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26201.

The oxyacid of phosphorous in which phosphorous has the lowest oxidation state is

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HYPOPHOSPHOROUS acid
orthopbosphoric acid
pyrophosphoric acid
metaphosphoric acid

Solution :Hypophophorous acid is `H_(3)PO_(2)` in which O.S. of P is `+1`.
26202.

The oxoacid of sulphur that contains a lone pair of electrons on sulphur is

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SULPHUROUS ACID
sulphuricacid
peroxodisulphuric acid,
Pyrosulphuric acid

Answer :A
26203.

The oxoacid of P having oxidation state +4 is :

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PHOSPHOROUS ACID
HYPOPHOSPHORIC acid
Phosphoricacid
METAPHOSPHORIC acid.

Answer :B
26204.

The oxime of which of the following species wukk exist in syn and anti geometrical forms.

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Benzophenone
Benzaldehyde
Acetone
Diethyl ketone.

Solution :ANTI & SYN. FORMS
26205.

The oxidtion state of Fe in k_(4)[Fe(CN)_(6)] is………. .

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ANSWER :II
26206.

The oxo acid of halogen with maximum acidic character is :

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`HClO_(4)`
`HClO_(3)`
`HClO_(2)`
`HCLO.`

ANSWER :A
26207.

The oxidizing property of nitric acid is due to:

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its concentration
the POSITIVE valency of N
its dilution
the unstability of its molecule and the PRESENCE of nitrogen in its HIGHEST state of oxidation

Answer :D
26208.

The oxidising property of H_(2)O_(2) is best explained by assuming that two oxygen atoms in its molecule are bonded

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DIFFERENTLY
similarly
covalently
by hydrogen bonds

Solution :It is believed that out of two hydrogens, one is associated with each oxygen atom at a bond ANGLE of `97^(@)`. All the FOUR atoms `H-O-O-H` do not lie in the same plane.
26209.

The oxidising property of nitric acid is due to

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its CONCENTRATION
the positive valency of N
its dilution
presence of nitrogen in its highest oxidation state.

Solution :`HNO_3` has nitrogen in its highest oxidation i.e., +5 state, due to this it reduces itself by OXIDISING other substances, HENCE is very strong oxidising AGENT.
26210.

The oxidising and reducing power of an element is measured in terms of ……

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SOLUTION :STANDARD ELECTRODE POTENTIAL
26211.

The oxidising agent used to convert toluene to benzaldehyde is ................

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SOLUTION :`CrO_2Cl_2`CHROMYL CHLORIDE
26212.

The oxidising agent used to prepare aldehyde (or) ketone from alcohol, the reagent usedis ………………… .

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acidified `Na_2Cr_2O_7`
alkaline `KMnO_4`
PYRIDINIUM CHLOROCHROMATE
CONC. `H_2SO_4`

Solution : Pyridinium chlorochromate
26213.

The oxides will decompose on heating even in the absence ofreducing agent is ……..

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`Ag_(2)O`
HgO
MgO
Both (a) and (B)

ANSWER :D
26214.

The oxides of non - metals are acidic and the oxides of the metals are basic .Would you expect boron to be more basic in its reactions than aluminium?.

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SOLUTION :Being nearer the top of the group, we would EXPECT BORON to show more non-metallic NATURE than aluminium, so, it should be more acidic in nature than aluminium..
26215.

The oxides of Ca, Sr and Ba are :

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Basic
Acidic
Amphoteric
Neutral

Answer :A
26216.

The oxide which shows transition from metal to insulation, i.e., semiconductors are:

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`V_2O_3`
`VO_2`
`Ti_2O_3`
All

Answer :D
26217.

The oxide which shows metallic conduction:

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`ReO_3`
VO
`CrO_2`
All

Answer :D
26218.

The oxide which is solid at room temperature is:

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`N_2O`
NO
`N_2O_4`
`N_2O_5`

ANSWER :D
26219.

The oxide which on strong heating evolves oxygen is :

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`Na_(2)O`
`Al_(2)O_(3)`
CaO
`BaO_(2)`

Solution :`2BaO_(2) overset(DELTA) to 2BaO+O_(2)`.
26220.

