A replacement reaction which is where the positively charged portion of one of the reactants takes on the negatively charge portion of the other reactant without giving any portion of itself to the other to form products.
The axial positions the bond angle is 120 degrees and in equatorial positions the bond angle is 90 degrees.
Functional groups on central atom gets reduce if lone pairs are added.
Explanation:
The number of lone pairs and base pairs of electrons tells the geometry of the molecule.
VSEPR Theory helps to know the lone pairs and bond pair electrons on the centre atom of the molecule.
Example of molecule containing 5 electron pairs can have four bond pairs and 1 lone pair.
eg: Cl
the repulsion is shown as
lp-lp> lp-bp>bp-bp
These are in equatorial position because of the repulsion of lone pairs.
It can have 2 lone pairs and 3 bond pairs. eg. Xe
Lone pairs in this is also in equatorial position as
lp-lp> lp-bp> bp-bp
In axial positions the bond angle is 120 degrees
in equatorial positions the bond angle is 90 degrees, due to the repulsion in lone pair of electrons.
If 1 lone pair is there it can be replaced by bonding with hydrogen.
If 2 lone pairs are there then bonding with oxygen is there. The covalent bond is formed.
Answer:
See Explanation
Explanation:
Let us consider the ionization of these compounds;
H3PO4 ⇔3H^+ + PO4^3-
H3BO3 ⇔3H^+ + BO3^3-
The next to consider is the type of electrolyte the both solution are; the both solutions are weak electrolytes and weak electrolytes do not ionize to a large extent.
The implication of this is that, not so much number of ions is added to the solution due to the poor ionization of these weak electrolytes. Hence, in spite of the subscript of 3, the conductivity of the solution does not significantly improve for the reason stated here quite unlike when strong electrolytes are used.
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
