In $$Ni{\left( {CO} \right)_4},Ni$$ has zero oxidation state. It is $$s{p^3}$$ hybridised.
$$Ni{\left( {CO} \right)_4} \to $$
Hence, it has no unpaired electron so, it shows the property of diamagnetism and has tetrahedral structure. NOTE
The valence shell electronic configuration of ground state $$Ni$$ atom is $$3{d^8}4{s^2}.$$ The two electrons of $$4s$$ are pushed into $$3d$$ orbitals and get paired up because of the presentce of strong ligand $$(CO).$$
62.
Among the following species the one which causes the highest $$CFSE,{\Delta _ \circ }$$ as a ligand is :
Following is the order of increasing value of $${\Delta _ \circ }$$ for the ligands.
$${I^ - } < B{r^ - } < C{l^ - } < {F^ - } < {H_2}O < Py = N{H_3} < en < NO_2^ - < < C{N^ - } < CO$$
63.
Tetraammineaquachloridocobalt (III) chloride can be written as
A
$$\left[ {CoCl\left( {{H_2}O} \right){{\left( {N{H_3}} \right)}_4}} \right]Cl$$
B
$$\left[ {CoCl\left( {{H_2}O} \right){{\left( {N{H_3}} \right)}_4}} \right]C{l_2}$$
C
$$\left[ {CoCl\left( {{H_2}O} \right){{\left( {N{H_3}} \right)}_4}Cl} \right]$$
$$\left[ {PtC{l_2}{{\left( {N{H_3}} \right)}_2}} \right]$$ has square planar geometry.
68.
Among the following complexes $$(K-P)$$
$${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\left( K \right),$$ $$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}\left( L \right),$$ $$N{a_3}\left[ {Co{{\left( {{\text{oxalate}}} \right)}_3}} \right]\left( M \right),$$ the $$\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]C{l_2}\left( N \right),$$ $${K_2}\left[ {Pt{{\left( {CN} \right)}_4}} \right]\left( O \right)$$ and $$\left[ {Zn{{\left( {{H_2}O} \right)}_6}} \right]{\left( {N{O_3}} \right)_2}\left( P \right)$$ the diamagnetic complexes are
70.
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]\left[ {CuC{l_4}} \right]$$ and $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{l_4}} \right]$$ are known as