Online Test: Structure of Atom | Class 11 Chemistry

Question 1 of 20

The charge on a single electron is approximately:

$1.602\times10^{-19}\,\text{C}$

$9.11\times10^{-31}\,\text{C}$

$1.602\times10^{-31}\,\text{C}$

$6.626\times10^{-34}\,\text{C}$

Explanation: Millikan's oil-drop experiment fixed the electron charge at $1.602\times10^{-19}\,\text{C}$.

Question 2 of 20

The neutron was discovered by:

Thomson

Rutherford

Chadwick

Bohr

Explanation: James Chadwick discovered the neutron in 1932 by bombarding beryllium with $\alpha$-particles.

Question 3 of 20

The number of neutrons in $^{23}_{11}\text{Na}$ is:

11

12

23

34

Explanation: Neutrons $=A-Z=23-11=12$.

Question 4 of 20

Atoms with the same mass number but different atomic numbers are called:

isotopes

isobars

isotones

isomers

Explanation: Same $A$, different $Z$ defines isobars (e.g. $^{40}\text{Ar}$ and $^{40}\text{Ca}$).

Question 5 of 20

Rutherford's $\alpha$-scattering experiment established the existence of the:

electron

nucleus

neutron

orbital

Explanation: Large-angle scattering revealed a tiny, dense, positively charged nucleus.

Question 6 of 20

The energy of the electron in the $n=1$ orbit of hydrogen is:

$-3.40\,\text{eV}$

$-13.6\,\text{eV}$

$-1.51\,\text{eV}$

$0\,\text{eV}$

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

Question 7 of 20

The radius of the $n$th Bohr orbit varies as:

$n$

$n^2$

$1/n$

$1/n^2$

Explanation: $r_n=0.529\,n^2\,\text{angstrom}$, so $r_n\propto n^2$.

Question 8 of 20

The visible lines of the hydrogen spectrum belong to the:

Lyman series

Balmer series

Paschen series

Brackett series

Explanation: The Balmer series ($n_1=2$) lies in the visible region.

Question 9 of 20

In the Rydberg equation $\bar{\nu}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, the constant $R_H$ is about:

$1.097\times10^{7}\,\text{m}^{-1}$

$6.626\times10^{-34}\,\text{m}^{-1}$

$1.602\times10^{-19}\,\text{m}^{-1}$

$3.0\times10^{8}\,\text{m}^{-1}$

Explanation: The Rydberg constant for hydrogen is $1.097\times10^{7}\,\text{m}^{-1}$.

Question 10 of 20

The de Broglie wavelength is given by:

$\lambda=mvh$

$\lambda=h/mv$

$\lambda=mv/h$

$\lambda=h/mv^2$

Explanation: $\lambda=h/mv=h/p$.

Question 11 of 20

Heisenberg's uncertainty principle is expressed as:

$\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$

$\Delta x\cdot\Delta p\le\frac{h}{4\pi}$

$\Delta x\cdot\Delta p=h$

$\Delta x+\Delta p\ge\frac{h}{4\pi}$

Explanation: $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$ — a lower bound on the product.

Question 12 of 20

The quantum number that determines the orientation of an orbital in space is:

$n$

$l$

$m_l$

$m_s$

Explanation: $m_l$ (values $-l$ to $+l$) fixes orbital orientation; there are $2l+1$ orbitals per sub-shell.

Question 13 of 20

For $n=3$, the number of orbitals is:

3

6

9

18

Explanation: Number of orbitals $=n^2=9$ ($3s+3p+3d=1+3+5$).

Question 14 of 20

The total number of nodes in a $3s$ orbital is:

0

1

2

3

Explanation: Total nodes $=n-1=3-1=2$ (both radial; angular $=l=0$).

Question 15 of 20

The shape of an $s$ orbital is:

dumb-bell

spherical

double dumb-bell

doughnut

Explanation: An $s$ orbital ($l=0$) is spherically symmetric.

Question 16 of 20

The maximum number of electrons in a shell with principal quantum number $n$ is:

$2n$

$2n^2$

$n^2$

$2l+1$

Explanation: Shell capacity is $2n^2$.

Question 17 of 20

According to the $(n+l)$ rule, the orbital filled before $3d$ is:

$3p$

$4s$

$4p$

$4d$

Explanation: $4s$ ($n+l=4$) is lower than $3d$ ($n+l=5$), so $4s$ fills first.

Question 18 of 20

The ground-state configuration of chromium ($Z=24$) is:

$[\text{Ar}]\,3d^4\,4s^2$

$[\text{Ar}]\,3d^5\,4s^1$

$[\text{Ar}]\,3d^6$

$[\text{Ar}]\,3d^3\,4s^2\,4p^1$

Explanation: A $4s$ electron shifts to $3d$ for the extra-stable half-filled $3d^5\,4s^1$.

Question 19 of 20

The number of unpaired electrons in $\text{Fe}^{2+}$ ($[\text{Ar}]\,3d^6$) is:

2

3

4

6

Explanation: In $3d^6$, five orbitals are filled with one pair and four singles, giving 4 unpaired electrons.

Question 20 of 20

Which configuration violates the Pauli exclusion principle?

two electrons in one orbital with opposite spins

two electrons in one orbital with the same spin

one electron in each of three $p$ orbitals

a full $2p^6$ sub-shell

Explanation: Two electrons in the same orbital must differ in $m_s$; identical spins would make all four quantum numbers equal.

Online Test — Structure of Atom