When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$, 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to (i) 1:3 electrolyte (ii) 1:2 electrolyte (iii) 1:1 electrolyte (iv) 3:1 electrolyte

When 0.1 mol CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of so

NCERT Exemplar · Q3

CoCl_3(NH_3)_5

(i) 1:3 electrolyte

(ii) 1:2 electrolyte

(iii) 1:1 electrolyte

(iv) 3:1 electrolyte

Uttarakhand UbseMCQ· 1mImportance★★★★★

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Concept understanding — Conductance And Conductivity

From Resistance to Conductance: Flipping the Idea

You already know resistance ( $R$ ) — it tells you how much a material opposes the flow of current. A high resistance means the wire fights the current; a low resistance means it lets current through easily.

Now flip that thought. Instead of asking "how much does it resist?", ask "how easily does it let current flow?" That's exactly what conductance measures.

Note

Conductance ( $G$ ) is the reciprocal of resistance:

$G = \frac{1}{R}$

Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).

If a wire has $R = 10\ \Omega$ , its conductance is $G = 0.1\ \text{S}$ . If $R = 0.5\ \Omega$ , $G = 2\ \text{S}$ — it conducts twice as well.

Ohm's Law in Conductance Form

You know $V = IR$ . Rearranging:

I = \frac{V}{R} = G V

So current = conductance × voltage. A high-conductance material draws a large current for the same voltage — it's a "good conductor."

Now, Conductivity: The Material's Intrinsic Property

Resistance depends on two things: the material itself (its "resistivity" $\rho$ ) and the geometry (length $L$ , cross-sectional area $A$ ):

R = \rho \frac{L}{A}

Conductance also depends on geometry. A thicker wire (larger $A$ ) or a shorter wire (smaller $L$ ) has higher conductance. To isolate the material's inherent ability to conduct, we define conductivity ( $\sigma$ ):

\sigma = \frac{1}{\rho}

Conductivity is the reciprocal of resistivity. It tells you how well the material itself conducts, independent of shape and size.

Unit: siemens per metre (S/m).
High $\sigma$ → good conductor (copper: $\approx 5.8 \times 10^7\ \text{S/m}$ ).
Low $\sigma$ → poor conductor / insulator (glass: $\approx 10^{-12}\ \text{S/m}$ ).

Watch out

Don't confuse conductance (property of a specific object, depends on geometry) with conductivity (property of the material, independent of geometry). A short thick copper wire has high conductance; a long thin copper wire has lower conductance — but both have the same conductivity.

The Big Picture in One Table

Quantity	Symbol	Definition	Depends on	Unit
Resistance	$R$	$V/I$	Material + geometry	$\Omega$
Resistivity	$\rho$	$R A / L$	Material only	$\Omega \cdot \text{m}$
Conductance	$G$	$1/R$	Material + geometry	S
Conductivity	$\sigma$	$1/\rho$	Material only	S/m

Intuitive Analogy

Think of a water pipe:

Resistance = how hard it is to push water through (narrow, long pipe).
Conductance = how easily water flows (wide, short pipe).
Resistivity = the "roughness" of the pipe's inner surface (material property).
Conductivity = the "smoothness" of the pipe's inner surface (material property).

A copper pipe is smooth (high conductivity). A rubber hose is rough (low conductivity). But a short, fat rubber hose might still have decent conductance — because geometry can compensate for poor material.

Key Takeaway for Exams

Important

$G = \frac{1}{R}$ and $\sigma = \frac{1}{\rho}$ .
$G = \sigma \frac{A}{L}$ for a uniform conductor.
Conductivity is an intrinsic material property; conductance is an extrinsic property of a specific object.
In circuits, you'll often use conductance when dealing with parallel resistors (total conductance = sum of individual conductances).

You now have the complete picture: from resistance to conductance, from resistivity to conductivity — and the clean relationship between them.

Conductance and conductivity are core quantitative ideas in the NCERT/CBSE Class 12 Chemistry Electrochemistry chapter, and ‘conductance vs conductivity formula’ is a regularly asked important question in board exams as well as JEE Main and NEET chemistry sections. This relationship also feeds directly into later topics like molar conductivity and Kohlrausch's law.

