$M o l e c u l e H y b r i d i s a t i o n S h a p e$

$A . P O_{4}^{- 3} 1. s p^{2} a) D i s t o r t e d o c t a h e d r a l$

$B . X e F_{6} 2. s p^{3} b) P y r a m i d$

$C . S O_{3}^{- 2} 3. s p^{3} d c) T e t r a h e d r a l$

$D . N O_{3}^{- 1} 4. s p^{3} d^{3} d) T r i a n g u l a r p l a n a r$

The correct match is

D - 1 - d

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C - 1 - d

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A - 2 - b

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B - 4 - c

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Solution

The correct option is A
D - 1 - d

$H y b r i d i s a t i o n o f N i n N O_{3}^{- 1} = 5 - \frac{(- 1)}{2} = \frac{6}{2} = 3 h y b r i d o r b i t a l s$

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Similar questions

$M o l e c u l e H y b r i d i s a t i o n S h a p e$

$A . P O_{4}^{- 3} 1. s p^{2} a) D i s t o r t e d o c t a h e d r a l$

$B . X e F_{6} 2. s p^{3} b) P y r a m i d$

$C . S O_{3}^{- 2} 3. s p^{3} d c) T e t r a h e d r a l$

$D . N O_{3}^{- 1} 4. s p^{3} d^{3} d) T r i a n g u l a r p l a n a r$

The correct match is

X e F_{4}

Math the compounds given in column I with hybridization and shape in column II

Column I Column II

(A) $X e F_{6}$ (1) $s p^{3} d^{3}$ – distorted octahedral

(B) $X e O_{3}$ (2) $s p^{3} d^{2}$ – square planar

(D) $X e F_{4}$ (4) $s p^{3} d^{2}$ – square pyramidal

Shape of the molecule having ${d s p}^{2}$ and $d^{2}$ ${s p}^{3}$ hybridisation is respectively:

In the molecule CH ≡ C -CH = $C H_{2}$ , the hybridisation of C -Cbond is

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