The oxide which is not reduced by hydrogen in

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<P>`Ag_(2)O`
`K_(2)O`
`Fe_(2)O_(3)`
`P_(4)O_(10)`

SOLUTION :On the BASIS of electrochemical series `K_(2)O` is not reduced by hydrogen.
26221.

The oxide which gives hydrogen peroxide on treatment with dilute acid is

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`PbO_(2)`
`Na_(2)O_(2)`
`MnO_(2)`
`TiO_(2)`

Solution :`Na_(2)O_(2)+H_(2)SO_(4)(dil).RARR H_(2)O_(2)+Na_(2)SO_(4)`
`PbO_(2),MnO_(2) and TiO_(2)` do not react with dil acids to GIVE HYDROGEN peroxide.
26222.

The oxide which gives H_(2)O_(2) with dil. HCl is :

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`PbO_(2)`
`Na_(2)O_(2)`
`MnO_(2)`
`SnO_(2)`.

Answer :B
26223.

The oxide which cannot act as reducing agent is

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`NO_2`
`SO_2`
`CO_2`
`CIO_2`

Solution :(C) CARBON has the MAXIMUM OXIDATION state of +4, therefore carbon dioxide `(CO_2)` cannot act as a REDUCING agent.
26224.

The oxide that possesses electrical conductivity:

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`V_2O_5`
`CrO_2`
NiO
MnO

Answer :B
26225.

The oxide which gives H_(2)O_(2) on treatment with dilute acid is

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`PbO_(2)`
`BaO_(2).8H_(2)O`
`MnO_(2)`
`TiO_(2)`

ANSWER :B
26226.

The oxide that is insulator is

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VO
COO
`ReO_(3)`
`Ti_(2)O_(3)`

Solution :Oxide that is INSULATOR is CoO.
26227.

The oxide that is not reduce by hydrogen in the hot is:

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`Ag_2O`
`Fe_2O_3`
CUO
`K_2O`

ANSWER :D
26228.

The oxide of which metal is most stable to heat:

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K
Ag
Hg
All of these

Answer :A
26229.

The oxide that does not act as an insulator even by a change in temperature

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`VO_(2)`
TIO
VO
`TiO_(2)`

ANSWER :B
26230.

The oxide of nitrogen which reacts with NaOH solution giving both sodium nitrite and sodium nitrate is:

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`NO_2`
`N_2O_5`
`N_2O_3`
`NO`

ANSWER :A
26231.

The oxide that gives H_2O_2 on treatment with a dilute acid is :

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`PbO_2`
`MnO_2`
`Na_2O_2`
`TiO_2`

ANSWER :C
26232.

The oxide of chlorine which is a mixed anhydride is

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`Cl_(2)O`
`Cl_(2)`
`Cl_(2)O_(3)`
`Cl_(2)O_(7)`

Answer :C
26233.

The oxide of an element possesses the molecular formula, M_(2)O_(3). If the equivalent mass of the metal is 9, the atomic mas sof the metal will be:

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27
18
9
4.5

Answer :A
26234.

The oxide of nitrogen existing in the solid state at room temperature is

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NO
`NO_(2)`
`N_(2)O_(3)`
`N_(2)O_(5)`

ANSWER :4
26235.

The oxide of an element whose electronic configuration is 1s^(2)2s^(2)2p^(6)3s^(1) is

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amphoteric
basic
acidic
neutral

Solution :The given element is SODIUM, which is an alkali metal and alkali metal OXIDES are STRONGLY basic.
26236.

The oxide of an element possesses the formula M_(2)O_(3). If the equivalent weight of the metal is 9, the molecular weight of the oxide will be

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54
102
120
200

Answer :B
26237.

The oxide of an element contains 32.33 per cent of the element and the vapour density of its chloride is 79. Calculate the atomic mass of the element.

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SOLUTION :MASS of the element=32.33parts
Mass of oxygen `=(100-32.33)=67.67` parts
Equivalent mass of the element`=(32.33)/(67.67)xx8=3.82`
Valency of the element `=(2VD)/(E+35.5)=(2xx79)/(3.82+35.5)=4`
HENCE, the atomic mass of the element `=3.82xx4=15.28`.
26238.

The oxide of a metal contains 60% of the metal. What will be percentage of bromine in the bromide of the metal. If the valency of the metal is the same in both, the oxide and the bromide?

