The formation of the dative covalent (or coordinate covalent) bond


The formation of a dative covalent bond, also known as a coordinate covalent bond, occurs when one atom donates both of its electrons to another atom to form a covalent bond. This type of bond is also known as a “dative” bond because the electrons are donated by one atom to another, rather than being shared equally.

For example, in the case of H3O+ (hydronium ion) and NH4+ (ammonium ion), the dative covalent bond forms between the oxygen atom in H3O+ and the nitrogen atom in NH4+. The electron diagram for this formation is as follows:

H3O+:

 
   O
  / \
 H   H

NH4+:

    N
   / \
  H   H
  |   |
  H   H

The oxygen atom in H3O+ has a lone pair of electrons, while the nitrogen atom in NH4+ has a partially filled outer shell. When these two atoms come into close proximity, the oxygen atom donates one of its lone pair of electrons to the nitrogen atom, forming a covalent bond. This results in the formation of H3O+ NH4+ molecule.

The electron diagram for this molecule is:

H3O+ NH4+ :

     
   O
  / \
 H   H
     |
     N
    / \
   H   H
   |   |
   H   H

In this case, the oxygen atom donates one of its electrons to the nitrogen atom, forming a dative covalent bond. As a result, the nitrogen atom now has a full outer shell of electrons, and the oxygen atom has a single lone pair of electrons.

It’s important to note that this type of bond formation can occur between other types of atoms and molecules as well, not just H3O+ and NH4+. Dative covalent bond formation can be observed in many chemical reactions and is an important concept in understanding the behavior of molecules and their interactions with other molecules.

Alkanes and alkenes react differently with bromine and potassium permanganate.


Reaction of Alkanes with Bromine:

Alkanes typically do not react with bromine in the absence of light or heat. However, when exposed to light or heat, alkanes can undergo a free radical substitution reaction with bromine, which results in the formation of a bromoalkane. The equation for this reaction is:

CnH2n+2 + Br2 → CnH2n+2Br

Reaction of Alkenes with Bromine:

Alkenes can react with bromine in the absence of light or heat. This reaction is called an electrophilic addition reaction, and it results in the formation of a dibromoalkane. The equation for this reaction is:

CnH2n + Br2 → CnH2nBr2

Reaction of Alkanes with Potassium Permanganate:

Alkanes do not react with potassium permanganate in the absence of heat or an acid catalyst. However, when heated with an acid catalyst, alkanes can undergo an oxidation reaction that results in the formation of carboxylic acids or alcohols. The equation for this reaction is:

CnH2n+2 + KMnO4 + H+ → CnH2n+1COOH or CnH2n+1OH

Reaction of Alkenes with Potassium Permanganate:

Alkenes do not react with potassium permanganate in the absence of heat or an acid catalyst. However, when heated with an acid catalyst, alkenes can undergo an oxidation reaction that results in the formation of carboxylic acids or alcohols. The equation for this reaction is:

CnH2n + KMnO4 + H+ → CnH2n-1COOH or CnH2n-1OH

It’s important to note that the reagents used and the conditions for the reaction can affect the products formed in these reactions.

Position vs. Time and velocity vs. time for a free-falling object:


The position of a free-falling object is measured downward from a certain starting point.

The position of the object increases with time as it falls.

The graph would be a straight line with a negative slope, starting at the initial position and ending at the final position.

Velocity vs. Time graph for a free-falling object:

The velocity of a free-falling object increases with time as it falls.

The graph would be a straight line with a positive slope, starting at zero and ending at the final velocity.

To determine the acceleration due to gravity, we can use the equation:

a = v^2/s or a = 2s/t^2

Where a is the acceleration, v is the final velocity, s is the final position, and t is the time of fall.

Since the acceleration due to gravity is constant, the slope of the velocity vs time graph is the acceleration.

Position vs. Time:

The graph for position vs. time for a free-falling object would be a straight line that starts at a non-zero position (the initial height of the object) and decreases as time goes on. The slope of this line would represent the velocity of the object. The equation for this line would be:

y = -g*t + h

where g is the acceleration due to gravity, t is the time, and h is the initial height of the object.

Velocity vs. Time:

The graph for velocity vs. time for a free-falling object would be a straight line that starts at a non-zero velocity (the initial velocity of the object) and decreases as time goes on. The slope of this line would represent the acceleration of the object. The equation for this line would be:

y = -g*t + v

where g is the acceleration due to gravity, t is the time, and v is the initial velocity of the object.

Determining the Acceleration due to Gravity:

To determine the acceleration due to gravity, we can use the slope of either the position vs. time or velocity vs. time graph. For the position vs. time graph, the slope is -g, and for the velocity vs. time graph, the slope is -g. Therefore, we can conclude that the acceleration due to gravity is -g.

Momentum


Momentum is a measure of an object’s inertia or its resistance to changes in its motion. It is calculated as the product of an object’s mass (m) and its velocity (v). So, the formula for momentum (p) is:

p = m x v

Where: p = momentum m = mass of the object (in kg) v = velocity of the object (in m/s)

Example: If an object has a mass of 5 kg and is moving at a velocity of 10 m/s, then its momentum would be: p = 5 kg x 10 m/s = 50 kg m/s

Momentum is a vector quantity, which means it has both magnitude and direction. The magnitude of momentum is simply the product of an object’s mass and velocity, as described by the formula p = mv. The direction of momentum is the same as the direction of an object’s velocity.

For example, consider a car moving east with a mass of 1000 kg and a velocity of 20 m/s. The magnitude of its momentum would be: p = mv = 1000 kg x 20 m/s = 20,000 kg m/s

The direction of the momentum would be in the east, the same as the direction of the car’s velocity.

Another example is a ball thrown upwards with a mass of 0.2 kg and a velocity of 10 m/s upward. The magnitude of its momentum would be: p = mv = 0.2 kg x 10 m/s = 2 kg m/s

The direction of the momentum would be upward, the same as the direction of the ball’s velocity.

It’s important to note that in collisions and other interactions between objects, the total momentum of a closed system (systems with no external forces) is conserved, which means that the momentum of all objects before and after the collision will be equal. This is the principle of conservation of momentum.

Oxidation and Reduction