

InterviewSolution
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.
1001. |
Which group elementof d-block do not form hydride at all?A. 7B. 8C. 9D. 10 |
Answer» Correct Answer - A::B::C | |
1002. |
Examples of polymeric hydrides areA. `BH_(3)`B. `BeH_(2)`C. `NaH`D. `CaH_(2)` |
Answer» Correct Answer - A::B | |
1003. |
Explain with examples Complex hydrides. |
Answer» Complex hydrides: In these hydrides, the hydride ion (H–) acts as the ligand and is attached to the central metal atom by coordinate bonds. These are formed both by transition elements and non-transition elements. Among the non-transition elements, the most important complex hydrides are formed by elements of group 13. These are sodium borohydride (NaBH4), lithium borohydride (LiBH4) and lithium aluminium hydride (LiAlH4). These are versatile reducing agents and are widely used for reduction of organic compounds. |
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1004. |
In what respects does hydrogen resemble alkali metals? How does it resemble halogen? |
Answer» (i) Resemble with alkali metals: 1. It has one valence electron in s-orbital like alkali metals 2. It can lose one electron to form H+ ion like alkali metals. 3. It is liberated at cathode during electrolysis of compounds like H2O, HCl etc. 4. It shows +1 oxidation state like alkali metals. 5. It is a strong reducing agent like other alkali metals. (ii) Resemblance with halogen: 1. Hydrogen is non-metal like halogens. 2. Hydrogen forms diatomic molecule like halogens. 3. It has ionization energy high like halogens. 4. Its electronegativity is similar to halogens. 5. It can gain one electron to form H- ion. 6. When NaH is electrolysed, hydrogen is liberated at anode like halogens. |
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1005. |
Which of the following explanation is best for not placing hydrogen with alkali metals or halogenA. The ionization energy of hydrogen is high for group of alkali metals or halogenB. Hydrogen can form compoundsC. Hydrogen is much lighter element than the alkali metals or halogensD. Hydrogen atoms does not contain any neutron. |
Answer» Correct Answer - C Hydrogen is much lighter than alkali metals. |
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1006. |
Which of the follwing terms is not correct for hydrogen ?A. Its molecule is diatomicB. It exists both as `H^(+)` and `H^(-)` in different chemical compoundsC. It is the only species which has no neutrons in the nucleusD. Heavy water is unstable because hydrogen is substituted by its isotope deuterium. |
Answer» Correct Answer - D Heavy water is not unstable. |
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1007. |
Knowing the properties of `H_(2)O` and `D_(2)O`, do you think that `D_(2)O` can be used for drinking purpose? |
Answer» Heavy water `(D_(2)O)` acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of `D_(2)O`, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and lead to a casualty. | |
1008. |
Knowing the properties of `H_(2)O and D_(2)O`, do you think that `D_(2)O` can be used for drinking purposes? |
Answer» Heavy water `(D_(2)O)` is quite injurious to living beings. Plants and animlas since it slows down the rates of reactions which occur in them. It fails to support to support life and has no utility in biopshere. | |
1009. |
Hydrogen acts as a reducing agent and thus resemblesA. halogenB. Noble gasC. Radioactive elementsD. Alkali metals |
Answer» Correct Answer - D | |
1010. |
The composition of tritium isA. 1 electron, 1 proton, 1 neutronB. 1 electron, 2 protons, 1 neutronC. 1 electron, 1 proton, 2 neutrons.D. 1 electron, 1 proton, 3 neutrons |
Answer» Correct Answer - C | |
1011. |
