NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Which has regular tetrahedral geometry?

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Correct option: C

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

The correct answer is C (

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

). Boron in

{\left[ {\mathrm{B}{\mathrm{F}_4}} \right]^ - }

has four bond pairs and no lone pairs, resulting in a regular tetrahedral geometry.

More Questions from CO-ORDINATION COMPOUNDS

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\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

\left[\mathrm{Cr}(\mathrm{Ox})_{2} \mathrm{ClBr}\right]^{3-}

\text{ox =}

x

{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^x}

\mathrm{M}{\mathrm{n}^{2 + }}

\mathrm{CuS}{\mathrm{O}_4}

{\mathrm{K}_2}\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]