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NEET CHEMISTRYStructure of AtomMedium

Question

The incorrect statement among the following is:

A

Total orbital angular momentum of an electron in 's' orbital is equal to zero.

B

An orbital is designated by three quantum numbers, while an electron in an atom is designated by four quantum numbers.

C

The electronic configuration of N atom is

D

The value of m for dz² is zero.

Step-by-Step Solution

We evaluate the validity of each statement:

  1. Option A: The orbital angular momentum of an electron is given by l(l+1)h2π\sqrt{l(l+1)}\frac{h}{2\pi}. For an 's' orbital, the azimuthal quantum number l=0l = 0, so the angular momentum is zero. This statement is Correct.
  2. Option B: An atomic orbital is defined by three quantum numbers: principal (nn), azimuthal (ll), and magnetic (mlm_l). An electron within that orbital is further distinguished by the spin quantum number (msm_s), requiring four quantum numbers total. This statement is Correct.
  3. Option D: By convention in quantum mechanics, the magnetic quantum number ml=0m_l = 0 is assigned to the dz2d_{z^2} orbital (which lies along the z-axis). This statement is Correct.
  4. Option C: Since options A, B, and D are correct, this option must be the Incorrect statement. In the context of this NEET question, this option typically presents an electronic configuration diagram for Nitrogen (1s22s22p31s^2 2s^2 2p^3) that violates Hund's Rule (e.g., by pairing electrons in the 2p orbitals before singly occupying them) or Pauli's Exclusion Principle.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomincorrectstatementfollowing

More Structure of Atom Questions

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In hydrogen atom, what is the de Broglie wavelength of an electron in the second Bohr orbit? [Given that Bohr radius, $a_0 = 52.9$ pm]

A.211.6 pm
B.211.6 $\pi$ pm
C.52.9 $\pi$ pm
D.105.8 pm
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Maximum number of electrons in a subshell with l = 3 and n = 4 is:

A.14
B.16
C.10
D.12
EasySolve

If the principal quantum number n=6, the correct sequence of filling of electrons will be:

A.ns → np → (n-1)d → (n-2)f
B.ns → (n-2)f → (n-1)d → np
C.ns → (n-1)d → (n-2)f → np
D.ns → (n-2)f → np → (n-1)d
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The measurement of the electron position is associated with an uncertainty in momentum which is equal to $1 \times 10^{-18} \text{ g cm s}^{-1}$. The uncertainty in velocity of the electron will be: (mass of an electron is $9 \times 10^{-28} \text{ g}$)

A.1 × 10⁹ cm s⁻¹
B.1 × 10⁶ cm s⁻¹
C.1 × 10⁵ cm s⁻¹
D.1 × 10¹¹ cm s⁻¹
EasySolve

What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l = 1, m = 0

A.1
B.2
C.3
D.4
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The relation between $n_m$ ($n_m$ = number of permissible values of magnetic quantum number, $m_l$) for a given value of azimuthal quantum number ($l$) is:

A.$n_m = l + 2$
B.$l = \frac{n_m - 1}{2}$
C.$l = 2n_m + 1$
D.$n_m = 2l^2 + 1$
EasySolve

Which one of the following ions has electronic configuration [Ar]3d⁶? (At. no: Mn = 25, Fe = 26, Co = 27, Ni = 28)

A.Ni³⁺
B.Mn³⁺
C.Fe³⁺
D.Co³⁺
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How many electrons can fit in the subshell for which n = 3 and l = 1?

A.2
B.6
C.10
D.14
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