Question

Among the following complexes, the diamagnetic complexes are:

$\left(K\right):K_3\left[Fe\right(CN{)}_6]$

$\left(L\right):\left[Co\right(N{H}_3{)}_6]C{l}_3$

$\left(M\right):N{a}_3\left[Co\right(oxalate{)}_3]$

$\left(N\right):\left[Ni\right(H_{2}O{)}_6]C{l}_2$

$\left(O\right):K_2\left[Pt\right(CN{)}_4]$

$\left(P\right):\left[Zn\right(H_{2}O{)}_6\left]\right(N{O}_3{)}_2$

• A. $K,L,M,N$
• B. $K,M,O,P$
• C. $L,M,O,P$
• D. $L,M,N,O$

Answer 1

The correct option is C $L,M,O,P$

$\left(K\right):K_3\left[Fe\right(CN{)}_6]\to 3d^5$, strong field ligand, paramagnetic (because d-orbital has odd no. of electrons)

$\left(L\right):\left[Co\right(N{H}_3{)}_6]C{l}_3\to 3d^6$strong field ligand, diamagnetic

$\left(M\right):N{a}_3\left[Co\right(oxalate{)}_3]\to 3d^6$strong field ligand, diamagnetic

$\left(N\right):\left[Ni\right(H_{2}O{)}_6]C{l}_2\to 3d^8$ weak field ligand, paramagnetic (because weak field ligand do not allow pairing of electrons in the d-orbitals eventually makes it a paramagnetic complex)

$\left(O\right):K_2\left[Pt\right(CN{)}_4]\to 5d^8$strong field ligand, diamagnetic

$\left(P\right):\left[Zn\right(H_{2}O{)}_6\left]\right(N{O}_3{)}_2\to 3d^{10}$ strong field ligand, diamagnetic (because d-orbital is completely occupied)
Except for $K$ and $N$ all are diamagnetic complexes.

Hence, option C is correct.