spins; but with asymmetric filling of t2g. Hence, orbital
contribution. So, greater magnetic moment.
(6) CoCl 42- & CoI4 2-
Both are tetrahedral.
Co2+ ;
d 7 ; e4 t 2 3
There is no orbital contribution is ground state. Orbital contribution in the excited state is
possible because of the following configuration: e3 t 2 4.
Mixing of ground state and excited state is more facile with the weak of the two each
ligands, viz I−. So, CoI42- has a higher magnetic moment.
(5) Dy2(SO4)3. 8H2O
Dy3+ = f 9
3 2 1 0 -1 -2 -3
L = 6+4+1-1-2-3= 5 S = 5/2
Since more than half –field J = L+S = 5+5/2
µ = g J(J+1)
g = 3/2 + S(S+1) – L(L+1) / 2J(J+1)
Substitute and work out. The answer is:
µ = 10.63 B.M
(8) Gd 3+ ≡ f 7 ≡
L = 0 (hence no orbital contribution, and spin-only formula is sufficient)
S = 7/2
µs = 4S(S+1) or n (n+2)
= 7.9 B.M
(9) At the Neel Temperature the antoferromagnetic substance becomes paramagentic. See
the graph below.
Or, when the temparature of a paramagentic system is lowered, at a critical temperature,
called as Neel Temperature, TN, this substance starts behaving antiferromagentically.
χm
TN
