MCQ
The complex ion which has no $ ‘d ’ $ electrons in the central metal atom is
- ✓${[Mn{O_4}]^ - }$
- B${[Co{(N{H_3})_6}]^{3 + }}$
- C${[Fe{(CN)_6}]^{3 - }}$
- D${[Cr{({H_2}O)_6}]^{3 + }}$
The electronic configuration of $Mn$ is $[ Ar ] 3 d^{5} 4 s ^{2} .$ In $MnO _{4}^{-}$, oxidation number of $Mn$ is +7 i.e. all
the $3 d$ and $4 s$ electrons are lost to form $\left[M n O_{4}\right]^{-}$ complex. Hence, it has no $d$ -electron.
The electronic configuration of $Mn ( V \|)$ is $[A r] 3 d^{0} 4 s^{0} .$
The complex ions $\left[ Co \left( NH _{3}\right)_{6}\right]^{3+},\left[ Fe ( CN )_{6}\right]^{3-}$ and $\left[ Cr \left( H _{2} O \right)_{6}\right]^{3+}$ have 6, 5 and 3 $d$ -electrons
respectively in the central metal atom.
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${\left[ Co \left( NH _{3}\right)\right]^{3+}}$
${\left[ CoCl \left( NH _{3}\right)_{5}\right]^{2+}}$
${\left[ Co ( CN )_{6}\right]^{3-}}$
${\left[ Co \left( NH _{3}\right)_{5}\left( H _{2} O \right)\right]^{3+}}$
The spin-only magnetic moment value of the complex that absorbs light with shortest wavelength is $B.M.$ (Nearest integer)