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Square Planar Complexes

Shape of Cl...

Question

Shape of $C l O_{4}^{-}$ ion is:

octahedral

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tetrahedral

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pyramid

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square planar

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Solution

The correct option is C tetrahedral
$C l O_{4}^{-}$ has $4$ bond pairs and zero lone pairs. Hence, it requires $4$ orbitals. The hybridization is $s p^{3}$ . Shape is tetrahedral.

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Similar questions

C l O_{4}^{-}

C l O_{4}^{-}

s p^{3}

C l O_{4}^{-}

Ion	$C l O_{4}^{-}$	$I O_{4}^{-}$	$B r O_{4}^{-}$
Reduction potential $E^{\circ}$ /V	$E^{\circ}$ = 1.19V	$E^{\circ}$ = 1.65V	$E^{\circ}$ = 1.74V

C l O_{4}^{-}

C l O_{4}^{-}

18^{\circ} C

N H_{4}^{+}

C l O_{4}

^{-}

6.6 \times 10^{- 4}

5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}

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