why do phosphorous pcl5 has ten electrons. justify
Phosphorus belonging to group V has five valence electrons. Chlorine has seven valence electrons. These five chlorine atoms uniformly distribute themselves around phosphorus atom to form five covalent bonds with P giving PCl5. The structure is given as:
But look at the phosphorus atom there are 2 x 5 = 10 valence electrons. This is referred to as an expanded octet. Atoms in period 3 and below in the periodic table may expand their octet because they have d orbitals. This is in contrast elements with atomic numbers of 1 – 10. That includes the elements in period 1 (H and He) and elements in period 2 (Li, Be, B, C, N, O, F and Ne). The first 10 elements of the periodic table do not have d orbitals and cannot expand their valence electrons outside of their duet (H and He) or octet for those in period 2.
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Most chemists agree that the bonding in molecules such as PCl5 is quite straightforward:
P in its atomic ground state is [Ne] 3s2 3p^3. In PCl5 an s e- is promoted to a 3d AO (actually 3dz^2) to give the configuration [Ne] 3s^1 3p^3 3d^1 You can at your level consider the P as sp^3d hybridized (cf sp^3 hybridization in C). The five AOs on the P can each overlap with a half-filled 3p AO on a Cl atom to form five P:Cl σ bonds. It comes from LCAO theory that overlap of five AOs gives five bonding MOs each of which holds 2e-s. Yes, the P atom has 10 e- in its valence shell and violates the octet rule, but there are numerous cmpds of the 3rd row and lower elements where this is the case. As in CCl4 the hybridization promotion energy is more than compensated for by the energy gained by forming two additional P:Cl bonds.
For N the e- configuration is [He] 2s^2 2p^3 but there are no low energy AOs available for hybridization so N and other 2nd period elements are restricted to employing only their 2s and 2p orbitals (as given by the Schrodinger wave equation) and maximum energy is gained if full use of these AOs is made. sp^3 hybridization corresponds to four AOs that give four bonding MOs that each hold 2 e- giving eight e- total. (Now let me think now - isn't there some rule that is called the octet rule?)