Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

The correct order of stoichiometries of `AgCI` formed when `AgNO_(3)` in excess is treated with the complex: `CoCI_(3) . 6NH_(3), CoCI_(3) . 5 NH_(3), CoCI_(3) . 4NH_(3)` respectively is:A. `3 AgCI, 1 AgCI, 2 AgCI`B. `3 AgCI, 2 AgCI,1 AgCI`C. `2 AgCI, 3 AgCI, 1 AgCI`D. `1 AgCI, 3 AgCI, 2 AgCI`

Answer» Correct Answer - B
`CoCl_(3). 6NH_(3) implies [Co (NH_(3))_(6)] Cl_(3) implies 3 AgCl`
`CoCl_(3). 5NH_(3) implies [Co(NH_(3))_(5) Cl] Cl_(2) implies 2 Ag Cl`
`CoCl_(3). 4NH_(3) implies [Co(NH_(3))_(4) Cl_(2)] Cl implies 1AgCl`
Note that only `Cl` present outside the coordination sphere is ionizable.
Discussion
2.

On which of the following properties does the coagulating power of an ion depend?A. The magnitude of the charge on the ion alone.B. Size of the ion alone.C. Both magnitude and sign of the charge on the ionD. The sign of charge on the ion alone

Answer» Correct Answer - C
According to Hardy-Sxhulze rule, the greater the valency (magnitude of charge) of the flocculating ion (decided by sign of charge), the greater is its power to cause precipitation.
Discussion
3.

Which of the following statements is not ture for halogens?A. All form monobasic oxyacids.B. All are oxidizing agents.C. All but fluorine show positive oxidation statesD. Chlorine has the highest electron-gain enthaply

Answer» Correct Answer - A
For a long time it was thought that `F` could not formk any oxacids. It is now known that `HOF` can be made under special conditions but it very unstable. No other oxacids of `F` are known.
Discussion
4.

Which ordering of compound is according to the decreasing order of the oxidation state of nitrogen?A. `HNO_(3), NO, N_(2), NH_(4) Cl`B. `HNO_(3), NO, NH_(4) Cl, N_(2)`C. `HNO_(3), NH_(4) Cl, NO, N_(2)`D. `NH_(4) Cl, N_(2), NO, HNO_(3)`

Answer» Correct Answer - A
The correct dereasing order of oxidation states is `Hoverset(+5)NO_(3), overset(+2)NO, overset(0)N_(2) gt NH_(4)^(-3) Cl`
Discussion
5.

Considering Ellingham diagram, which of the following metals can be used to reduce alumina?A. `Fe`B. `Zn`C. `Mg`D. `Cu`

Answer» Correct Answer - C
Metal whose oxide formation has more `- ve Delta G` value can reduce oxides of those metals whose formation has less `-ve Delta G`. Among the given metals only `(Mg rarr Mg O)` curve lies below `(Al rarr Al_(2) O_(3))` curve. Thus, `Mg` metal can be used to reduce. alumina.
Discussion
6.

Which fo the following is not correct?A. Ovalbumin is a simple food reserve in egg-whiteB. Blood proteins thrombin and fibrinogen are involved in blood clottingC. Denaturation makes the proteins more activeD. Insulin maintains sugar level in the blood of a human body

Answer» Correct Answer - C
Due to denaturation, the native shape of the protein is destryoed and biological activity is lost
Discussion
7.

The correct increasing order fo basic strength for the following compound is: A. `III lt I lt II`B. `III lt II lt I`C. `II lt I lt III`D. `II lt III lt I`

Answer» Correct Answer - C
In general, electron - donating groups such as `- CH_(3)` increase the basic strength of aniline by decreasing the delocalization of lone pair of electrons on the `N` atom while electron - withdrawing groups such as `NO_(2) ` decrease the basic strength of aniline by increasing the delocalization of lone pair of electrons on the `N` atom. Thus, the correct order of basic strength is : `(II) lt (I) lt (III)`
Discussion
8.

An example of a sigma bonded organometallic compound is:A. GrignardB. FerroceneC. CobaltoceneD. Ruthenocene

Answer» Correct Answer - A
In a sigma bonded organometallic compound such as Grignard reagent the carbon is bonded to a metal by means of a sigma bond.
Discussion
9.

