The Mistake Bank
Chapter 2: Structure of Atom
Quantum mechanics is weird. Your mistakes shouldn't be.
The "Iron Ion" Trap
Electronic ConfigurationScenario: Write the electronic configuration for \( Fe^{2+} \) (Atomic No. 26).
Student writes Fe config: \( [Ar] 3d^6 4s^2 \).
Removes last added electrons (from 3d):
$$ [Ar] 3d^4 4s^2 $$
(Wrong! The 4s electrons leave first.)
4s Fills First, but Empties First!
Electrons are removed from the outermost shell (highest n value).
Fe: \( [Ar] 3d^6 \mathbf{4s^2} \)
Remove 2 electrons from 4s:
$$ Fe^{2+}: [Ar] 3d^6 $$
Orbital Angular Momentum
Quantum NumbersScenario: Calculate the Orbital Angular Momentum of an electron in a 4s orbital.
Student uses Bohr's formula:
$$ mvr = \frac{nh}{2\pi} $$
For n=4: \( \frac{4h}{2\pi} = \frac{2h}{\pi} \)
(This is for Orbit, not Orbital!)
Use the Quantum Mechanics formula:
$$ L = \sqrt{l(l+1)} \frac{h}{2\pi} $$
For s-orbital, \( l = 0 \).
$$ L = \sqrt{0(0+1)} \frac{h}{2\pi} = \mathbf{0} $$
Counting Nodes
Shapes of OrbitalsScenario: Calculate the number of Radial Nodes in a 3p orbital.
Student calculates Total Nodes instead:
$$ \text{Nodes} = n - 1 $$
\( 3 - 1 = 2 \)
(This is the sum of radial + angular nodes.)
Radial Nodes Formula:
$$ n - l - 1 $$
For 3p: \( n=3, l=1 \).
$$ 3 - 1 - 1 = \mathbf{1} $$
(Note: Angular nodes = \( l = 1 \))
Bohr's Energy Formula
Bohr ModelScenario: Calculate the energy of the 2nd orbit of \( Li^{2+} \).
Student forgets the Atomic Number (Z):
$$ E_n = -13.6 \frac{1}{n^2} \text{ eV} $$
\( -13.6 / 4 = -3.4 \text{ eV} \)
(This is only true for Hydrogen!)
Include \( Z^2 \) in the numerator!
$$ E_n = -13.6 \frac{Z^2}{n^2} \text{ eV} $$
For \( Li^{2+} \), \( Z=3 \).
$$ E = -13.6 \times \frac{3^2}{2^2} = -13.6 \times \frac{9}{4} = \mathbf{-30.6 \text{ eV}} $$
De Broglie Units
Dual NatureScenario: Find the wavelength of a 10g ball moving at 100 m/s.
Student uses mass in grams directly:
$$ \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{10 \times 100} $$
(Wrong! Planck's constant is in Joules·s, which implies kg.)
Convert Mass to Kilograms!
\( m = 10g = 0.01 \text{ kg} \)
$$ \lambda = \frac{6.626 \times 10^{-34}}{0.01 \times 100} $$
$$ \lambda = 6.626 \times 10^{-34} \text{ meters} $$
The "Impossible" Quantum Number
Quantum NumbersScenario: Which set of quantum numbers is NOT possible?
\( n=3, l=3, m=0, s=+1/2 \)
Student checks \( m \) (0 is inside -3 to +3) and \( s \) and thinks it looks fine.
Often overlooks the relationship between \( n \) and \( l \).
l must always be less than n!
Range of \( l \): \( 0 \) to \( n-1 \).
If \( n=3 \), max \( l \) is 2 (3d orbital).
Therefore, \( l=3 \) is Impossible.
Chromium Configuration
ExceptionsScenario: Write the electronic configuration of Chromium (Cr, Z=24).
Student follows standard Aufbau principle:
$$ [Ar] 4s^2 3d^4 $$
(Technically correct by energy rules, but experimentally wrong due to stability.)
Half-Filled Stability Rule!
One electron jumps from 4s to 3d to make the d-subshell half-filled (stable).
$$ [Ar] 4s^1 3d^5 $$
(Same logic applies to Copper: \( 4s^1 3d^{10} \))
Confess Your Sins!
"The atom is mostly empty space, but your brain shouldn't be."
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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