Hydrogen does not combine withA. heliumB. bismuthC. antimonyD. sodium |
Answer» Correct Answer - A Helium is a noble gas and does not combine with hydrogen. |
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1012. |
The composition of tritium isA. 1 electron, 1 proton, 1 neutronB. 1 electron, 2 protons, 1 neutronC. 1 electron, 1 proton, 2 neutronD. 1 electron, 1 proton, 3 neutrons |
Answer» Correct Answer - c An atom of tritium contains 1 proton, 1 electron and 2 neutrons. |
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1013. |
The possible form of ` H_2` molecule on the basis of three isotopes can be :A. 3B. 6C. 9D. 12 |
Answer» Correct Answer - b The number of possible diatomic molecules of three isotopes of hydrogen is six. |
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1014. |
Which of the following is not a water softener?A. calgonB. permutitC. `Na_(2)SO_(4)`D. `Na_(2)CO_(3)` |
Answer» Correct Answer - C | |
1015. |
As its melting point, ice is lighter than water because:A. `H_(2)O` molecules are more closely packed in the solid stateB. ice crystals have follow hexagonal arrangement of `H_(2)O` moleculesC. on melting of ice `H_(2)O` molecules shrink in sizeD. ice forms mostly heavy water on first melting. |
Answer» (b) Ice crystals have `H_(2)O` molecules arranged hexxagonally and linked by hydrogen bonds. | |
1016. |
Calgon used as water softener isA. `Na_(2)[Na_(4)(PO_(3))_(6)]`B. `Na_(4)[Na_(4)(PO_(3))_(6)]`C. `Na_(2)[Na_(4)(PO_(4))_(5)]`D. `Na_(4)[Na_(2)(PO_(4))_(6)]`. |
Answer» Correct Answer - A Calgon or sodium hexametaphosphate is `Na_(2)[Na_(4)(PO_(3))_(6)` |
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1017. |
which of the following on oxidation gives `H_(2)O_(2)` ?A. 2-EthylanthraquinolB. 2-EthylanthraquinoneC. AnthraceneD. 2-Ethylanthracene. |
Answer» Correct Answer - A 2-Ethylanthraquinol on oxidation gives `H_(2)O_(2)`. |
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1018. |
Which of the following statement(s) is/are correct in the case of heavy water ?A. Heavy water is used as a moderator in nuclear reactorB. Heavy water is more effective as solvent than ordinary water.C. Heavy water is more associated than ordinaryD. Heavy water has lower boiling point than ordinary water. |
Answer» Correct Answer - A::C Heavy water is used as moderator in nuclear reactor and is more associated than water. |
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1019. |
Which of the following statement(s) is/are correct in the case of heavy water?(i) Heavy water is used as a moderator in nuclear reactor.(ii) Heavy water is more effective as solvent than ordinary water.(iii) Heavy water is more associated than ordinary water.(iv) Heavy water has lower boiling point than ordinary water. |
Answer» (i, iii) Heavy water is used as a moderator in nuclear reactor and is more associated than ordinary water. |
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1020. |
One would expect proton to have very largeA. ChargeB. Ionisation potentialC. Hydration energyD. Radius. |
Answer» Correct Answer - C A proton or `H^(+)` has very high hydration energy. |
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1021. |
When hydrogen peroxide is added to acidified potassium dichromate, a blue colour is produced due to formation of `:`A. `CrO_(3)`B. `Cr_(2)O_(3)`C. `CrO_(5)`D. `CrO_(4)^(2-)` |
Answer» Correct Answer - c c |
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1022. |
The pure hydrogen (99.9%) can be made by which of the following processes ?A. Mixing natural hydrocarbons of high molecular massB. Electrolysis of waterC. Reaction of salt like hydrides with waterD. Reaction of methane with steam |
Answer» Correct Answer - B | |
1023. |
