Why is $[C o (C N)_{6}]^{3 +}$ and $[M n (C N)_{6}]^{3 +}$ diamagnetic $?$

Answer

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Hint :Diamagnetic materials are repelled by a magnetic field; an applied magnetic field creates an induced magnetic field in them in the opposite direction, causing a repulsive force. In contrast, paramagnetic and ferromagnetic materials are attracted by a magnetic field.

Complete Step By Step Answer:
We know that, the

C o^{+ 3}

and

M n^{+ 3}

are present in the compound

[C o (C N)_{6}]^{+ 3}

and

[M n (C N)_{6}]^{3 +}

.
So, the electronic configuration of

C o^{3 +}

[A r] 4 s^{3} d^{5}

and

M n^{3 +}

[A r] 3 d^{4}

.
So, in the excited state, and undergoing hybridization,

C N^{-}

being a strong field ligand forces the inner

d

electrons to pair up. Cyanide ion is a strong field ligand because it is a pseudo halide ion. Pseudo halide ions are stronger coordinating ligands & they have the ability to form σ bond and π bond.
Due to this,

C o^{3 +}

and

M n^{3 +}

have electrons in multiples of two and have no unpaired electron. That’s why

[C o (C N)_{6}]^{3 +}

and

[M n (C N)_{6}]^{3 +}

are diamagnetic.

Additional Information:
Hund's rule states that every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.

Note :
Cyanide is an extremely simple and narrow ligand, so the coordination sphere of metal ions is readily saturated. Ligand, in chemistry, any atom or molecule bound in coordination or complex compound to a central atom, usually a metallic element.