The hybridisation of C in diamond,graphite an | vedclass

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(B) The hybridization of carbon in the given compounds is determined by the number of sigma bonds and lone pairs attached to the carbon atom.$1$. In d

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$sp^3, sp^2, sp$

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(B) The hybridization of carbon in the given compounds is determined by the number of sigma bonds and lone pairs attached to the carbon atom.$1$. In diamond,each carbon atom is bonded to four other carbon atoms via single bonds,resulting in $sp^3$ hybridization.$2$. In graphite,each carbon atom is bonded to three other carbon atoms in a planar hexagonal structure,resulting in $sp^2$ hybridization.$3$. In ethyne $(C_2H_2)$,each carbon atom is bonded to one hydrogen atom and one other carbon atom via a triple bond,resulting in $sp$ hybridization.Therefore,the order is $sp^3, sp^2, sp$.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

The hybridisation of $C$ in diamond,graphite and ethyne is in the order

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In the compound $CH_3COCl$,which type of orbitals have been used by the carbonyl carbon in bond formation?

The hybridization of the central atom of $ClF_3$,$NH_3$,and $SO_3$ are respectively:

How many $sp^{2}$ and $sp-$ hybridised carbon atoms are present respectively in the following compound?

Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$ $sp^3$ tetrahedral
$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

The species having bond angles of $120^{\circ}$ is

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