NEET UG CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Analyzing +4 | -1

Which have octahedral shape

({d^2}s{p^3})

hybridization of central atom?

View answer and explanation

Correct option: D

All are correct

Option D (All are correct) is correct because all the given complexes,

{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}

{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}

, and

{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^ + }

, possess octahedral shape with

d^2sp^3

hybridization.

More Questions from CO-ORDINATION COMPOUNDS

Practice All Chapter MCQs →

\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]