$x+4(0)=0 $
$x=0$
The electronic configuration of $Ni$ is $[ Ar ] 3 d ^{8} 4 s ^{2} 4 p ^{0} .$ The orbital diagram of nickel in the presence of strong filed ligand $CO$ is shown below.
As no unpaired electron is present in this complex, so it is diamagnetic.
$(P)\, Fe(CO)_5\,\,\, (Q)\,CO\,\,\, (R)\, H_3B \leftarrow CO\,\,\,(S)\, [Mn(CO)_5]^-$
સંકીર્ણ : $\left[{CoF}_{6}\right]^{3-},\left[{Co}\left({H}_{2} {O}\right)_{6}\right]^{2+},\left[{Co}\left({NH}_{3}\right)_{6}\right]^{3+}$ અને $\left[{Co}({en})_{3}\right]^{3+}$
$\quad\quad\quad\quad\quad\quad A\quad\quad\quad\quad \quad B\quad\quad\quad\quad\quad\quad C\quad\quad\quad\quad\quad\quad D$
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