Solution

A. Carbon and Silicon belong to the fourth group . Hence both Carbon and Silicon is supposed to have a maximum co ordination no. of $$4$$ in commonly occurring carbon or Silicon compounds. But, although the maximum co- ordination number of Carbon in Carbon compounds is $$4$$; in case of Silicon, it is $$6$$. Availability of low lying d-orbitals in case of Silicon allows it to accept lone pair of electrons in the pressure of highly electronegative elements, due to which it has a co-ordination number of $$6$$. Hence statement A is False. B. Bond energy of $${ F }_{ 2 }$$ is less than $${ Cl }_{ 2 }$$ Bond energy, also known as Bond enthalpy (H) determines how strong the chemical bonds are between the $$2$$ atoms. When the Bond energy/Bond enthalpy is less, the bonds can be cleaved easily. So when we are comparing two halogens, Florine and Chlorine. Both have $$7$$ valence electrons, but the size of Florine is smaller than chlorine;due to the fact that atomic radius increases down the group. Now, due to smaller size and higher electron density, there is a repulsion between the electrons which is known as inter-electronic repulsion. Due to this repulsion between the electrons, both the atoms move apart. Hence the bond can be cleaved easily and the bond dissociation energy or bond enthalpy is less. Hence the statement B is true. C. Mn(III) oxidation state is more stable than Mn(II) in aqueous state. We know that atomic no. of Mn$$=25$$ $$\therefore $$Electronic configuration$$=1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 }3{ s }^{ 2 }3{ p }^{ 6 }3{ d }^{ 5 }4{ s }^{ 2 }$$ Now after removal of $$2$$ outermost electrons, it will become Mn(II) and the remaining d-orbitals will be half filled which is stable. And as we already know the most stable oxidation states have completely filled or half filled orbitals. In case of Mn(III), the electronic configuration $$=3{ d }^{ 4 }4{ s }^{ 0 }$$ (Image) As per the rules to obtain maximum stability, the orbitals should either be completely filled or half filled. Hence statement C is false. A. Elements of $${ 15 }^{ th }$$ group shows only $$+3$$ and $$+5$$ oxidation state. Group $$15$$ elements consist of nitrogen, Phosphorous, Arsenic, Antimony and Bismuth, these group of elements not only shows$$+3$$ and $$+5$$ oxidation state but also exhibits $$+1,+2,+4$$ and $$-3$$ oxidation states. In Nitrogen, since it has $$4$$ valence electrons, the oxidation states vary from $$+1$$ to $$+4$$. In case of Phosphorous, the intermediate oxidation states disproportionate into $$+5$$ and $$-3$$ in both acids and alkalis. Hence statement D is false. The correct answer:-The only true statement is B, i.e., Bond energy of $${ F }_{ 2 }$$ is less than $${ Cl }_{ 2 }$$