explain the structure and magnetic properties of
[Mn(NH3)6] on the basis of VBT
Answers
Answer:
Mn(CN)
6
]
3−
Electronic configuration is Mn
3+
=[Ar]3d
4
hence box electronic structure
(i) Type of hybridisation d
2
sp
3
(ii) Inner orbital complex
(iii) paramagnetic , due to presence of three unpaired electrons.
(iv) Spin only magnetic moment is calculated using the formula : n=2 in this case , we get spin only magnetic moment in BMas =
2(2+2)
=
8
=2.87BM
[Co(NH
3
)
6
]
3+
Electronic configuration of Co
3+
=[Ar]3d
6
(i) Hyb As shown in the above box electronic structure the type of hybridisation is ............d
2
sp
3
(ii) Inner orbital complex
(iii) Diamagnetic
(iv) Zero( Since no unpaired electrons are present, as depicted above)
[Cr(H
2
O)
6
]
3+
(i) type of hybridisation........d
2
sp
3
(ii) Inner orbital complex
(iii) paramagnetic
(iv) 3.87BM
[Fe(cl)
6
]
4−
electronic configuration is of Fe
2+
=[Ar]3d
6
(i) type of hybridisation ......sp
3
d
2
(ii) Outer orbital complex
(iii) Paramagnetic
(iv) 4.9BM