Syngas is a mixture ofA. `CO_(2) + H_(2)`B. `CO+H_(2)`C. `CO + CO_(2)`D. `CO + N_(2)` |
Answer» Correct Answer - B | |
1024. |
a. Would you except the hydrides of `N,O` and `F` to have lower boiling points than the hydrides of their subsequent group members? Give reason. b. Can phosphorous with outer electronic configuration `3s^(2)3p^(3)` form `PH_(5)`? c. How many hydrogen-bonded water molecules(s) are associated with `CuSO_(4).5H_(2)O`? |
Answer» Although phosphorus exhibits +3 and +5 oxidation states. It cannot form `PH_(5)`. Besides some other cosiderations, high `Delta_(a)H` value of dihydrogen and `Delta_(eg)H` value of hydrogen do not favour to exhibit the highest oxidation state of P and consequently the formation of `PH_(5)`. | |
1025. |
Arrange the following (i) `CaH_(2), BeH_(2) and TiH_(2)` in order of increasing electrical conductance. (ii) LiH, NaH and CsH in order of increasing ionic character. (iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy. (iv) `NaH, MgH_(2) and H_(2)O` in order of increasing reducing property. |
Answer» (i) The electrical conductance of a molecule depends upon its ionic or covalent nature. Ionic compounds conduct, whereas covalent compounds do not. `BeH_(2)` is a covalent hydride. Hence, it does not conduct. `CaH_(2)` is an ionic hydride, which conducts electricity in the molten state. Titanium hydride, `TiH_(2)` is metallic in nature and conducts electricity at room temperature. Hence, the increasing order of electrical conductance is as follows: `BeH_(2) lt CaH_(2) lt TiH_(2)` (ii) The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the smaller is the ionic character. Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase (as shown below). `LiH lt NaH lt CsH` Hence, the increasing order of the reducing property is `H_(2)O lt MgH_(2) lt NaH`. (iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the attractive and repulsive forces present in a molecule. The bond pair in D–D bond is more strongly attracted by the nucleus than the bond pair in H–H bond. This is because of the higher nuclear mass of `D_(2)` stronger the attraction, the greater will be the bond strength and the higher is the bond dissociation enthalpy. Hence, the bond dissociation enthalpy of D–D is higher than H–H. However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre. Therefore, the increasing order of bond dissociation enthalpy is as follows: `F-F lt H-H lt D-D` (iv) Ionic hydrides are strong reducing agents. NaH can easily donate its electrons. Hence, it is most reducing in nature. Both, `MgH_(2) and H_(2)O` are covalent hydrides. `H_(2)O` is less reducing than `MgH_(2)` since the bond dissociation energy of `H_(2)O` is higher than MgH2. |
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1026. |
Status coated with white lead on long exposure to atmosphere turn black and the original colour can be restored on treatment with `H_(2)O_(2)` . Why ? |
Answer» On long exposure to atmosphere , white lead is converted into black PbS due to the action of `H_(2)S` present in the atmosphere . As a result , statues turn black . `PbO_(2) + 2H_(2)S to PbS + 2H_(2)O` On treatment of these blackened statues with `H_(2)O_(2)` , black PbS gets oxidised to white `PbSO_(4)` and the colour is restored. `PbS + 4H_(2)O_(2) to PbSO_(4) + 4H_(2)O` |
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1027. |
Why do alcohol (a covalent compound) dissolves in water (ionic)? |
Answer» Alcohol being a polar covalent compound, dissolves in water because of the formation of hydrogen bonds between them. | |
1028. |
Does water gets oxidised in the process of photosynthesis? |
Answer» Yes, water gets oxidised to `O_(2)` in the process of photosynthesis `6CO_(2)(g)+12H_(2)O(l)toC_(6)H_(12)O_(6)(aq)+6H_(2)O(l)+6O_(2)(g)` |
