The electronic configuration of valence shell of Cu is 3d10 4s1 and not 3d9 4s2. How is this configuration exp

NCERT Exemplar · Q31

Q. The electronic configuration of valence shell of Cu is 3d10 4s1 and not 3d9 4s2. How is this configuration explained?

Rajasthan RbseShort· 2mImportance★★★★★

83% · 116/140 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Standard Reduction Potential

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V. …

Why this formula?

Standard Reduction Potential: Why the Formula Holds

Let's build this from first principles — understanding why before memorising what.

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

A standard reduction potential ( $E^\circ$ ) measures how strongly a species wants to gain electrons (be reduced) under standard conditions (1 M concentration, 1 atm pressure, 25°C).

But why can't we measure this directly? Because every reduction must be paired with an oxidation — you can't have electrons flowing without a complete circuit.

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

Consider a Daniell cell:

Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu

Experimentally, we measure the cell potential as +1.10 V.

Now, we define the standard hydrogen electrode (SHE) as exactly 0.00 V:

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00 \text{ V}

The reasoning step-by-step:

We can only measure differences — like measuring height difference between two points.
If we connect the SHE to the copper half-cell, we measure +0.34 V (Cu²⁺ is reduced).
If we connect the SHE to the zinc half-cell, we measure −0.76 V (Zn²⁺ is reduced less readily than H⁺).

Now, the cell potential is the difference in their tendencies:

E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = (+0.34) - (-0.76) = +1.10 \text{ V}

Key insight: The formula uses subtraction because we're comparing two half-cells against the same reference (SHE). The cathode is where reduction happens (higher $E^\circ$ ), the anode is where oxidation happens (lower $E^\circ$ ).

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

This is the thermodynamic derivation — the real "why."

From Gibbs free energy:

\Delta G = \Delta G^\circ + RT \ln Q

For an electrochemical cell:

\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ

Substituting:

-nFE = -nFE^\circ + RT \ln Q

Rearranging:

E = E^\circ - \frac{RT}{nF} \ln Q

Why this makes physical sense:

$RT \ln Q$ represents the entropy penalty of non-standard concentrations
$nF$ converts charge to energy (Faraday's constant × number of electrons)
The minus sign means: as products accumulate ( $Q$ increases), the cell potential drops — the reaction is approaching equilibrium

At equilibrium ( $Q = K$ ), $E = 0$ — the battery is "dead."

--- …

Start your 14-day free trial to unlock the full solution →

Copper's valence electron configuration is $3d^{10} 4s^1$ instead of the expected $3d^9 4s^2$ because a fully filled $d$ -subshell ( $d^{10}$ ) is significantly more stable than a partially filled one, leading to a lower overall energy state for the atom.

When determining the electronic configuration of an atom, we typically follow the Aufbau principle, Hund's rule, and Pauli's exclusion principle. These rules help us predict how electrons fill atomic orbitals in increasing order of energy. For copper (Cu), with atomic number $Z=29$ , the expected configuration based on these rules would be $[Ar] 3d^9 4s^2$ . However, copper, along with some other transition metals, exhibits an anomalous configuration. This deviation is a direct consequence of the enhanced stability associated with completely filled or half-filled subshells.

The core concept here is that certain electron arrangements within subshells lead to greater stability for the atom. Specifically:

A half-filled subshell (e.g., $p^3, d^5, f^7$ )
A fully filled subshell (e.g., $p^6, d^{10}, f^{14}$ ) These configurations are exceptionally stable due to a combination of factors, primarily:

Symmetry: A half-filled or fully filled subshell has a highly symmetrical distribution of electrons, which minimizes electron-electron repulsion.
Exchange Energy: For degenerate orbitals (orbitals of the same energy, like the five $d$ -orbitals), electrons with parallel spins can exchange their positions. Each such exchange contributes to a stabilization energy called exchange energy. The more parallel spins possible (as in half-filled subshells) or the more paired electrons in a fully filled subshell, the greater the exchange energy and thus the greater the stability.

Let's apply this to copper:

Expected Configuration:

Based on the Aufbau principle, electrons fill orbitals in increasing order of energy. For copper ( $Z=29$ ), the expected electronic configuration would be:

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9$

This can be written more compactly using the noble gas core as $[Ar] 4s^2 3d^9$ .
Stability of Fully Filled Subshells:

The $3d$ subshell can hold a maximum of 10 electrons. A $3d^{10}$ configuration represents a fully filled $d$ -subshell, which is a highly stable arrangement. In contrast, $3d^9$ is one electron short of being fully filled.
Energy Trade-off: …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ30 — Wavelengths of different radiations are given below : λ(A) = 300 nm λ(…→Next questionQ32 — The Balmer series in the hydrogen spectrum corresponds to the transiti…

Q.The electronic configuration of valence shell of Cu is 3d10 4s1 and not 3d9 4s2. How is this configuration explained?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days