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Standard XII

Chemistry

Diagonal Relationship

Metal oxides ...

Question

Metal oxides are formed during the reaction between a metal cation and oxygen gas. Since transition metals can have two or more valence shell configurations, there can be more than one metal oxide formula for a given transition metal. Based on this knowledge, what is the correct empirical formula for chromium (IV) oxide?

C r_{2} O

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C r O_{2}

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C r_{2} O_{4}

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C r_{4} O_{2}

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Solution

The correct option is D $C r O_{2}$
In Chromium (IV) Oxide, the Oxidation state of Cr is 4+. Oxygen is present here $O^{2 -}$
So, Cr O
$↘$ $↙$
4 2
=CrO $_{2}$
So, the correct formula is CrO $_{2}$
Option B is the correct empirical formula of Chromium (IV) Oxide.

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Similar questions

Q. A transition metal M can exist in two oxidation state

+ 2

and

+ 3

. It forms an oxide whose experimental formula is given by

M_{x} O

where

x < 1

. Then the ratio of metal ions in

+ 3

state to those in

+ 2

state in oxide is given by:

On being heated in oxygen 5.72 g of red metallic oxide A was converted to 6.36 g Black metallic oxide B. When 4.77 g of B was heated in a stream of H₂ gas, 3.81 g of metal M was formed. ( Atomic weight of metal is 63.50 g) Find

The formula of Red metallic oxide A .....

The formula if Black metallic oxide .....

The equivalent weight of metal M in B .....

Q. A metal oxide has the formula

A_{2} O_{3}

. It can be reduced by hydrogen to give free metal and water.

0.1596 g

of this metal oxide requires

6 m g

of hydrogen for complete reduction. What is the atomic weight of metal?

Q. Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series).
(iii)

M n^{2 +}

is much more resistant than

F e^{2 +}

towards oxidation.

Q. A metal oxide contains

70 %

metal. If atomic weight of the metal is 56, empirical formula of the oxide is:

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The correct option is D CrO2In Chromium (IV) Oxide, the Oxidation state of Cr is 4+. Oxygen is present here O2−So, Cr O ↘ ↙ 4 2 =CrO2So, the correct formula is CrO2Option B is the correct empirical formula of Chromium (IV) Oxide.

The correct option is D $C r O_{2}$
In Chromium (IV) Oxide, the Oxidation state of Cr is 4+. Oxygen is present here $O^{2 -}$
So, Cr O
$↘$ $↙$
4 2
=CrO $_{2}$
So, the correct formula is CrO $_{2}$
Option B is the correct empirical formula of Chromium (IV) Oxide.