NEET UG 2017 — Chemistry Inorganic Chemistry

2017mcqeasy

It is because of the inability of ${\mathrm{ns}}^{2}$ electrons of the valence shell to participate in bonding that

Official previous-year question

Held on 30 Apr 2017 · Verified 9 Jul 2026.

Options

A
${\mathrm{Sn}}^{2+}$ is reducing while ${\mathrm{Pb}}^{4+}$ is oxidizing.
B
${\mathrm{Sn}}^{2+}$ is oxidizing while ${\mathrm{Pb}}^{4+}$ is reducing.
C
${\mathrm{Sn}}^{2+}$ and ${\mathrm{Pb}}^{2+}$ are both oxidizing and reducing.
D
${\mathrm{Sn}}^{4+}$ is reducing while ${\mathrm{Pb}}^{4+}$ is oxidizing.

Did you get this right?

Solution

The inability of ${\mathrm{ns}}^{2}$ electrons of the valence shell to participate in bonding on moving down the group in heavier $p-$ block elements is called inert pair effect.

As a result, $\mathrm{Pb}(\mathrm{II})$ is more stable than $\mathrm{Pb}(\mathrm{IV})$

$\mathrm{Sn}(\mathrm{IV})$ is more stable than $\mathrm{Sn}(\mathrm{II})$

$\therefore$ $\mathrm{Pb}(\mathrm{IV})$ is easily reduced to $\mathrm{Pb}(\mathrm{II})$ .

$\therefore$ $\mathrm{Pb}(\mathrm{IV})$ is an oxidizing agent.

$\mathrm{Sn}(\mathrm{II})$ is easily oxidized to $\mathrm{Sn}(\mathrm{IV})$

$\therefore$ $\mathrm{Sn}(\mathrm{II})$ is reducing agent.

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NEET UG Inorganic Chemistry in other years

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Work through every NEET UG Inorganic Chemistry PYQ, year by year.

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NEET UG 2017 — Chemistry Inorganic Chemistry

2017mcqeasy

It is because of the inability of ${\mathrm{ns}}^{2}$ electrons of the valence shell to participate in bonding that

Official previous-year question

Held on 30 Apr 2017 · Verified 9 Jul 2026.

Options

A
${\mathrm{Sn}}^{2+}$ is reducing while ${\mathrm{Pb}}^{4+}$ is oxidizing.
B
${\mathrm{Sn}}^{2+}$ is oxidizing while ${\mathrm{Pb}}^{4+}$ is reducing.
C
${\mathrm{Sn}}^{2+}$ and ${\mathrm{Pb}}^{2+}$ are both oxidizing and reducing.
D
${\mathrm{Sn}}^{4+}$ is reducing while ${\mathrm{Pb}}^{4+}$ is oxidizing.

Did you get this right?

Solution

The inability of ${\mathrm{ns}}^{2}$ electrons of the valence shell to participate in bonding on moving down the group in heavier $p-$ block elements is called inert pair effect.

As a result, $\mathrm{Pb}(\mathrm{II})$ is more stable than $\mathrm{Pb}(\mathrm{IV})$

$\mathrm{Sn}(\mathrm{IV})$ is more stable than $\mathrm{Sn}(\mathrm{II})$

$\therefore$ $\mathrm{Pb}(\mathrm{IV})$ is easily reduced to $\mathrm{Pb}(\mathrm{II})$ .

$\therefore$ $\mathrm{Pb}(\mathrm{IV})$ is an oxidizing agent.

$\mathrm{Sn}(\mathrm{II})$ is easily oxidized to $\mathrm{Sn}(\mathrm{IV})$

$\therefore$ $\mathrm{Sn}(\mathrm{II})$ is reducing agent.

Your note

NEET UG Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

Work through every NEET UG Inorganic Chemistry PYQ, year by year.

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NEET UG 2017 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

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NEET UG Inorganic Chemistry in other years

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Physical Chemistry

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NEET UG 2017 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

NEET UG Inorganic Chemistry in other years

More NEET UG Inorganic Chemistry questions

Other Chemistry topics for NEET UG

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry