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 chemical bonding and structure formal charge basics of chemical bonding types of bonds

Effective charge on each oxygen is $-\frac{1}{4}$ in?

  1. $SO _{3}^{2-}$
  2. $SO _{3-}^{4}$
  3. $ClO _{3}^{-}$
  4. $ClO _{4}^{-}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the perchlorate ion (ClO4-), there are four equivalent oxygen atoms sharing a single negative charge delocalized over the entire ion, meaning the effective charge on each oxygen atom is -1/4. In sulfite (SO3 2-), the charge per oxygen is -2/3, and in chlorate (ClO3-), it is -1/3.

Multiple choice chemistry chemical bonding and structure formal charge basics of chemical bonding types of bonds

Among ${ KO } _{ 2 },{ AIO } _{ 2 }^{ - },{ BaO } _{ 2 }\ and\ { NO } _{ 2 }^{ + }$, unpaired electron is present in :

  1. ${ NO } _{ 2 }^{ + }\quad and\quad { BaO } _{ 2 }$
  2. ${ KO } _{ 2 }\quad and\quad AI{ O } _{ 2 }^{ - }$
  3. ${ KO } _{ 2 }$ only
  4. $Ba{ O } _{ 2 }$ only
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This can be done by counting number of valence electrons


$KO _2=1+2\times 6=17$-Odd number of electrons and hence unpaired.


$AlO^{2-} _2=3+2\times 6+1=14$-Even number of electrons and hence paired.

$BaO _2=2+2\times6=14$-Even number of electrons and hence paired.

$NO^{+} _{2}=5+2\times6 -1=16$-Even number of electrons and hence paired.

Hence option C is the correct answer.

Multiple choice chemistry changes around us can all changes be reversed? substances and objects classification of changes

 Iron adopts changes slowly and gradually by process of :

  1. rusting

  2. moldings

  3. fermentation

  4. explosions
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Iron adopt changes slowly and gradually when it comes in contact with air (moisture, oxygen etc)
When Iron reacts with $'n'$ molecules of water in the pressure of oxygen, it forms rust. Hence the process is known as a rusting.
$4Fe+3{ O } _{ 2 }+2n{ H } _{ 2 }O\longrightarrow 2{ Fe } _{ 2 }{ O } _{ 3 }.n{ H } _{ 2 }O$
Multiple choice chemistry occurrence of carbon compounds in nature importance of carbon covalent bonding in carbon compounds carbon and its forms

Carbon has zero oxidation number in

  1. $CH _{4}$
  2. $\mathrm { CH } _ { 3 } \mathrm { Cl}$
  3. $C C l _ { 4 }$
  4. $\mathrm { CH } _ { 2 } \mathrm { Cl } _ { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In CH2Cl2, the oxidation state of hydrogen is +1 and chlorine is -1. Calculating for carbon: x + 2(+1) + 2(-1) = 0, which gives x = 0.

Multiple choice chemistry atoms, molecules and radicals molecular formula and valency valency and chemical formulae combining capacity and chemical formulae

Two oxides of a metal contain 30.0% and 27.6 % of oxygen respectively. If the formula of the first oxide is $M _3O _4$, the formula of the second oxide, is

  1. $M _2O _3$
  2. $M _4O _3$
  3. $M _3O _4$
  4. $M _3O _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Formula of first oxide = $M _3O _4$ 
let mass of the metal be = x
percentage of metal in $\displaystyle M _3O _4 = \frac{3x}{ 3x+64}\times100$
but % age = (100-27.6) = 72.4 %
so, $\displaystyle \frac{3x}{ 3x+64}\times100$
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, ratio :--
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide =$ M _2O _3$

Multiple choice chemistry atoms, molecules and radicals molecular formula and valency valency and chemical formulae combining capacity and chemical formulae

The formula for potassium permanganate is:

  1. ${ K } _{ 2 }{ MnO } _{ 4 }$
  2. ${ K }{ MnO } _{ 4 }$
  3. ${ K } _{ 2 }{ Mn } _{ 2 }{ O } _{ 4 }$
  4. ${ K }{ Mn } _{ 2 }{ O } _{ 4 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potassium permangnate is a strong oxidizing agent and an inorganic chemical compound which is used in titrations. It has formula ${ KMnO } _{ 4}\ consisting\ of\ { K }^{ + },{ MnO } _{ 4 }^{ - }\quad ions.$ In this compound manganese is in $+7$ oxidation state.

Multiple choice chemistry atoms, molecules and radicals molecular formula and valency valency and chemical formulae combining capacity and chemical formulae

The salt of an oxyacid of a metal M contains one sulphur and four oxygen atoms

The oxyacid of the salt is

  1. $H _{2}SO _{4}$
  2. $H _{2}SO _{3}$
  3. $H _{2}S _{2}O _{7}$
  4. $H _{2}S$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The salt of an oxyacid of a metal M contains one sulphur and four oxygen atoms. The oxyacid of the salt is $H _{2}SO _{4}$. One sulphur and four oxygen atoms indicate sulphate ion $(SO _4^{2-})$. Hence, the salt is metal sulphate and the oxyacid is sulphuric acid.

Multiple choice chemistry atoms, molecules and radicals molecular formula and valency valency and chemical formulae combining capacity and chemical formulae

Predict the formulae of the binary compounds formed by combination of the following pairs of elements:
(i)    Magnesium and nitrogen
(ii)   Silicon and oxygen

  1. $MgN _{2}, SiO _{2}$
  2. $Mg _{3}N _{2}, SiO _{2}$
  3. $Mg _{2}N _{3}, Si _{2}O _{3}$
  4. $MgN, SiO _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Three Mg ( $+2$ oxidation state) combine with two N ($-3$ oxidation state) to form $Mg _3N _2$.

One Si ( $+4$ oxidation state) combine with two O ($-2$ oxidation state) to form $SiO _2$.

Note: In a neutral molecule, the total charge on cations is numerically equal to the total charge on anions.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Different proportions of oxygen in the various oxides of nitrogen, prove the law of:

  1. reciprocal proportions

  2. multiple proportions

  3. constant proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Different proportion of oxygen in the various oxides of nitrogen proves the law of multiple proportions, which states:

when two elements combine in more than one proportion to form one or more compounds, the weight of one element that combine with the given weight of other elements are in the ratio of small whole number.


Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Two oxides of metal contain $27.6$% and $30$% of oxygen respectively. if the first one is $M _3O _4$ then which of the following will be the other one?

  1. $M _3O _3$
  2. $M _2O _3$
  3. $M _4O _2$
  4. $M _1O _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, ${ M } _{ 3 }{ O } _{ 4 }$; Let mass of the metal$=x$
% of metal in ${ M } _{ 3 }{ O } _{ 4 }=\cfrac { 3x }{ 3x+64 } \times 100$
But as given % age$=\left( 100-27.6 \right) =72.4$
So, $\left( \cfrac { 3x }{ 3x+64 }  \right) \times 100=72.4$
$\Rightarrow x=56$
In 2nd oxide, oxygen$=30$%, So metal$=70$%
So, the ratio is $M:O$
$\cfrac { 70 }{ 56 } :\cfrac { 30 }{ 16 } $
$1.25:1.875$
$2:3\Rightarrow $ oxide is ${ M } _{ 2 }{ O } _{ 3 }$