The state of hybrid orbitals of carbon in CO_ | vedclass

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(C) The hybridization state is calculated using the formula: $	ext{Hybridization} = \frac{1}{2}(V + M - C + A)$Where $V$ is the number of valence ele

Vedclass · Accepted Answer

$sp, sp^{3}$ and $sp^{2}$

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(C) The hybridization state is calculated using the formula: $	ext{Hybridization} = \frac{1}{2}(V + M - C + A)$Where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $CO_{2}$: $V=4, M=0, C=0, A=0$. $	ext{Hybridization} = \frac{4+0}{2} = 2 \Rightarrow sp$.For $CH_{4}$: $V=4, M=4, C=0, A=0$. $	ext{Hybridization} = \frac{4+4}{2} = 4 \Rightarrow sp^{3}$.For $CO_{3}^{2-}$: $V=4, M=0, C=0, A=2$. $	ext{Hybridization} = \frac{4+0+2}{2} = 3 \Rightarrow sp^{2}$.Thus,the states are $sp, sp^{3}, sp^{2}$ respectively.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

The state of hybrid orbitals of carbon in $CO_{2}$,$CH_{4}$ and $CO_{3}^{2-}$ respectively is

Similar Questions

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

Which of the following hybridisation has the highest percentage of $s-$character?

$C-H$ bond length is greatest in

Which of the following pairs have the same type of hybridization but different shapes?

What is the hybridisation of each carbon in $H_2C = C = CH_2$?

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