Give reasons for the following
(a) `PF_(5)` is know but `NF_(3)` is not
(b) `H_(2)O` is a good solvent
(c ) `BF_(3)` is non- polar but planar
(e) Carbon -oxygen `(C-O)` bond lengths are equal in `Na_(2)CO_(3)`
(f) `MgF_(2)` is more soluble in water than `MgCI_(2)` .

Explanation

Valence electronic configuration of `N =2s^(2) 2p^(3)` N-atom does not have vacant d-orbitals, so it cannot form five bonds with `F` (although `EN` of `F` is very high) So `NH_(3)` is not know
(ii) Valence electronic configuration of `P=3s^(2) 3p^(3) 3d^(0)` P-atom have vacant 3d-orbitals and `EN` of `F` is very high so F-atom can excite `3se^(-)s` to 3d orbitals thus forming five bonds with `F.So PF_(5)` is known
(b) `H_(2)O` is a good solvent because
(i) It is highly polar
(ii) It is capable of forming H-bonds among their molecules
(iii) For ionic solutes, it acts as a dielectric (substances which lower the strenght of eletric field)
(c) In `BF_(3)` B-atom is `sp^(2)` hybridised hence the shape of molecule is (triangular) planar But due symmertic structure `mu_("net") =0`
(d) N-atom in `overset(..)NF_(3)` is `sp^(3)` hybridised and the strure is pyramidal due to the presence of one lp of N-atom which gives a net `mu` (dipole moment) to molecule i.e `mu_("net") ne 0` which results in a polar molecule
(e) `na_(2) CO_(3) rarr 2Na^(o+) + CO_(3)^(2-)`
The `CO_(3)^(2-)` ion shows resonance as follows Due to delocalisation of `,'pi' e^(-') s` all `C-O` bonds are of similar nature and hence they have equal bond lenghts
(f) `MgFI_(2)` is more ionic than `MgCI_(2)` (Fajans' rule) hence `MgF_(2)` is more soluble in `H_(2)O` than `MgCI_(2)`
[Small cation, large anion `implies` more covalent `implies` less soluble in `H_(2)O)]`


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