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

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

1. The Core Idea: How Easily Does Current Flow?

Think of a conductor (like a copper wire). When you apply a voltage across it, electrons drift through the material. Two questions arise:

How much current flows for a given voltage? → This is conductance ( $G$ ).
How well does the material itself allow current? → This is conductivity ( $\sigma$ ).

The key distinction: Conductance depends on the size and shape of the object. Conductivity is an intrinsic property of the material.

2. Ohm's Law in Terms of Conductance

You know Ohm's law:

V = IR

But we can rewrite it as:

I = \frac{V}{R}

Define conductance $G$ as the reciprocal of resistance:

G = \frac{1}{R}

So:

I = G V

Why this makes sense:

A larger $G$ means more current for the same voltage — the conductor "conducts" better.
$G$ has units of siemens (S) = $\text{A/V}$ .

3. From Resistance to Conductivity: The Geometry Factor

Resistance of a uniform conductor depends on:

Length $L$ (longer → more resistance)
Cross-sectional area $A$ (thicker → less resistance)
Material property $\rho$ (resistivity)

The formula:

R = \rho \frac{L}{A}

Now, conductivity $\sigma$ is the reciprocal of resistivity:

\sigma = \frac{1}{\rho}

So:

R = \frac{1}{\sigma} \cdot \frac{L}{A}

Why this form?

If you double the length, electrons have to travel twice as far, colliding more → resistance doubles.
If you double the area, there's twice as many "lanes" for electrons → resistance halves.

4. The Key Formula: Conductance in Terms of Conductivity

Since $G = 1/R$ , we get:

G = \sigma \frac{A}{L}

This is the central relationship. Let's see why it holds:

$\sigma$ tells you how well the material conducts (intrinsic).
$A/L$ tells you how the geometry amplifies or reduces that.

Intuition:

A fat, short wire ( $A$ large, $L$ small) has high conductance.
A thin, long wire ( $A$ small, $L$ large) has low conductance.
A material with high $\sigma$ (like copper) gives higher $G$ than one with low $\sigma$ (like iron), for the same shape.

5. Microscopic Derivation (Why $\sigma$ Exists)

At the microscopic level, conductivity arises from electron motion:

\sigma = n e \mu

Where:

$n$ = number of free electrons per unit volume
$e$ = electron charge
$\mu$ = electron mobility (how fast they drift per unit electric field)

Why this works:

More free electrons ( $n$ large) → more charge carriers → higher conductivity.
Higher mobility ( $\mu$ large) → electrons move faster for the same push → higher conductivity.

This explains why metals (high $n$ ) are good conductors, and why heating reduces $\sigma$ (more collisions → lower $\mu$ ).

6. Summary: The Logical Chain

Step	Concept	Formula	Why
1	Ohm's law	$V = IR$	Voltage drives current against resistance
2	Conductance	$G = 1/R$	Measures ease of flow
3	Resistivity	$R = \rho L/A$	Geometry + material
4	Conductivity	$\sigma = 1/\rho$	Material's intrinsic ability
5	Key result	$\boxed{G = \sigma \frac{A}{L}}$	Combines material + geometry

Final takeaway:

Conductance $G$ is not just a number — it's the product of how good the material is ( $\sigma$ ) and how the shape helps ( $A/L$ ). This is why a thick copper wire conducts far better than a thin iron wire of the same length.

The complex is $[Co(NH_3)_5Cl]Cl_2$ , which dissociates into 3 ions (1 Co complex cation + 2 Cl⁻ anions), so it behaves as a 1:2 electrolyte. The correct option is (ii).

Why Conductivity and Precipitation Tell the Same Story

When you dissolve a coordination compound in water, its conductivity depends on how many ions it releases. A 1:1 electrolyte (like NaCl) gives 2 ions, a 1:2 electrolyte (like CaCl₂) gives 3 ions, and so on. The precipitation experiment with AgNO₃ is a clever way to count how many chloride ions are free — because only free Cl⁻ ions react with Ag⁺ to form AgCl. Chloride ions that are tightly bound inside the coordination sphere (as ligands) do not precipitate.