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1029. |
Which properties of hydrogen as responsible for moderation of the climate and body temperature of living beings? |
Answer» High heat of vaporisation and high heat capacity. | |
1030. |
Compare the density of ice and water. |
Answer» Water is denser that ice. Ice forms an open cage like structure because of extensive hydrogen bonding which results in the large number of vacant spaces between them. Because of these vacant spaces the density of ice is lesser than that of water. | |
1031. |
What is the percentage strength of a solution of 100 volume `H_(2)O_(2)`? |
Answer» 100 volume solution means that 1L of this `H_(2)O_(2)` solution will give 100 L of oxygen at STP `underset(2xx34g=68g)(2H_(2)O_(2))to2H_(2)O+underset(22.7L" at STP")(O_(2))` From equation, 22.7 of `O_(2)` at STP is obtained from `H_(2)O_(2)=68g` 100 L of `O_(2)` at STP is obtained from `H_(2)O_(2)=(68)/(22.7)xx100` `=299.55g//L` `=30%" "H_(2)O_(2)` solution |
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1032. |
Why `H_(2)O_(2)` is kept away from dust? |
Answer» `H_(2)O_(2)` is kept away from dust because dust induces explosive explosive decomposition of the compound. | |
1033. |
`PbS(s)+4H_(2)O_(2)(aq)toPbSO_(4)(s)+4H_(2)O(l)` In the above reaction, `H_(2)O_(2)` acts as a/an_____agent. |
Answer» Correct Answer - Oxidising. | |
1034. |
To a 25 mL `H_(2)O_(2)` solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N solution thiosulphate solution. Calculate the volume strength of `H_(2)O_(2)` solution. |
Answer» Step 1. To determine the normality of `H_(2)O_(2)` solution. Let the normality of the `H_(2)O_(2)` solution be `N_(1)` . According to the question. 25 mL of `H_(1)H_(2)O_(2)-= 20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution. or `25xxN_(1)=20xx0.3` or `N_(1)=(20xx0.3)/(25)=0.24N` Thus, the normality of the given `H_(2)O_(2)` solution=0.24 N Step 2. To determine the amount of `H_(2)O_(20` in 25 mL solution 1000 mL of 1 `H_(2)O_(2)` solution contain `H_(2)O_(2)` =17 g `" "` (`because ` Eq. wt. of `H_(2)O_(2)` =17) `therefore ` 25 mL of 0.24 `H N_(2)O_(2)` solution will contain `H_(2)O_(2)=(17xx25xx24)/(1000xx100)=0.102 g` Step 3. To determine the volume strength of `H_(2)O_(2)` solution. Consider the chemical equation `underset(2xx34=68 g)(2H_(2)O_(2))to 2H_(2)O + underset(22.4 litres at N.T.P.)(O_(2))` 68 g of `H_(2)O_(2)` give `O_(2)` =22.4 litres at N.T.P. `therefore 0.102 g H_(2)O_(2)` will give `O_(2)=(22.4xx1000xx0.102)/(68)=33.6` mL at N.T.P. Now, 25 mL of `H_(2)O` solution give `O_(2)=33.6 mL` at N.T.P. `therefore ` 1 mL of `H_(2)O_(2)` solution will give `O_(2)=(33.6)/(25)=1.344` Thus, the volume strength of the given `H_(2)O_(2)` solution =1.344 |
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1035. |
To a 25 mL `H_(2)O_(2)` solution excess of an acidified solution of potassium iodide was added. The iodine liberated required 20 " mL of " 0.3 N sodium thiosulphate solution Calculate the volume strength of `H_(2)O_(2)` solution. |
Answer» Correct Answer - 1.34 | |
1036. |
Which one of the following is not a crystalline hydrate?(a) CH4 . 20H2O (b) Na2,CO3 . 10H2O (c) CuSO4 . 5H2O (d) FeSO4 . 7H2O |
Answer» (a) CH4 . 20H2O |
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1037. |
Hydrogen readily combines with non`-` metals and thus it shows itsA. electronegativity characterB. electropositive characterC. Both (A) and (B)D. None of these. |
Answer» Correct Answer - B `H_(2)+Cl_(2)tooverset(+)overset(-)Cl`. In this hydrogen has positive oxidation state. |
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1038. |