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0.93
0.87
0.7
0.77

Answer :B
26239.

The oxide obtained in the roasting of ironpyrites

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`SO_(2)`
`SO_(3)`
FEO
`SO_(2) and SO_(3)`

ANSWER :A
26240.

Which is insoluble in water ?

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`TeO_2`
`SO_2`
`PoO_2`
`SeO_2`

ANSWER :C
26241.

The oxide involved in the formationof brown ring for testing nitrate ion

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`N_(2)O`
NO
`NO_(2)`
`N_(2)O_(3)`

ANSWER :2
26242.

The oxide formed in the maximum oxidation state is ZrO_(x).x is

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SOLUTION :N//A
26243.

The oxide containing chloride in the highest oxidation state is :

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`ClO_(2)`
`ClO_(3)`
`Cl_(2)O_(7)`
`Cl_(2)O`.

Solution :In `Cl_(2)O_(7)`, the OXIDATION state of Cl is `+7`.
26244.

The oxide bridges through which a large number of monosaccharides are called

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PEPTIDE BOND
hydrogen bond
NITROGEN bond
glycosidic LINKAGE

Solution :glycosidic linkage
26245.

The oxidation states of V and Br in V(BrO_(2))_(2)are respectively

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2 and 2
2 and 1
4 and 2
2 and 3.

Solution :`V(+2),Br+3" in "(BrO_(2))^(-)" so it EXISTS as "V^(II)(Br^(III)O_(2))_(2)`
26246.

The oxidation states of the most electronegative element in the products of the reaction betweeen BaO_(2) with dil H_(2)SO_(4) are

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`-1`
`+1`
`-2`
0

Solution :`BaO_(2)+H_(2)SO_(4)rarrBaSO_(4)+H_(2)O_(2)`
Most electronegative element in products is oxygen. In `H_(2)O_(2)` oxidation state of O is `-1` and in `BaSO_(4), -2`.
26247.

The oxidation states of the metal atoms in [Fe(CO)_(5)] and Na[Co(CO)_(4)] complexesare respectively :

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5 and 4
0 and +3
`+3` and `+4`
0 and -1.

ANSWER :D
26248.

The oxidation states of sulphur in the anions SO_(3)^(2-)S_(2),O_(4)^(2-) and S_2 O_(6)^(2-)follow the order

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`S_2 O_(4)^(2-)ltSO_(3)^(2-) lt S_2 O_(6)^(2-)`
`SO_(3)^(2-)ltS_2 O_(4)^(2-) lt S_2 O_(6)^(2-)`
`S_2 O_(4)^(2-)ltSO_(6)^(2-) lt S_2 O_(3)^(2-)`
`S_2 O_(6 )^(2-)ltS_2 O_(6)^(2-) lt S O_(3)^(2-)`

Solution :`OVERSET(x-2)((SO_3 )^(2-)) ,x-6=-2 ,x=+4`
` overset(x-2)((S_2O_4 )^(2-), 2x - 8 =- 2, x= +3 `
` overset(x-2) (S_2 O_(6))^(2-) .2x - 12 =-2, x = +5`
26249.

The oxidation states of S atoms in S_(4)O_(6)^(2-) from left to right respectively are {:("OO"),("||||"),(O^(-)-S-S-S-S-O^(-)),("||||"),("OO"):}

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`+6, 0, 0, +6`
`+3, +1, +1, +3`
`+5, 0, 0, +5`
`+4, +1, +1, +4`

Solution :LEFT S atom : +2 each from `=O+4`
`+1` from `- O^(-)""+1`
0 from `-S""0/(+5)`
26250.

The oxidation states of sulphur in the anions SO_(3)^(-2),S_(2)O_(4)^(2-)andS_(2)O_(6)^(2-)follow the order :

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`S_(2)O_(4)^(2-)ltSO_(3)^(2-)ltS_(2)O_(6)^(2-)`
`SO_(3)^(2-)ltS_(2)O_(4)^(2-)ltS_(2)O_(6)^(2-)`
`S_(2)O_(4)^(2-)ltS_(2)O_(6)^(2-)ltSO_(3)^(2-)`
`S_(2)O_(6)^(2-)ltS_(2)O_(4)^(2-)ltSO_(3)^(2-)`

Answer :A