IMO Practice Test: Structure of Atom | Class 11 Chemistry

Question 1 of 14

If the radius of the first Bohr orbit of hydrogen is $r_1$, the radius of the orbit in which the electron has energy $-1.51\,\text{eV}$ is:

$4r_1$

$9r_1$

$3r_1$

$16r_1$

Explanation: $E=-13.6/n^2=-1.51\Rightarrow n^2=9\Rightarrow n=3$, so $r_3=9r_1$.

Question 2 of 14

The shortest-wavelength line of the Lyman series corresponds to a transition from:

$n=2\to1$

$n=\infty\to1$

$n=3\to1$

$n=\infty\to2$

Explanation: Shortest $\lambda$ (series limit) comes from the largest energy gap, $n=\infty\to1$.

Question 3 of 14

For a one-electron species, the energy is $E_n=-13.6\,\frac{Z^2}{n^2}\,\text{eV}$. The ground-state energy of $\text{He}^+$ ($Z=2$) is:

$-13.6\,\text{eV}$

$-27.2\,\text{eV}$

$-54.4\,\text{eV}$

$-6.8\,\text{eV}$

Explanation: $E_1=-13.6\times2^2/1=-54.4\,\text{eV}$.

Question 4 of 14

The ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference (so equal kinetic energy times charge) — for equal momentum the lighter particle has the longer $\lambda$. For equal speed, $\lambda_{\text{proton}}:\lambda_{\alpha}$ equals:

$1:1$

$4:1$

$1:4$

$2:1$

Explanation: At equal speed $\lambda\propto1/m$; $m_\alpha\approx4m_p$, so $\lambda_p:\lambda_\alpha=4:1$.

Question 5 of 14

An electron and a proton have the same de Broglie wavelength. Which has the greater speed?

the proton

the electron

both equal

cannot be decided

Explanation: Equal $\lambda$ means equal momentum $mv$; the lighter electron must move faster.

Question 6 of 14

The minimum uncertainty in the velocity of an electron confined to $\Delta x=1\,\text{angstrom}$ is of the order:

$10^{5}\,\text{m s}^{-1}$

$10^{6}\,\text{m s}^{-1}$

$10^{3}\,\text{m s}^{-1}$

$10^{9}\,\text{m s}^{-1}$

Explanation: $\Delta v\ge\frac{h}{4\pi m\Delta x}\approx\frac{5.3\times10^{-35}}{9.1\times10^{-31}\times10^{-10}}\approx5.8\times10^{5}\,\text{m s}^{-1}$.

Question 7 of 14

The maximum number of electrons in an atom with $n=4$ and $m_s=+\tfrac{1}{2}$ is:

8

16

32

10

Explanation: $n=4$ holds $2n^2=32$ electrons; half have $m_s=+\tfrac{1}{2}$, so $16$.

Question 8 of 14

The number of electrons in an atom that can have the quantum numbers $n=3$, $l=2$ is:

2

6

10

14

Explanation: $n=3,l=2$ is the $3d$ sub-shell: $2(2l+1)=2\times5=10$ electrons.

Question 9 of 14

Which set of quantum numbers is NOT allowed?

$n=3,l=2,m_l=-2$

$n=2,l=2,m_l=0$

$n=4,l=0,m_l=0$

$n=3,l=1,m_l=+1$

Explanation: $l$ cannot equal $n$; for $n=2$, $l$ can only be $0$ or $1$, so $n=2,l=2$ is forbidden.

Question 10 of 14

The element with electronic configuration $[\text{Ar}]\,3d^{10}\,4s^1$ is:

potassium

chromium

copper

zinc

Explanation: Copper ($Z=29$) gains stability from the fully-filled $3d^{10}$, giving $[\text{Ar}]\,3d^{10}\,4s^1$.

Question 11 of 14

How many spectral lines appear when electrons return from the $n=4$ level to the ground state in a sample of hydrogen atoms?

3

4

6

10

Explanation: Number of lines $=\frac{n(n-1)}{2}=\frac{4\times3}{2}=6$.

Question 12 of 14

A $4d$ orbital has how many radial nodes?

0

1

2

3

Explanation: Radial nodes $=n-l-1=4-2-1=1$.

Question 13 of 14

Two electrons occupying the same orbital differ only in their value of:

$n$

$l$

$m_l$

$m_s$

Explanation: Same orbital means same $n,l,m_l$; the Pauli principle forces different $m_s$.

Question 14 of 14

The wavenumber of the series limit of the Balmer series ($n_2=\infty\to n_1=2$) equals:

$R_H$

$R_H/4$

$R_H/2$

$4R_H$

Explanation: $\bar{\nu}=R_H(1/4-0)=R_H/4$.

IMO Practice Test — Structure of Atom