Co Ordination Compounds Question 290
Question: Match Column I with Column II and select the con- answer with respect to hybridisation using the codes gr below:
| Column I (Complex) | Column II (Hybridisation) |
|---|---|
| (I) $ {{[AuF_4]}^{-}} $ | (P) $ dsp^{2} $ hybridisation |
| (II) $ {{[Cu{{(CN)}_4}]}^{3-}} $ | (Q) $ dsp^{3} $ hybridisation |
| (III) $ {{[Cu{{(C_2O_4)}_3}]}^{3-}} $ | (R) $ sp^{3},d^{2} $ hybridisation |
| (IV) $ {{[Fe{{(H_2O)}_5}NO]}^{2+}} $ | (S) $ d^{2}sp^{3} $ hybridisation |
Codes:
Options:
A) (I) $ \to $ Q, (II) $ \to $ P, (III) $ \to $ R, (IV) $ \to $ S
B) (I) $ \to $ P, (II) $ \to $ Q, (III) $ \to $ S, (IV) $ \to $ R
C) (I) $ \to $ P, (II) $ \to $ Q, (III) $ \to $ R, (IV) $ \to $ S
D) (I) $ \to $ Q, (II) $ \to $ P, (III) $ \to $ S, (IV) $ \to $ R
Show Answer
Answer:
Correct Answer: B
Solution:
(I) Au in +3 oxidation state with $ 5d^{8} $ configuration has higher CFSE.
So complex has $ dsp^{2} $ hybridisation and is diamagnetic. (II) Cu is in +1 oxidation state with $ 3d^{10} $ configuration and no $ ( n-1 )d $ orbital is available for $ dsp^{2} $ hybridisation and complex is diamagnetic. (III) Co is in +3 oxidation state and $ 3d^{6} $ configuration has higher CFSE.
So complex is diamagnetic and has $ d^{2}sp^{3} $ hybridisation. (IV) Fe is in +1 oxidation state and the complex is paramagnetic with three unpaired electrons. $ {{[Fe{{(H_2O)}_5}NO]}^{2+}}; $
BETA