Hydrogen readily combines with metals and thus show itsA. electropositive characterB. electronegative characterC. Both (A) and (B)D. None of these. |
Answer» Correct Answer - B `2Na+H_(2)to2Na^(+)H^(-)` Hydrogen has -1 oxidation state. |
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1039. |
Hydrogen readily combines with non`-` metals and thus it shows itsA. Electronegativity characterB. Electropositive characterC. Bothe `(a)` and `(b)`D. None of these |
Answer» Correct Answer - b `H_(2)+Clrarr H^(+)Cl^(-)` . In this, hydrogen has positive oxidation state. |
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1040. |
Hydrogen directly combines withA. AuB. CuC. NiD. Ca. |
Answer» Correct Answer - D `H_(2)` does not react with Au, Cu or Ni With Ca it gives `CaH_(2)` `Ca+H_(2)toCaH_(2)` |
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1041. |
When the same amount of zinc is treated separately with excess of sulphric acid and excess of sodium hydroxide, the ratio of volume of hydrogen evolved isA. `1:1`B. `1:2`C. `2:1`D. `9:4`. |
Answer» Correct Answer - A `Zn+H_(2)SO_(4)toZnSO_(4)+H_(2)` `Zn+2NaOHtoNa_(2)ZnO_(2)+H_(2)` `:.` Ratio of volumes of `H_(2)` evolved is `1:1`. |
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1042. |
It is possible to remove completely the temporary hardness due to `Mg(HCO_(3))` by boiling? |
Answer» Yes, it is possible on boiling, `Mg(HCO_(3))_(2))` will readily decompose to form a precipitate of `MgCO_(3)` which can be removed by filtration. `Mg(HCO_(3))_(2) overset("Heat")to underset("ppt")(MgCO_(3))+H_(2)O+CO_(2)` As a result, water will become free from temporary hardness. |
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1043. |
Chemical A is used for water softening to remove temporary hardness. A reacts with sodium carbonate to generate caustic soda. When `CO_(2)` is bubbled through a solution of A, it turns cloudy. What is the chemical formula of A ?A. `CaCO_(3)`B. CaOC. `Ca(OH)_(2)`D. `Ca(HCO_(3))_(2)`. |
Answer» Correct Answer - C `Ca(OH)_(2)` is used for the softening of temporary hard water. `Ca(OH)_(2)(aq)+CO_(2)(g)tounderset("Cloudiness")(CaCO_(3)(s))+H_(2)O(l)` |
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1044. |
The reagent(s) used for softening the temporary hardness of water is (are):A. `Ca_(3)(PO_(4))_(3)`B. `Cu(OH)_(2)`C. `Na_(2)CO_(3)`D. NaOCl. |
Answer» Correct Answer - B::C | |
1045. |
The reagent(s) used for softening the temporary hardness of water is (are):A. `Ca_(3)(PO_(4))_(2)`B. `Ca(OH)_(2)`C. `Na_(2)CO_(3)`D. `NaOCl` |
Answer» Correct Answer - B::C | |
1046. |
Chemical `(X)` is used for water softening to remove temporary hardness. `(X)` reacts with sodium carbonate to generate caustic soda. When `CO_(2)` is bubbled through `(X)`?A. `CaCO_(3)`B. `CaO`C. `Ca(OH)_(2)`D. `Ca(HCO_(3))_(2)` |
Answer» Correct Answer - c `Ca(OH)_(2)` is used for the softening of temporary hard water . `Ca(OH)_(2)(aq) +CO_(2)(g) rarr underset("cloudiness" )(CaCO_(3)(s)+H_(2)O(l)` |
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1047. |
Name the method used to remove temporary hardness of water? |
Answer» Clark s method OR By Boiling | |
1048. |
How many hydrogen-bonded water molecule(s) are associated in CuSO4.SH2O ? |
Answer» One hydrogen-bonded water molecule is associated in CuSO4. 5H2O. |
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1049. |
Write the chemical formula of “Calgon”. |
Answer» Na2P6O16 Or Na2 [Na4(PO3)6] |
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1050. |
Name the compound used in Clark’s method to remove temporary hardness of water. |
Answer» Lime is used in Clark’s method to remove temporary hardness of water. |
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