Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Please solve This Question sir Coordinaton compound question Once asked in aipmt also

Please solve This Question sir
Coordinaton compound question 
Once asked in aipmt also

Question Image
Grade:12th pass

1 Answers

tarun rooniy
9 Points
2 years ago
Option (A)
Greater the number of unpaired electrons, larger is the paramagnetism.

 

[V(gly)2 (OH)2(NH3)2]+ V23 = [Ar] 4s2, 3d3 Oxidation state of V in [V(gly)2 (OH)2(NH3)2]+  is

x + (-l)x2 + (-l)x2 + (0)x2 = +1

x =+5

Vs+ =[Ar]3d° (No unpaired electron)

[Fe(en)(bpy)(NH3)2)2+  Fe26 =[Ar]4s2, 3d6

Oxidation state of Fe in [Fe(en) (bpy) (NH3)2]2+ is

x + (0) + (0) + (0) x 2 = +2

x=+2

Fe2+ = [Ar]3d6

But en, bpy and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons. [Co(OX)2(OH)2]-

CO27 = [Ar]4s2, 3d7 

Oxidation state of Co in[CO(OX)2(OH)2]- is

x + (- 2) x 2+ (-1) x 2 = -1

x-6=-1

X =+ 5

CO5+ = [Ar]3d4 (4 unpaired electrons)

[Ti(NH3)6]3+

Ti22 =[Ar]4s2, 3d2

Oxidation state of Ti in [Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.

Hence [Co(OX)2(OH)2]- has highest paramagnetic behaviour.

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free