Chemistry

Oxidation States and Oxides

308 Questions

Oxidation states represent the degree of oxidation of an atom in a chemical compound. This topic covers calculating valency, classifying oxides as acidic or basic, and understanding thermal decomposition. These concepts are frequently tested in chemistry sections across multiple competitive examination platforms.

calculating oxidation statesclassification of oxidesvalency determinationthermal decompositionstability of metal oxides

Oxidation States and Oxides Questions

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The oxides of a certain (hypothetical) element contain 27.28 %, 42.86% and 52.94% oxygen. What is the ratio of the valances of the element in the 3 oxides? 

  1. 2 : 3 : 4

  2. 1 : 3 : 4

  3. 1 : 2 : 4

  4. 1 : 2 : 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The oxygen percentages are 27.28%, 42.86%, and 52.94%. Assuming 100g of oxide, the mass of the element is 72.72g, 57.14g, and 47.06g. The ratio of oxygen to element mass (normalized) leads to valency ratios of 1:2:3.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Two different oxides of manganese are compare in the table shown below.

Color % Ms (by mass) % O (by mass)
Oxide # $1$Oxide # $2$ BlackDk Green $63.19$$77.50$ $36.81$$22.50$

When substituted X and Y in the fraction below, which pair CORRECTLY gives a result that illustrates the Law of Multiple Proportions?
$\underline {63.19}$
$X$
$\overline {77.50}$
$\overline {Y}$

  1. $X = 54.94$

    $Y = 54.94$
  2. $X = 36.81$

    $Y = 22.50$
  3. $X = 16.00$

    $Y = 16.00$
  4. $X = 22.50$

    $Y = 36.81$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To illustrate the Law of Multiple Proportions with mass percentages, one element's mass is held constant while the ratio of the masses of the other element in the two oxides is compared. Using the oxygen percentages directly as X = 36.81 and Y = 22.50 satisfies this ratio when the mass of manganese is scaled or compared.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Two different oxides of manganese are compared in the table shown below.

Color % Mn (by mass) % O (by mass)
Oxide # $1$ Black $63.19$ $36.81$
Oxide # $2$ Dk Green $77.50$ $22.50$

When substituted for X and Y in the fraction below, which pair CORRECTLY gives a result that illustrates the Law of Multiple Proportions?
$63.19$
$\overline {X}$
$\overline {77.50}$
$\overline {Y}$

  1. $X = 54.94$

    $Y = 54.94$
  2. $X = 36.81$

    $Y = 22.50$
  3. $X = 16.00$

    $Y = 16.00$
  4. $X = 22.50$

    $Y = 36.81$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

Which of the following statement regarding $D _{2}O$ is correct?

  1. It is an isotope of hydrogen

  2. It is used as a bleaching agent

  3. It is used as a moderator in atomic reactors

  4. It is used as a coolant in atomic reactors
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

$D _2O$  is used as a moderator in atomic reactors and as coolant in atomic reactors.

The chain reaction in a reactor is sustained by the free neutrons coming from one fission being absorbed by a Uranium nucleus, which then fissions itself. The probability of this absorbtion is increased if the neutrons are moving slowly. The purpose of the moderator is to slow the neutrons down. 
Lighter nuclei make better moderators than heavy nuclei. If a neutron were to hit an infinitely heavy nucleus, it would bounce off in a different direction with the same velocity. When a neutron hits a nucleus of atomic mass one head on, it comes to a complete stop. 
So, we want to use light nucleus. The reason water $(H _2O)$ is unsuitable is that it is likely to absorb the neutrons instead of slowing them down. The probability of water absorbing a neutron is more than $200$ times higher than the probability of deuterium (which already has a neutron) doing so.
Hence, options C and D are correct.

Multiple choice chemistry hydrogen and its compounds heavy water study of heavy water hard water and soft water

$D _{2}O$ is used more in:

  1. chemical industry

  2. nuclear moderator

  3. pharmaceutical preparation

  4. insecticide preparations
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heavy water is used in nuclear reactors, where it acts as a neutron moderator to slow down neutrons so that they are more likely to react with the fissile uranium-235 than with uranium-238, which captures neutrons without fissioning.

Multiple choice chemistry reactivity series and electrochemistry reactivity series and displacement reactions chemical properties of metals chemical properties of metals and non metals

Red hot carbon will remove oxygen from the oxides $XO$ and $YO$ but not from $ZO. Y$ will remove oxygen from $XO$. Use this evidence to deduce the order of activity of the three metals $X, Y$ and $Z$ putting the most active first.

  1. $XYZ$
  2. $ZYX$
  3. $YXZ$
  4. $ZXY$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since carbon can't remove oxygen from $ZO , Z $ is the most active metal.

In $X$ and $Y$, $Y$ can replace $X$ so $Y$ is more reactive and $X$ is the least reactive. 
Hence , order of activity $ZYX$.

Multiple choice chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

KMnO$ _4$ oxidises X$^{n+}$ ion to XO$^- _3$ itself changing into Mn$^{+2}$ in acid solution. $2.68$ $\times$ $10^{-3}$ mole of X$^{n+}$ requires $1.61$ $\times$ $10^{-3}$ mole of MnO$ _4^-$. The value of n is :

  1. $1$
  2. $2$
  3. $4$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the redox equivalence: (n_final - n_initial) * moles_X = (5 - 2) * moles_MnO4. (5 - n) * 2.68e-3 = 3 * 1.61e-3. 5 - n = 1.8. n = 2.

Multiple choice chemistry d- and f-block elements some important compounds of transition elements general properties of transition metals the d-and f-block elements
Mass (g)
Empty crucible $22.01$
Crucible and copper oxide $42.01$
Crucible and copper product $(1^{st}mass)$ $39.87$
Crucible and copper product $(2^{nd}mass)$ $38.01$
Crucible and copper product $(3^{rd}mass)$ $38.01$

Copper ions can have multiple oxidative states. The data table above was obtained during a lab in which an unknown copper oxide was heated in a crucible.
From this data, what is the name of the unknown copper oxide?

  1. Copper(I) oxide.

  2. Copper (II) oxide.

  3. Copper (I) peroxide.

  4. Copper (II) peroxide.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Subtract the mass of the empty crucible (22.01 g) from the final copper product mass (38.01 g) to find the mass of copper, which is 16.00 g. Subtract the copper mass from the mass of the crucible and copper oxide (42.01 g - 22.01 g = 20.01 g total oxide minus copper gives oxygen mass), yielding 4.00 g of oxygen. Calculating the mole ratio of copper to oxygen gives 16.00/63.55 = 0.25 moles of Cu and 4.00/16.00 = 0.25 moles of O, resulting in a 1:1 empirical formula of CuO, which is copper(II) oxide.

Multiple choice chemistry d- and f-block elements some important compounds of transition elements general properties of transition metals the d-and f-block elements
State True or False.
Fe$ _3$O$ _4$ is mixed oxide of FeO and Fe$ _2$O$ _3$. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Iron (II,III) oxide is the chemical compound with formula $Fe _3O _4$. It occurs in nature as the mineral magnetite. It is one of a number of iron oxides, the others being iron(II) oxide ($FeO$), which is rare, and iron(III) oxide ($Fe _2O _3$) also known as hematite. $Fe _3O _4$ contains both $Fe^{2+}$ and $Fe^{3+}$ ions and is sometimes formulated as FeO ∙ Fe2O3.