Single Choice

Which of the following has $$sp^3$$ hybridization?

A$$XeO_3$$
Correct Answer
B$$BCl_3$$
C$$XeF_4$$
D$$BBr_3$$

Solution

To calculate the hybridization, we have: $$s = \dfrac{\text{valence shell electron of central metal + number of mono atomic atoms}}{2}$$ In $$XeO_3$$, we have, $$s = \dfrac{8+0}{2} = 4$$ if $$s=4$$, we have hybridization as $$sp^3$$ In $$BCl_3$$, we have, $$s = \dfrac{3+3}{2} = 3$$ if $$s = 3$$, we have hybridization as $$sp^2$$ In $$XeF_4$$, we have $$s=\dfrac{8+4}{2} = 6$$ if $$s=6$$, we have hybridization as $$sp^3d^2$$ In $$BBr_3$$, we have, $$s=3$$ and we have hybridization as $$sp^2$$. Hence, option A is correct.

SIMILAR QUESTIONS

p - block Elements

The compound in which the number of $$d\pi-p\pi$$ bonds are equal to those present in $$ClO^{-}_{4}$$ is:

p - block Elements

The F-Xe-F bond angle in $$XeF_{4}$$ is:

p - block Elements

The type of hybridisation and number of lone pair(s) of electrons of $$Xe$$ in $$XeOF_4$$ ,respectively, are:

p - block Elements

Choose the incorrect formula out of the four compounds for an elements $$X$$ below:

p - block Elements

In $$XeO_3F_2$$, the number of bond pair(s), $$\pi$$-bond(s) and lone pair(s) on Xe atom respectively are.

p - block Elements

A compound of $$Xe$$ and $$F$$ is found to have $$53.5$$% of $$Xe$$. What is oxidation number of $$Xe$$ in this compound?

p - block Elements

Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code: Column-I Column-II (a) $$XeF_4$$ (i) Pyramidal (b) $$XeF_6$$ (ii) Square planer (c) $$XeOF_4$$ (iii) Distorted octahedral (d) $$XeO_3$$ (iii) Square pyramidal

p - block Elements

The shape of $$XeO_{2}Fe$$ molecule is

p - block Elements

The four sets of shape and hybridization of xenon oxyfluorides are given. Choose the wrong set.

p - block Elements

In the clathrates of xenon with water the nature of bonding in $$Xe$$ and $$H_{2}O$$ molecule is:

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