Among the following complexes (K to P),
$\begin{array}{l} K_{3} [Fe {(CN)}_{6}] (K), [Co {({NH}_{3})}_{6}] {Cl}_{3} (L), \\ {Na}_{3} [Co {(oxalate)}_{3}] (M), [Ni {(H_{2} O)}_{6}] {Cl}_{2} (N), \\ K_{2} [Pt {(CN)}_{4}] (O), and [Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} (P) \end{array}$
The diamagnetic complexes are:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

K, L, M, N

L, M, O, P

K, M, O, P

L, M, N, O

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

[K] : $K_{3} [Fe {(CN)}_{6}] : {Fe}^{3 +}$ has $3 d^{5}$ configuration; ${CN}^{-}$ is strong field ligand and thus four electrons are paired leaving one unpaired electron with $d^{2} s p^{3}$ hybridization and thus paramagnetic.

[L]: $[Co {({NH}_{3})}_{6}] {Cl}_{3} : {Co}^{3 +}$ has configuration; ${NH}_{3}$ is strong field ligand thus all six electrons are paired with $d^{2} {sp}^{3}$ hybridization and therefore diamagnetic.

[M]: $N a_{3} [Co {(ox)}_{3}] : {Co}^{3 +}$ has $3 d^{6}$ configuration: $C_{2} O_{4}^{2 -}$ is strong field ligand and thus all the six electrons are paired with $d^{2} {sp}^{3}$ hybridisation and therefore diamagnetic.
[N]: $[Ni {(H_{2} O)}_{6}] {Cl}_{2} : {Ni}^{2 +}$ has $3 d^{8}$ configuration with two unpaired electrons. $H_{2} O$ is weak field ligand and thus ${sp}^{3} d^{2}$ hybridisation and paramagnetic.

[O]: $K_{2} [Pt {(CN)}_{4}] : {Pt}^{2 +}$ has $3 d^{8}$ configuration; ${CN}^{-}$ is strong field ligand and thus all the eight electrons are paired with ${dsp}^{2}$ hybridisation and therefore diamagnetic.
[P]: $[Zn {(H_{2} O)}_{6}] {({NO}_{3})}_{2} : {Zn}^{2 +}$ has $3 d^{10}$ configuration and thus all paired showing ${sp}^{3} d^{2}$ hybridization and so diamagnetic.