Co Ordination Compounds Question 136
Question: Among the following complexes (K-P)
$ K_3[ Fe{{( CN )}_6} ] $ (K), $ [ Co{{( NH_3 )}_6} ]Cl_3( L ), $ $ Na_3[ Co{{( oxalate )}_3} ] $ (M), $ [Ni{{(H_2O)}_6}]Cl_2)(N), $ $ K_2[Pt{{( CN )}_4}( O ) $ and $ [Zn{{(H_2O)}_6}]{{(NO_3)}_2}(P) $ the diamagnetic complexes are
Options:
A) K, L, M, N
B) K, M, O, P
C) L, M, O, P
D) L, M, N, O
Show Answer
Answer:
Correct Answer: C
Solution:
| Complex | No.of electrons in outer d orbital | No.of unpaired electrons (s) |
|---|---|---|
| $ {{[Fe{{(CN)}_6}]}^{3-}} $ | $ 3d^{5} $ | 1 ( $ C{N^{-}} $ causes pairing of electrons) |
| $ {{[Co{{(NH_3)}_6}]}^{3-}} $ | $ 3d^{6} $ | 0 |
| $ {{[Co{{(OX)}_3}]}^{3-}} $ | $ 3d^{6} $ | 0 |
| $ {{[Ni{{(H_2O)}_6}]}^{2+}} $ | $ 3d^{8} $ | 2 |
| $ {{[Pt{{(CN)}_4}]}^{2-}} $ | $ 5d^{8} $ | 0 ( $ C{N^{-}} $ cause pairing of electrons) |
| $ {{[Zn{{(H_2O)}_6}]}^{2+}} $ | $ 3d^{10} $ | 0 |
Thus L, M, O and P are diamagnetic.
BETA
