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M.Sc Chemistry Class9 Atom Structure — practice questions

Explore 39 free Class9 Atom Structure questions in M.Sc Chemistry Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Class9 Atom Structure questions and exercises with answers

Showing questions 1–30 of 39.

The maximum number of electrons that can be accommodated in the $n$-th shell of an atom is:Read the answer and worked solution →The four quantum numbers describing an electron in an atom are:Read the answer and worked solution →According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \geq$:Read the answer and worked solution →Two electrons in the same orbital must have:Read the answer and worked solution →Calculate the de Broglie wavelength of an electron moving at $10^6$ m/s. (Take $h = 6.6 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg)Read the answer and worked solution →Bohr's first postulate: electrons revolve in ___ orbits around the nucleus.Read the answer and worked solution →In Bohr's model, angular momentum of an electron in orbit n equals:Read the answer and worked solution →Energy of electron in n-th orbit of hydrogen:Read the answer and worked solution →Principal quantum number n indicates:Read the answer and worked solution →Number of orbitals in subshell of azimuthal quantum number l:Read the answer and worked solution →How many electrons can fit in a single orbital?Read the answer and worked solution →Electron configuration of carbon (Z = 6):Read the answer and worked solution →For hydrogen, wavelength emitted in n=3 to n=2 transition (Balmer):Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →True or false: Bohr's modelRead the answer and worked solution →Complete the reason Bohr's model failsRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Match each quantum number to what it is calledRead the answer and worked solution →True or false: the quantum mechanical modelRead the answer and worked solution →Order the structure: from atom down to electronRead the answer and worked solution →Complete the (n + l) ruleRead the answer and worked solution →True or false: orbital energiesRead the answer and worked solution →Match each rule to what it forbids or requiresRead the answer and worked solution →Tap every statement Hund's rule requiresRead the answer and worked solution →The fundamental particlesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →

Worked practice questions

The maximum number of electrons that can be accommodated in the $n$-th shell of an atom is:

  1. $n^2$
  2. $n$
  3. $2n$
  4. $2n^2$
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Correct answer: D. $2n^2$

Bohr-Bury rule. Each shell of principal quantum number $n$ contains $n^2$ orbitals, and each orbital holds at most 2 electrons (Pauli). So shell capacity is $2n^2$. Quick check: K shell ($n=1$) holds 2, L ($n=2$) holds 8, M ($n=3$) holds 18, N ($n=4$) holds 32. All powers-of-two-friendly numbers.

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The four quantum numbers describing an electron in an atom are:

  1. $n, l, m_l, m_s$
  2. $n, l, m_l, m_l$
  3. $n, p, d, f$
  4. $n, l, m, s$
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Correct answer: A. $n, l, m_l, m_s$

**Principal $n$** sets the shell and energy. **Azimuthal $l$** sets the subshell shape (s, p, d, f). **Magnetic $m_l$** sets the orbital's orientation in space. **Spin $m_s$** picks $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$. First three come from the Schrödinger equation the spin comes from electron's intrinsic angular momentum. Option D lists subshell labels (s, p, d, f), not quantum numbers.

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According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \geq$:

  1. $\dfrac{h}{2\pi}$
  2. $h$
  3. $\dfrac{h}{4\pi}$
  4. $\dfrac{h}{4}$
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Correct answer: C. $\dfrac{h}{4\pi}$

$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} = \dfrac{\hbar}{2}$, where $\hbar = h/2\pi$ is the reduced Planck constant. **Meaning:** you cannot simultaneously know an electron's position and momentum with arbitrary precision. Trying to pin one down makes the other less knowable. This forces us to talk about probability clouds (orbitals) rather than precise electron trajectories. Option B is often a near miss in memory because $\hbar = h/2\pi$, but the uncertainty bound has the factor of 2 baked into the denominator twice over.

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