So the two experiments — conductivity and precipitation — are two sides of the same coin. If you know how many Cl⁻ ions are free, you know the ionic formula of the complex, and from that you can predict how many total ions it will produce in solution.

Step-by-Step Reasoning

Interpret the precipitation data.

The reaction is:

$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow$

From 0.1 mol of the complex, we get 0.2 mol of AgCl. That means 0.2 mol of free Cl⁻ ions were present in the solution.

So per mole of complex, the number of free Cl⁻ ions is:

$\frac{0.2}{0.1} = 2$

Therefore, the complex must have two chloride ions outside the coordination sphere (as counter-ions) and the rest inside as ligands.
Deduce the formula of the complex.

The molecular formula is $CoCl_3(NH_3)_5$ . We have 3 Cl atoms total. Two are free (ionic), so the remaining one must be coordinated to cobalt inside the coordination sphere.

Hence the correct structural formula is:

$[Co(NH_3)_5Cl]Cl_2$

The square brackets enclose the coordination sphere: one Co³⁺ ion, five NH₃ molecules, and one Cl⁻ ligand. Outside the sphere, two Cl⁻ ions balance the +3 charge on the cobalt (since Co³⁺ + 5 neutral NH₃ + 1 Cl⁻ ligand gives a net +2 charge on the complex ion, which is neutralized by the two free Cl⁻ ions).
Determine the number of ions in solution.

When $[Co(NH_3)_5Cl]Cl_2$ dissolves, it dissociates completely into:

$[Co(NH_3)_5Cl]^{2+} + 2Cl^-$

That is 3 ions per formula unit: one complex cation with charge +2, and two chloride anions.
Classify the electrolyte type.

An electrolyte that gives one cation of charge +2 and two anions of charge -1 is called a 1:2 electrolyte (the ratio of cation charge to anion charge is 1:2, but more commonly it refers to the number of ions: 1 cation : 2 anions).

Option (ii) matches this.

Watch out

A common mistake is to think that all 3 chlorides are free because the formula shows 3 Cl atoms. Remember: only chlorides outside the coordination sphere precipitate with Ag⁺. The coordinated Cl⁻ is "invisible" to AgNO₃.

Tip

You can also work backwards from the conductivity: a 1:2 electrolyte gives a molar conductivity roughly double that of a 1:1 electrolyte (for similar ions), but here the precipitation data alone is sufficient — no need for actual conductivity values.

✓Final answer

The correct option is (ii) 1:2 electrolyte .

Common Mistakes & How to Avoid Them

Mistake 1: Counting All Chlorines as Precipitable

The error: Students see the formula $CoCl_3(NH_3)_5$ and assume all 3 chlorine atoms will precipitate with $AgNO_3$ , expecting 0.3 mol of $AgCl$ .

Why it's wrong: Only chloride ions outside the coordination sphere (free ions) react with $AgNO_3$ . Chlorine atoms inside the coordination sphere (bonded to the metal) do not precipitate.

How to avoid: Always distinguish between:

Ionizable chlorines (outside coordination sphere) → precipitate with $Ag^+$
Coordinated chlorines (inside coordination sphere) → do not precipitate

Mistake 2: Misinterpreting the Given Data

The error: Not connecting "0.2 mol $AgCl$ obtained" to the number of free chloride ions.

Why it's wrong: The reaction is:

CoCl_3(NH_3)_5 + AgNO_3 \rightarrow AgCl \downarrow + \text{other products}

0.2 mol $AgCl$ means only 2 out of 3 chlorines are free $Cl^-$ ions.

How to avoid: Write the precipitation equation:

Cl^- + Ag^+ \rightarrow AgCl

So moles of $AgCl$ = moles of free $Cl^-$ = 2 (from 0.2 mol obtained from 0.1 mol complex).

