thode aur dbts...

Q1) Should be True..

Is tht NaCl in Q3 ?.....My guesses - b) & d)

Q2) Er...yes M Pt. generally decreases dwn the group, due to an increase in ionic character..

Q4) Backdonation should be possible...(as far as the eye indicates :p)
Arshad bhai, why did u say no ?!?

@debo ..whats reason for Q1 ???

Lithium is the most powerful reducing agent because of its largest value of hydration energy....AND HYDRATION ENERGY is high bcos-Lithium is the smallest atom among alkali metals. The degree of hydration is greater if the size of the ion is smaller. Hence Li+ ions are hydrated to the maximum extent. ...

Ans 1) TRUE
Lithium, the first element of the group is expected to be the least reducing agent due to its very high ionization energy. However, it is the strongest reducing agent as indicated by the reduction potential value.

Ionization is the property of isolated atoms in the gaseous state.
M (g) → M ⁺ (g) + e- Ionization energy
Oxidation potential is the property when the metal goes into the solution as M⁺(aq) ions:
M (s) → M ⁺ (aq) + e- Oxidation Potential
The process of oxidation potential could proceed as:
(i) M (s) → M (g) Sublimation
(ii) M (g) → M ⁺ (g) + e- Ionization energy
(iii) M ⁺ (g) + H₂O → M + (aq) Hydration
The overall tendency for the change depends upon the net effect of the three steps. Since, Li+ has the smallest size and is hydrated to maximum extent, a very large amount of energy (called hydration energy) is released in the third step. This compensates the higher energy needed to remove electron in second step. The net effect is that it has greater tendency to lose electrons in solution than other alkali metals and the reaction:
Li (s) → Li ⁺ (aq) + e- Occurs easiy
Therefore, lithium is the strongest reducing agent because of its greater hydration energy.

Ans 4) YES
PF₃ can act as donor molecules using their lone pair to form a coordinate bond, for eg in Ni (PF₃) ₄. In addition to this sigma bond, there is pie backbonding frm a filled orbital on the metal to an empty d orbital on P, similar to the way CO acts as a ligand.

Ni (C0) ₄ + 4 PF₃ → Ni (PF3)₄ + 4 CO

@uttara NaCl

@tushar thx

for q4) no
in PF3 p has vacant 3d orbitals and lone pairs of F are in 2p
the extent of 2p-3d bonding is very less due to large energy
gap

q5) total number of electrons to be accomodated in hybrid orbitals is odd in both => either we should have a singlet bond or a unpaired electron both of which is not possible

for q3 it is just ratta...

thx all

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