Only those $$d$$ orbitals whose lobes are directed along $$X, Y$$ and $$Z$$ directions hybridise with $$s$$ and $$p$$ orbitals. In other three, $$d$$ orbitals namely $${d_{xy}},{d_{yz}}$$ and $${d_{xz}},$$ the lobes are at an angle of $${45^ \circ }$$ from both axis, hence the extent of their overlap with $$s$$ and $$p$$ orbitals is much lesser than $${d_{{x^2} - {y^2}}}$$ and $${d_{{z^2}}}$$ orbitals.
33.
How many $$sigma$$ and $$pi$$ bonds are present in toluene?
A
$$10\,\sigma $$ and $${\text{3}}\,\pi $$ bonds
B
$$12\,\sigma $$ and $${\text{3}}\,\pi $$ bonds
C
$$15\,\sigma $$ and $${\text{3}}\,\pi $$ bonds
D
$$6\,\sigma $$ and $${\text{3}}\,\pi $$ bonds
Answer :
$$15\,\sigma $$ and $${\text{3}}\,\pi $$ bonds
$$\,H_2^{2 + } = \sigma 1{s^0}{\sigma ^*}1{s^0}$$
bond order for $$H_2^{2 + } = \frac{1}{2}\left( {0 - 0} \right) = 0\,\,\,$$
$$H{e_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}$$
bond order for $$H{e_2} = \frac{1}{2}\left( {2 - 2} \right) = 0$$
so both $$H_2^{2 + }$$ and $$H{e_2}$$ does not exist
36.
Element $$X$$ is strongly electropositive and element $$Y$$ is strongly electronegative. Both are univalent. The compound formed would be
$$SC{l_4}$$ is not isostructural with $$SiC{l_4}$$ because it shows square planar structure due to involvement of
repulsion between lone pair and bond pair of electrons.
$$SO_4^{2 - }$$ shows tetrahedral structure due to $$s{p^3}$$ hybridisation.
$$PO_4^{3 - }$$ shows tetrahedral structure due to $$s{p^3}$$ hybridisation.
$$NH_4^ + $$ shows tetrahedral structure due to $$s{p^3}$$ hybridisation.