Structure Of Atom

The correct answer is Option C)

Fluorine, a halogen, is the most electronegative atom, has the electronic configuration (valence shell).

Electronegativityrefers to the tendency of attracting electrons.

The electronic configuration of the most electronegative element is ns²np⁵. As after an electron is gained, the element becomes more stable.
Fluorine, with atomic number 9, is the most electronegative element. Its electronic configuration is 1s²2s²2p⁵.

Therefore the correct answer is Option C.

Discover More From Chapter: P-Block Elements

Structures and Oxidation States:

The first molecule shows a bromine (Br) atom with oxidation states of +6 and +4, and the oxygen atoms (O) have oxidation states of -1. The fluorine (F) atom has an oxidation state of -1.

The second structure features sulfur (S) atoms with oxidation states of +5. The oxygen atoms (O) in the sulfonic acid groups have a zero oxidation state, while the hydrogen atoms (H) are in their usual +1 state.

The third structure has two carbon (C) atoms each in the +2 oxidation state, with oxygen atoms in zero oxidation states.

Summary of Oxidation States:

• In the first structure, atoms of zero oxidation number: 0.
• In the second structure, the sulfur atoms (S) have +5 oxidation states, and oxygen atoms (O) have 0 oxidation states, totaling 2 atoms with zero oxidation states.
• In the third structure, the carbon atoms (C) each have a +2 oxidation state, while the oxygen atoms (O) are in zero oxidation state, giving 4 atoms with zero oxidation states.

Conclusion: The total number of atoms with zero oxidation state is 6. Thus, the total number of atoms with zero oxidation state in the given structures is 6.

The correct answer is 6

Total atom with zero oxidation number state are 6.

Kinetic Energy Calculation for Various Orbits

The total energy (T.E.) of an electron in Bohr’s nth orbit is given by:

The kinetic energy (K.E.) of the electron is the negative of the total energy:

Thus, is proportional to .

Calculate the K.E. for Each Option:

• For the first orbit of H atom ( ):
• For the first orbit of ( ):
• For the second orbit of ( ):
• For the second orbit of ( ):

Conclusion:

Comparing these values, the highest K.E. is associated with the first orbit of , with a value of 4.

Step 1: Evaluate Statement (A)

• The uncertainty principle states that the exact position and momentum of an electron cannot be simultaneously determined.
• This implies that electrons do not have well-defined paths or trajectories.
• Hence, statement (A) is correct.

Step 2: Evaluate Statement (B)

• Electrons in orbitals closer to the nucleus (e.g., 2s) have lower energy compared to electrons at a large distance (infinitely far away) from the nucleus.
• This is because the potential energy of attraction between the electron and the nucleus decreases as the electron moves closer.
• Hence, statement (B) is correct.

Step 3: Evaluate Statement (C)

According to Bohr’s model, the energy of an electron in the -th orbit is given by:

where is the atomic number.

• For , the energy is most negative, corresponding to the most stable orbit.
• Hence, statement (C) is correct.

Step 4: Evaluate Statement (D)

According to Bohr’s model, the velocity of an electron in the -th orbit is given by:

where is a constant.

• As increases, the velocity of the electron decreases because it becomes less tightly bound to the nucleus.
• Hence, statement (D) is incorrect.

Conclusion:

The correct statements are: (A, B, C).