Structure Of Atom

The correct answer is (C) : (A), (C) and (D)
Possible values of l for a given ‘n’ = 0,1,2 …(n – 1)
For l = 5, total orbitals = 2l + 1
= 2(5) + 1 = 11 orbital
Therefore A, C and D are correct statements

The uncertainty principle in quantum mechanics can be stated as Δx · Δp ≥ ħ/2, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and ħ is the reduced Planck's constant (h/2π). Here, the region of confinement of the electron is 2a₀, and thus, Δx = 2a₀.

Step 1: Calculate Δp. Using the uncertainty relation Δx · Δp ≥ ħ/2, we have:

Δp ≥ ħ/(2Δx)

Substitute Δx = 2a₀ and a₀ = 52.9 × 10^-12 m.

ħ = h/(2π) = (6.63 × 10^-34 Js)/(2π)

Δp ≥ (6.63 × 10^-34)/(2π × 2 × 52.9 × 10^-12).

Step 2: Calculate Δv (uncertainty in speed). The relationship between momentum and speed is p = m·v. Thus, Δp = m·Δv. Then:

Δv = Δp/m

Substitute the values: m = 9.1 × 10^-31 kg.

Δv ≥ [(6.63 × 10^-34)/(4π × 52.9 × 10^-12)]/(9.1 × 10^-31)

After calculating, Δv ≥ 547.952 km/s.

Step 3: Verification:

The computed value is approximately 548 km/s, which is within the expected range (548,548), validating our solution.

The correct answer is (A) : 1 and 2

Explanation:

(i) In 4d,
n = 4 l = 2

Radial nodes = n – l – 1
= 4 – 2 – 1
= 1
Angular nodes = l
= 2

• Statement I: Bohr’s model quantizes the angular momentum as , which is correct.
• Statement II: Bohr’s model is not based on the uncertainty principle; it assumes fixed orbits.
• The uncertainty principle applies to orbitals, where the probability of finding an electron is calculated.

The correct answer is :
The radii of the Bohr orbits in hydrogen atom are given by the formula:

where r_n is the radius of the nth Bohr orbit, n is the principal quantum number, h is the Planck's constant, ε₀ is the permittivity of free space, m is the mass of the electron, and e is the charge of the electron.
The ratio of radii of the second and third Bohr orbits can be calculated by substituting n=2 and n=3 in the above formula and taking the ratio:


Therefore, the ratio of radii of second and third Bohr orbit of H-atoms is

425

Significant figures represent all the digits in a number that are meaningful in terms of precision. They include all non-zero digits, zeros between non-zero digits, and trailing zeros in decimal numbers. Here is the breakdown:

• 0.00253 : Contains 3 significant figures (leading zeros are not counted).
• 1.0003 : Contains 5 significant figures (includes all digits, including zeros between non-zero digits).
• 15.0 : Contains 3 significant figures (the trailing zero is significant in decimal numbers).
• 163 : Contains 3 significant figures (all digits are non-zero).

Thus, 0.00253 (option A), 15.0 (option B), and 163 (option C) all have the same number of significant figures, which is 3.

Therefore, the correct answer is option (4).

Determining Lone Pairs on Xenon (Xe):

Using VSEPR theory, the number of lone pairs on the central atom Xenon (Xe) is calculated for each species:

1. XeF₅⁺: Xenon has 8 valence electrons. It forms 5 bonds with fluorine, and 1 electron is removed due to the positive charge.
   Lone pairs = 8 − 5 − 1 = 2 (not a single lone pair).
2. XeO₃: Xenon forms 3 bonds with oxygen atoms.
   Lone pairs = 8 − 6 = 2 (not a single lone pair).
3. XeO₂F₂: Xenon forms 4 bonds (2 with oxygen and 2 with fluorine).
   Lone pairs = 8 − 4 = 1 (single lone pair).
4. XeF₅: Xenon forms 5 bonds with fluorine atoms. One extra electron is added due to the negative charge.
   Lone pairs = 8 − 5 + 1 = 2 (not a single lone pair).
5. XeO₃F₂: Xenon forms 5 bonds (3 with oxygen and 2 with fluorine).
   Lone pairs = 8 − 5 = 1 (single lone pair).
6. XeOF₄: Xenon forms 5 bonds (1 with oxygen and 4 with fluorine).
   Lone pairs = 8 − 5 = 1 (single lone pair).
7. XeF₄: Xenon forms 4 bonds with fluorine atoms.
   Lone pairs = 8 − 4 = 2 (not a single lone pair).

Species with a Single Lone Pair:

XeO₂F₂, XeO₃F₂, XeOF₄, XeF₅

Total Number of Such Species:

4

A. For 1s orbital, the probability density is maximum at the nucleus.

• Correct. The 1s orbital has a spherically symmetrical distribution, and the probability density is highest at the nucleus, decreasing as the distance from the nucleus increases.

B. For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero.

• Incorrect. For the 2s orbital, the probability density is maximum at the nucleus, decreases to zero at a radial node, increases again to a smaller maximum, and finally decreases asymptotically to zero. It does not simply increase to a maximum before decreasing.
  

C. Boundary surface diagrams of the orbitals enclose a region of 100% probability of finding the electron.

• Incorrect. Boundary surface diagrams typically enclose a region where there is a high (usually 90%) probability of finding the electron, not 100%. Electrons can still be found outside this region with a small probability.

D. p and d-orbitals have 1 and 2 angular nodes respectively.

• Correct. The number of angular nodes is equal to the azimuthal quantum number ( ). For p orbitals, , and for d orbitals, .

E. Probability density of p-orbital is zero at the nucleus.

• Correct. p-orbitals have a nodal plane passing through the nucleus, making the probability density at the nucleus zero.

Final Answer:

The number of correct statements is 3 (A, D, and E).

Analyzing the lone pairs of electrons in the given molecules.
: Oxygen has 2 lone pairs, and each hydrogen atom has 0 lone pairs. Total = 2 lone pairs.
: Nitrogen in has no lone pairs in the molecular structure. Total = 0 lone pairs.
: Carbon in CO has no lone pairs, but oxygen has 2 lone pairs. Total = 2 lone pairs.
: Xenon in has 2 lone pairs, and each fluorine atom has 3 lone pairs. Total = 2 lone pairs.
: Nitrogen in has 1 lone pair, and each hydrogen atom has 0 lone pairs. Total = 1 lone pair.
: Nitrogen in NO has 1 lone pair, and oxygen has 2 lone pairs. Total = 3 lone pairs.
: Carbon in has no lone pairs, and oxygen has 2 lone pairs on each oxygen atom. Total = 4 lone pairs.
: Each fluorine atom in has 3 lone pairs. Total = 3 lone pairs. From the analysis, the molecules that contain only 2 lone pairs are , , and .
Therefore, the correct answer is 4 molecules.

Statement Analysis:

• (A): Incorrect—NF₃ has a trigonal pyramidal structure due to the lone pair on the nitrogen atom, not a planar structure.
• (B): Correct—N₂ has a triple bond, resulting in a shorter bond length compared to O₂, which has a double bond.
• (C): Correct—Isoelectronic species such as O₂⁻, F⁻, and N₂ have identical bond orders because they share the same number of electrons and similar electronic configurations.
• (D): Incorrect—Water (H₂O) has a higher dipole moment due to the bent geometry and the high electronegativity of oxygen, whereas NF₃ has a lower dipole moment due to the partial cancellation of dipoles in its pyramidal structure.

Conclusion:

Correct Statements: (B) and (C).

Incorrect Statements: (A) and (D).