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NEET UG Unit 1B1: Atomic Structure & Quantum Mechanics — practice questions

Explore 4 free Unit 1B1: Atomic Structure & Quantum Mechanics questions in NEET UG Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1B1: Atomic Structure & Quantum Mechanics questions and exercises with answers

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A photon emitted in a Bohr-model transition between levels n_i and n_f has frequency given by:Read the answer and worked solution →Which of the following is incorrect?Read the answer and worked solution →Find the formula of halide of a metal whose successive ionization energies are 10, 16, 25, 400, 500 eV mol$^{-1}$ respectively:Read the answer and worked solution →The distance between 4th and 3rd Bohr orbits of $He^{+}$ is:Read the answer and worked solution →

Worked practice questions

A photon emitted in a Bohr-model transition between levels n_i and n_f has frequency given by:

  1. hν = E_n_i E_n_f
  2. hν = E_n_i + E_n_f
  3. hν = |E_n_i − E_n_f|
  4. ν = h/(E_n_i − E_n_f)
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Correct answer: C. hν = |E_n_i − E_n_f|

Bohr's frequency condition: hν = |ΔE|.

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Which of the following is incorrect?

  1. $\mathrm{IE}_1$ of $\mathrm{Li} < \mathrm{IE}_1$ of Be
  2. $\mathrm{IE}_1$ of $\mathrm{Be} < \mathrm{IE}_1$ of B
  3. $\mathrm{IE}_1$ of $\mathrm{Li} > \mathrm{IE}_1$ of Na
  4. $\mathrm{IE}_1$ of $\mathrm{He} > \mathrm{IE}_1$ of Ne
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Correct answer: B. $\mathrm{IE}_1$ of $\mathrm{Be} < \mathrm{IE}_1$ of B

Across a period, first ionization energy generally increases, but there are known exceptions due to subshell stability. In period $2$, $\mathrm{Be}$ has configuration $1s^2 2s^2$ and $\mathrm{B}$ has configuration $1s^2 2s^2 2p^1$. Removing a $2p$ electron from $\mathrm{B}$ is easier than removing a filled $2s$ electron from $\mathrm{Be}$ so $$\mathrm{IE}_1(\mathrm{Be}) > \mathrm{IE}_1(\mathrm{B}).$$ Therefore statement $\mathrm{Be} < \mathrm{B}$ is incorrect. Also, $$\mathrm{IE}_1(\mathrm{Li}) < \mathrm{IE}_1(\mathrm{Be})$$ is correct, $$\mathrm{IE}_1(\mathrm{Li}) > \mathrm{IE}_1(\mathrm{Na})$$ is correct because ionization energy decreases down the group, and $$\mathrm{IE}_1(\mathrm{He}) > \mathrm{IE}_1(\mathrm{Ne})$$ is also correct. After checking all options, the incorrect statement is $\text{(B)}$.

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Find the formula of halide of a metal whose successive ionization energies are 10, 16, 25, 400, 500 eV mol$^{-1}$ respectively:

  1. $MX$
  2. $MX_2$
  3. $MX_3$
  4. $M_2X$
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Correct answer: C. $MX_3$

The large jump in ionization energy occurs between the 3rd and 4th ionizations: $25 \to 400$. This means the atom loses three valence electrons relatively easily, but removing the fourth would disturb a stable inner shell. **Hence the metal commonly forms $M^{3+}$, so its halide is $MX_3$.**

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