Chemical Bonding
The hybridisation of carbon in diamond, graphite and acetylene are respectively:
The hybridisation of the central atom in the following species are respectively:$$[Ni (CN)_4]^{2-},\, XeO_4,\, SF_4\,$$ and $${NO_3}^-$$
The order of hybridisation of the central atom in the following species; $$[Ni (CN)_4]^{2-} \ - dsp^2$$
$$XeO_4\ - sp^3$$
$$SF_4\ - sp^3d$$
$$NO_3^\ - sp^2$$
The hybridisation of carbon in diamond, graphite and acetylene are respectively:
An octahedral complex is formed when hybrid orbitals of the following types are involved:
Pair of species having identical shapes for molecules is:
Which type of hybridization Is shown by carbon atoms from left to right in the given compound? $${CH_2 = CH - C \equiv N}$$
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.
The species in which the N atom is in a state of sp hybridization is :
The hybridization of orbitals of $${N}$$ atom in $${N}{O}_{3}^{-},\ {N}{O}_{2}^{+}$$ and $${N}{H}_{4}^{+}$$ are respectively:
In which of the following molecules S atom does not assume $$\displaystyle { sp }^{ 3 }$$ hybridsation?
Which of the following organic compounds has same hybridization as its combustion product, $$CO_2$$?
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 $$Xe = 54 ]$$