If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).
<h3>What is base dissociation constant? </h3><h3 />
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 6.2× 10^(-10)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{6.2×10^(-10) }
= 1.6× 10^(-5)
Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).
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Answer:
more negative
Explanation:
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Answer: 2 lone pairs, square planar
Explanation:
Using the VSEPR ( Valence Shell Electron Pair Repulsion)Theory
To calculate the number of lone pairs electron can be done using the formula;
Number of electrons = ½ (V+N-C+A)
V mean valency of the central atom
N means number of monovalent bonding atoms
C means charge on cation
A means charges on anion
Therefore, to calculate the number of lone pair electron C=A=0;
Number of electrons = ½ (8+4) = 12/2 = 6
Number of bonding pair = 4
Number of lone pairs of electron = 6-4 = 2
The hybridrization of the compound is sp3d2 because the number of electrons around the central atom is 6.
The geometry of the compound is square planar and this is because of the repulsion between the bonding pair of electrons and lone pair of electrons which causes the lone pair of electrons to lie in a perpendicular plane in order to acquire stability.
Metals are on the left side of the table and nonmetals are on the left with metalloids between them. And the noble gases are all in group 18 of the periodic table.