Mistake 3: Confusing Conductivity with Number of Ions

The error: Thinking conductivity depends only on the number of ions, ignoring charge.

Why it's wrong: Conductivity depends on both number and charge of ions. A 1:2 electrolyte (like $CaCl_2$ ) gives 3 ions total, but the conductivity pattern differs from a 1:3 electrolyte.

How to avoid: Remember:

1:1 electrolyte → 2 ions (e.g., $NaCl$ )
1:2 electrolyte → 3 ions (e.g., $MgCl_2$ )
1:3 electrolyte → 4 ions (e.g., $AlCl_3$ )

Mistake 4: Forgetting the Complex Cation's Charge

The error: Determining the complex formula but not calculating the charge on the complex ion.

Why it's wrong: The complex is $[Co(NH_3)_5Cl]Cl_2$ (since 2 $Cl^-$ are free). You must find the charge on $[Co(NH_3)_5Cl]^{n+}$ .

How to avoid: Use charge balance:

$Co$ is in +3 oxidation state (common for cobalt complexes)
$NH_3$ is neutral
$Cl$ inside sphere is $-1$
So charge on complex = $+3 + (-1) = +2$

Thus the complex is $[Co(NH_3)_5Cl]^{2+}$ with 2 $Cl^-$ ions → 1:2 electrolyte.

Mistake 5: Picking 1:3 Electrolyte Out of Habit

The error: Seeing $CoCl_3$ and automatically choosing 1:3 without checking coordination.

Why it's wrong: Coordination chemistry changes the number of free ions. The formula $CoCl_3(NH_3)_5$ is not the same as $CoCl_3$ in solution.

How to avoid: Always check:

How many $Cl^-$ precipitate with $AgNO_3$ ?
That number = number of free $Cl^-$ ions
The rest are inside the coordination sphere

Quick Summary Checklist

Step	Action
1	Moles of $AgCl$ = moles of free $Cl^-$
2	Subtract free $Cl^-$ from total $Cl$ to get coordinated $Cl$
3	Write correct complex formula: $[Co(NH_3)_5Cl]Cl_2$
4	Find charge on complex ion using oxidation states
5	Count total ions and their charges → determine electrolyte type

Final answer: The complex gives $[Co(NH_3)_5Cl]^{2+}$ and $2Cl^-$ → 1:2 electrolyte → Option (ii).

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Q.When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$ , 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to

Concept understanding — Conductance And Conductivity

From Resistance to Conductance: Flipping the Idea

Ohm's Law in Conductance Form

Now, Conductivity: The Material's Intrinsic Property

The Big Picture in One Table

Intuitive Analogy

Key Takeaway for Exams

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

1. The Core Idea: How Easily Does Current Flow?

2. Ohm's Law in Terms of Conductance

3. From Resistance to Conductivity: The Geometry Factor

4. The Key Formula: Conductance in Terms of Conductivity

5. Microscopic Derivation (Why $\sigma$ Exists)

6. Summary: The Logical Chain

Why Conductivity and Precipitation Tell the Same Story

Step-by-Step Reasoning

Method: Conductance & Coordination Compound Analysis

Step-by-step reasoning

Final Answer

Common Mistakes & How to Avoid Them

Mistake 1: Counting All Chlorines as Precipitable

Mistake 2: Misinterpreting the Given Data

Mistake 3: Confusing Conductivity with Number of Ions

Mistake 4: Forgetting the Complex Cation's Charge

Mistake 5: Picking 1:3 Electrolyte Out of Habit

Quick Summary Checklist

Q.When 0.1 mol CoCl3(NH3)5CoCl_3(NH_3)_5CoCl3​(NH3​)5​ is treated with excess of AgNO3AgNO_3AgNO3​, 0.2 mol of AgClAgClAgCl are obtained. The conductivity of solution will correspond to

Why Conductivity and Precipitation Tell the Same Story

Step-by-Step Reasoning

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Q.When 0.1 mol $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$ , 0.2 mol of $AgCl$ are obtained. The conductivity of solution will correspond to

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