Which of the underlined atoms in the molecule | vedclass

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(D) The hybridization of an atom is determined by its steric number,which is the sum of the number of sigma bonds and lone pairs. $(u) H_2\underline{C

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$v, x$ and $y$

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(D) The hybridization of an atom is determined by its steric number,which is the sum of the number of sigma bonds and lone pairs. $(u) H_2\underline{C}=CH-CH_3$: The underlined carbon has $3$ sigma bonds and $0$ lone pairs. Steric number = $3$,so it is $sp^2$ hybridized. $(v) CH_2=\underline{C}=CHCl$: The central carbon has $2$ sigma bonds and $0$ lone pairs. Steric number = $2$,so it is $sp$ hybridized. $(w) CH_3-\underline{C}H_2^+$: The carbocation carbon has $3$ sigma bonds and $0$ lone pairs. Steric number = $3$,so it is $sp^2$ hybridized. $(x) H-C \equiv C-H$: Each carbon has $2$ sigma bonds and $0$ lone pairs. Steric number = $2$,so it is $sp$ hybridized. $(y) CH_3-\underline{C} \equiv N$: The carbon has $2$ sigma bonds and $0$ lone pairs. Steric number = $2$,so it is $sp$ hybridized. $(z) (CH_3)_2C=\underline{N}-NH_2$: The nitrogen has $2$ sigma bonds and $1$ lone pair. Steric number = $3$,so it is $sp^2$ hybridized. Therefore,$v, x$ and $y$ have $sp$-hybridization.

Which of the underlined atoms in the molecules shown below have $sp$-hybridization?
$(u) H_2\underline{C}=CH-CH_3$
$(v) CH_2=\underline{C}=CHCl$
$(w) CH_3-\underline{C}H_2^+$
$(x) H-C \equiv C-H$
$(y) CH_3-\underline{C} \equiv N$
$(z) (CH_3)_2C=\underline{N}-NH_2$

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Which of the underlined atoms in the molecules shown below have $sp$-hybridization? $(u) H_2\underline{C}=CH-CH_3$ $(v) CH_2=\underline{C}=CHCl$ $(w) CH_3-\underline{C}H_2^+$ $(x) H-C \equiv C-H$ $(y) CH_3-\underline{C} \equiv N$ $(z) (CH_3)_2C=\underline{N}-NH_2$