The difference between amylose and amylopectin isA. Amylopectin have `1 rarr 4 alpha`- linkage and `1 rarr 6 alpha`-linkageB. Amylose have `1 rarr 4 alpha`-linkage and `1 rarr 6 beta`- linkageC. Amylopectin have `1 rarr 4 alpha`- linkage and `1 rarr 6 beta`-linkageD. Amylose in made up of glucose and galactose

Answer» Correct Answer - A
Starch is a polymer of `alpha`-glucose and consists of two components: Amylose and Amylopectin.
Amylose is water soluble component which constitutes about `15 - 20%` of starch. Chemically, amylose is a long unbranched chain held by `C_(1) - C_(4)` glycosidic linkage. Amylopection is insoluble in water and constitutes about `80 - 85%` fo starch. It is a branched chain polymer of `alpha-D-` glucose units in which chain is formed by `C_(1) - C_(6)` gycosidic linkage.
Discussion
10.

Regarding cross-linked or network polymers, which of the following statements is incorrect?A. They contain covalent bonds between various linear polymer chains.B. They are formed from bi- and tri - functional monomers.C. Examples are bakelite and melamine.D. They contain strong covalent bonds in their polymer chains.

Answer» Correct Answer - D
Cross linked of network polymers are usually formed from bi-functional and tri-functional monomers and contain strong covlent bonds between various linear polymer chains. Typical examples are bakelite, melamine. Option 4 is not related to cross-linkage.
Discussion
11.

The number of lone pairs of electrons present on the central atom of `CIF_3` isA. oneB. twoC. fourD. three

Answer» Correct Answer - B
The central atom `Cl` (Group 17) has even seven valence electrons. Three of these are used up to form thrice covalent bonds with three `F` atoms. This leaves behind four electrons which constitute two lone pair
Discussion
12.

Consider the change in oxidation state of Bromine corredponding to different emf values as shown in the diagram below : . The the species undergoing dispropprtionation is .A. `BrO_(3)^(-)`B. `BrO_(4)^(-)`C. `Br_(2)`D. `HBrO`

Answer» Correct Answer - D
disproprotionation involves oxidation as well as reaction of the same element
`{:("Species","O.N. of Br"),( B r O_(4)^(-),+7),(BrO_(3)^(-),+5),(HBrO,+1),(Br_(2),0),(Br^(-),-1):}`
`E_(cell)^(@)` corresopinding to compound undergoing disproporitonation reactioin should be positive so that the reaction is spontaneous `(DeltaG^(@)=-2FE^(@))`
`overset(+5)(BrO_(3))underset(1.5V)overset("oxidation")larroberset(+1)(HBrO)underset(1.59V)overset("reduction")tooverset(0)(Br_(2))`
`E_(cell)^(@)=E_(R)^(@)-E_(L)^(@)`
`=(1.595V)-(1.5V)`
`=0.095V`
Since `E_(cell)^(@)` is +ve, `DeltaG^(@)` will be negative i.e., HBrO will undergo disproportionation reaction.
Discussion
13.

Correct increasing order for the wavelengths of absorption in the visible region by the complexes of `Co^(3+)` is:A. `[Co(H_(2)O)_(6)]^(3+), [Co(en)_(3)]^(3+), [Co(NH_(3))_(6)]^(3+)`B. `[Co(H_(2)O)_(6)]^(3+), [Co(NH_(3))_(6)]^(3+), [Co(en)_(3)]^(3+)`C. `[Co(H_(2)O)_(6)]^(3+), [Co(en)_(3)]^(3+), [Co(H_(2)O)_(6)]^(3+)`D. `[Co(en)_(3)]^(3+), [Co(NH)_(3))_(6)]^(3+), [Co(H_(2)O)_(6)]^(3+)`

Answer» Correct Answer - D
`Delta E prop` field of ligand strength `prop (1)/(lambda)`
correct order of field strength of ligand:
`H_(2) O lt NH_(3) lt en`
Discussion
14.

Extraction of gold and silver involves leaching with `CN^(-)`ion.silver is later recovered by:A. distillationB. zone refiningC. displacement by `Zn`D. liquation

Answer» Correct Answer - C
`4 Ag + 8 NaCN + 2 H_(2) O + O_(2)`
`rarr 4 Na [Ag (CN)_(2)] + 4 NaOH`
`2[Ag (CN)_(2)]^(-) (aq) + Zn (s)`
`rarr 2 Ag (s) + [Zn (CN)_(4)]^(2-) (aq)`
Discussion