Question
Given below are two statements:
Statement-I Hybridisation, shape and spin only magnetic moment of $K_{3}\left\lbrack Co\left( {CO}_{3} \right)_{3} \right\rbrack$ is ${sp}^{3}{\text{ }d}^{2}$, octahedral and 4.9 BM respectively.
Statement-II Geometry, hybridisation and spin only magnetic moment values (BM) of the ions $\left\lbrack Ni(CN)_{4} \right\rbrack^{2 -},\left\lbrack {MnBr}_{4} \right\rbrack^{2 -}$ and $\left\lbrack {CoF}_{6} \right\rbrack^{3 -}$ respectively are square planar, tetrahedral, octahedral : ${dsp}^{2},{sp}^{3}$, ${sp}^{3}{\text{ }d}^{2}$ and 0, 5.9, 4.9.
In the light of the above statements, choose the correct answer from the options given below
Options
Solution
In $K_{3}\left\lbrack Co\left( {CO}_{3} \right)_{3} \right\rbrack \Rightarrow {sp}^{3}{\text{ }d}^{2}$ hybridized, octahedral
$$\begin{matrix} & \ \Rightarrow 4\text{~}\text{unpaired electron}\text{~} \\ & \ \Rightarrow 4.9\text{~}\text{B.M.}\text{~} \\ & \ \Rightarrow {\text{~}\text{dsp}\text{~}}^{2}\text{~}\text{hybridized, square planar}\text{~} \\ & \ \Rightarrow 0\text{~}\text{unpaired electron}\text{~} \\ & \ \Rightarrow 0\text{~}\text{B.M.}\text{~} \end{matrix}$$
$\left\lbrack {MnBr}_{4} \right\rbrack^{2 -}\ \Rightarrow {sp}^{3}$ hybridized , tetrahedral
$\Rightarrow 5$ unpaired electron
⇒ 5.9 B.M.
$\left\lbrack {CoF}_{6} \right\rbrack^{3 -}\ \Rightarrow {sp}^{3}{\text{ }d}^{2}$ hybridized, octahedral
⇒ 4 unpaired electron
⇒ 4.9 B.M.
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