**iii.** How is dioxygen prepared in laboratory from KClO₃ ?

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✓ Free question

Step 1. Section 7.9.1's laboratory preparation of dioxygen states that oxygen-containing salts such as chlorates can be heated to release O2.

Step 2. Specifically, potassium chlorate (KClO3) is heated in the presence of manganese dioxide (MnO2) as a catalyst, decomposing to potassium chloride and oxygen gas: 2KClO3(s), with MnO2 and heat, gives 2KCl(s) + 3O2(g).

Step 3. MnO2 here is a true catalyst -- it speeds up the decomposition and lowers the temperature needed, but is not itself consumed in the reaction.

✓Final answer

Dioxygen is prepared from KClO3 in the laboratory by heating it with MnO2 as catalyst: 2KClO3(s) -> 2KCl(s) + 3O2(g).

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$Q. Which of the following cannot act as a reducing agent ?$
(a) NO2
(b) SO2
(c) ClO2
(d) CO2
›Reveal solutionSolution

A reducing agent must be oxidisable; carbon in CO2 is at its maximum +4 state, so CO2 cannot act as a reducing agent.

A species acts as a reducing agent only if the central atom can be oxidised to a higher oxidation state. In CO2 carbon is already at its maximum oxidation state, +4 (group 14), so it cannot lose electrons further and cannot reduce anything. In NO2 (N is +4, can go to +5), SO2 (S is +4, can go to +6) and ClO2 (Cl is +4, can go higher), the central atoms have higher states available, so those oxides can act as reducing agents.

✓Final answer
$(d) CO2 — cannot act as a reducing agent (carbon already at +4).$
CBSE 2025Set D1 markMCQ
$Q. Which of the following oxides is paramagnetic?$
(a) CO2
(b) ClO2
(c) SO2
(d) SiO2
›Reveal solutionSolution
ClO2 is an odd-electron (free-radical) molecule and is therefore paramagnetic.

A species is paramagnetic if it has one or more unpaired electrons. Counting valence electrons:

ClO2: Cl(7) + 2xO(6) = 19 valence electrons — odd number, so at least one electron is unpaired. ClO2 is a stable odd-electron radical and is paramagnetic.

CO2, SO2, SiO2: all have an even electron count with fully paired electrons, so they are diamagnetic.

✓Final answer
$(B) ClO2.$
CBSE 2025Set D1 markMCQ
$Q. The number of unpaired electrons in one molecule of liquid oxygen is$
(a) 0
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

O2 has 2 unpaired electrons (in the pi* orbitals), making it paramagnetic.

By molecular orbital theory, the 12 valence electrons of O2 fill the orbitals up to the two pi* (antibonding) orbitals, each of which gets one electron with parallel spins (Hund's rule):

... (pi 2p)4 (pi* 2p)2

The two pi* electrons remain unpaired. This gives O2 two unpaired electrons and explains why oxygen (including liquid oxygen) is paramagnetic and is attracted by a magnetic field.

✓Final answer
$(B) 2.$
CBSE 2021Set A1 markMCQ
$Q. The substance which exhibits paramagnetism is$
(a) H2O
(b) O2
(c) NaCl
(d) C6H6
›Reveal solutionSolution
A species is paramagnetic if it has unpaired electrons; among the options only O2 does.

Paramagnetism arises from the presence of unpaired electrons in a species. Let us examine each option:

H2O: all electrons are paired (covalent O–H bonds and lone pairs) → diamagnetic.

O2: molecular orbital theory gives the configuration with two unpaired electrons in the degenerate pi*(2p) antibonding orbitals. These two unpaired electrons make O2 paramagnetic — a classic result that explains why liquid oxygen is attracted to a magnet.

NaCl: an ionic solid of Na+ and Cl-, both closed-shell ions → diamagnetic.

C6H6 (benzene): all electrons paired in sigma and delocalised pi bonds → diamagnetic.

Hence only O2 has unpaired electrons and is paramagnetic.

✓Final answer
$(b) O2 — it has two unpaired electrons in its pi* molecular orbitals, making it paramagnetic.$

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Q.iii. How is dioxygen prepared in laboratory from KClO $_3$ ?

Concept understanding — Dioxygen and Simple Oxides

Q.iii. How is dioxygen prepared in laboratory from KClO3_33​ ?

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Q.iii. How is dioxygen prepared in laboratory from KClO $_3$ ?

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