Single Choice

The hybridisation of carbon in diamond, graphite and acetylene are respectively:

A$$sp^{3},\;sp,\;sp^{2}$$
B$$sp^{3},\;sp^{2},\;sp$$
Correct Answer
C$$sp,\;sp^{2},\;sp^{3}$$
D$$sp^{2},\;sp^{3},\;sp$$

Solution

Hybridization of carbon in:

Diamond: $$sp^3$$.In diamond each carbon combine with 4 other carbon atoms with four sigma bonds. so $$sp^3$$ hybridization.

Graphite: $$sp^2$$.In graphite each carbon combine with 3 other carbon atoms with three sigma bonds. so $$sp^2$$ hybridization.

Acetylene ($$C_2H_2$$) : $$sp$$.In acetylene one carbon combine with another carbon atom with three bonds(1 sigma and 2 pi bonds). so $$sp^2$$ hybridization.

Hence option B is correct.

SIMILAR QUESTIONS

Chemical Bonding

The hybridisation of the central atom in the following species are respectively:$$[Ni (CN)_4]^{2-},\, XeO_4,\, SF_4\,$$ and $${NO_3}^-$$

Chemical Bonding

An octahedral complex is formed when hybrid orbitals of the following types are involved:

Chemical Bonding

Pair of species having identical shapes for molecules is:

Chemical Bonding

Which type of hybridization Is shown by carbon atoms from left to right in the given compound? $${CH_2 = CH - C \equiv N}$$

Chemical Bonding

Whay is that $$SF_4$$ molecule , the lone pair of $$e^{\circleddash }$$. occupy equatorial positions in preference to axial position? What is the shape of the molecule.

Chemical Bonding

The species in which the N atom is in a state of sp hybridization is :

Chemical Bonding

The hybridization of orbitals of $${N}$$ atom in $${N}{O}_{3}^{-},\ {N}{O}_{2}^{+}$$ and $${N}{H}_{4}^{+}$$ are respectively:

Chemical Bonding

In which of the following molecules S atom does not assume $$\displaystyle { sp }^{ 3 }$$ hybridsation?

Chemical Bonding

Which of the following organic compounds has same hybridization as its combustion product, $$CO_2$$?

Chemical Bonding

Among the triatomic molecules/ions, $$\displaystyle BeCl_{2},\ {N_{3}}^{-},\ N_{2}O,\ NO_{2}^{+},\ O_{3},\ SCl_{2},\ {ICl_{2}}^{-},\ {I_{3}}^{-}$$ and $$\displaystyle XeF_{2},$$ the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is [Atomic number : $$S = 16, Cl = 17,I=53$$ and $$Xe = 54 ]$$

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