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.
| 7701. |
Which element was discovered by a team of astronomers who had camped near Bellary, Karnataka in 1868? |
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Answer» Correct answer is Helium |
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| 7702. |
Which national park in north-east is famous for the dancing dear? |
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Answer» Keibul Lamjao in Manipur |
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| 7703. |
Sachin Tendulkar’s first world record was for his school Shardashram Vidyamandir in a school cricket tournament in Mumbai. Which tournament? |
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Answer» Harris Shield |
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| 7704. |
What ist the coordination number of each ion present in the closed packed sturctures of (a) `Na_(2) O` (b) `CaF_(2)` at ordinary temperature and pressure? |
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Answer» a. `Na^(o+) = 4, O^(2) = 8`. b. `Ca^(2+) = 8, F^(ɵ) = 4` |
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| 7705. |
When diethyl ether is treated with `CI_(2)` in the presence of dark. Then the product formed isA. `CH_3CCI_2-O-CH_2CH_3`B. `CH_3CHCI-O-CHCICH_3`C. `CCI_3CCI_2OCCI_2CCI_3`D. `CH_2CCI_2-O-CHCICH_3` |
| Answer» Correct Answer - B | |
| 7706. |
What is the eye irritating compound when cutting onions?A. CarbonB. NitrogenC. SulphurD. Hydrogen1. B2. C3. D4. A |
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Answer» Correct Answer - Option 2 : C The correct answer is Sulphur.
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| 7707. |
Two grams of dihydrogen diffuse from a container in 10 minutes. How many gram of dioxygen would diffuse through the same container in thhe same time under similar conditions?A. 4B. 8C. 6D. 0.5 |
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Answer» Correct Answer - B `((n_(H_(2)))/(10))/((n_(O_(2)))/(10))=sqrt((M_(O_(2)))/(M_(H_(2))))` or `(2//2)/(10)xx(10)/(n_(O_(2)))=sqrt((32)/(2))=4` `therefore(1)/(n_(O_(2)))=4impliesn_(O_(2))=(1)/(4)impliesW_(O_(2))=(1)/(4)xx32=8g` |
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| 7708. |
The electronegativity of phosphorus is maximum inA. `PCl_(5)`B. `PClF_(4)`C. `PCl_(2)F_(3)`D. `PCl_(3)F_(2)` |
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Answer» Correct Answer - B Phosphorus contains maximum positive charge in `PClF_(4)` |
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| 7709. |
Hyper conjugation involves overlap of the following orbitals (a)σ−σ(b)σ−π(c)p−p(d)π−π |
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Answer» Hyperconjugation involves delocalization of σ and π bonds orbitals. ie it undergoes σ−π conjugation. The kind of delocalization involving sigma electrons of single bond and π- electrons of multiple bond is called hyperconjugation. Hence b is the correct answer. |
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| 7710. |
The maximum oxidation state shown by `V(Z=23),Cr(Z=24),Co(Z=27),Sc(Z=21)`, are respectivelyA. `+5, +6, +3,+3`B. `+3,+4,+5,+2`C. `+5,+3,+2,+1`D. `+4` in each case. |
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Answer» Correct Answer - A (A)Electron configuration of V is [Ar] `3d^3 4s^2` and thus maximum 5 electrons participate in bonding. (B)Electron configuration of Cr is [Ar] `3d^5 4s^1` and thus maximum 6 electrons participate in bonding. ( C)Electron configuration of Co is [Ar] `3d^7 4s^2`.In octahedral splitting in presence of ligands , half filled `t_(2g)^6` has higher CFSE and thus +3 oxidation state is most stable. (D)Electron configuration of Sc is [Ar] `3d^1 4s^2` and thus maximum 3 electrons participate in bonding. |
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| 7711. |
`{:("Column-I","Column-II"),("Species","Characteristics"),((A)Co^(3+)(Z=27),(p)"Total number of fully filled orbitals is nine"),((B)Sc^(3+)(Z=21),(q)"The value of magnetic moment (spin only) is greater than or equal to 3.87 BM"),((C )Cr^(3+)(Z=24),(r)"Number of electrons with (n+l=3) is eight"),((D)Ni^(2+)(Z=28),(s) "Number of electrons with (m=0) may be either 11 or 12 "),(,(t)"No unpaired electron"):}` |
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Answer» Correct Answer - A-q,r,s ; B-p,r,t ; C-p,q,r,s ; D-r,s (A)`Co^(3-):1s^2 2s^2 2p^6 3s^2 3p^6 3d^6` so no of fully filled orbitals is 10. no of unpaired electrons =4 So, magnetic moment is greater than 3.87 no of electrons with n+l=3 i.e. 3s, 2p are eight. (B)`Sc^(3+):1s^2 2s^2 2p^6 3s^2 3p^6 ` no of fully filled orbitals is 9. no of unpaired electrons =zero no of electrons with n+l=3 i.e. 3s, 2p are eight. no of electrons with m=0 are ten. (C )`Cr^(3+):1s^2 2s^2 2p^6 3s^2 3p^6 3d^3` no of fully filled orbitals is 9. no of unpaired electrons =3 no of electrons in 3s & 2p are eight no of electrons with m=0 may be 11 (D)`Ni^(2+):1s^2 2s^2 2p^6 3s^2 3p^6 3d^8` so no of fully filled orbitals is 12. no of unpaired electrons =2 no of electrons in 3s & 2p is eight no of electrons with m=0 may be 11 or 12 |
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| 7712. |
Select from the following the statement which is true for bases. (a) Bases are bitter and turn blue litmus red. (b) Bases have a pH less than 7 (c) Bases are sour and change from red litmus to blue. (d) Bases turn pink when a drop of phenolphthalein is added to them |
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Answer» (d) Bases turn pink when a drop of phenolphthalein is added to them. |
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| 7713. |
Examine these compounds available. Find more colored compounds and extend the list. |
Answer»
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| 7714. |
1. List out the dements of the s block?2. Which elements shows+2 oxidation state?3. Which elements contains 5 electrons in the outermost shell?4. Which is the element that has 5 p electrons in the outermost shell?5. Which are the elements in which the last electron enters the d subshell?6. Which element has the highest ionization energy?7. Which is the highly reactive nonmetal?8. Which elements show -2 oxidation state? |
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Answer» 1. A,B 2. B, C, D 3. E 4. G 5. C, D 6. H 7. G 8. F |
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| 7715. |
Which of the following statements about the reaction given below are correct?MnO2 + 4HCI → MnCl2 +2H2O + Cl2 (i) HCl is oxidized to Cl2 (ii) MnO2 is reduced to MnCl2(iii) MnCl2 acts as an oxidizing agent (iv) HCl acts as an oxidizing agent (a) (i), (iii) and (iv) (b) (i), (ii) and (iii) (c) (i) and (ii) only (d) (iii) and (iv) only |
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Answer» Correct option is (c) (i) and (ii) only |
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| 7716. |
Match the following. BlockOuter electronic configurationCharacteristicss3p5Most of the compounds are coloured.p3d44s2Includes lanthanoids (6thperiod)d4f15d1622Highest atomic redius in the respective periodf3s1High electro negativity |
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| 7717. |
Complete the tableCompoundOxidation state of MnSub shell electronic configuration of Mn ionsMnCl2 .....1s22s22p63s23p63d5MnO2 +4 ....Mn2O3 .... .....Mn2O7 .... ..... |
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| 7718. |
Complete the TableShell number123 4Sub shellsspspdspdfRepresentation of sub shells1s......3p......4d ... |
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| 7719. |
The atomic number of four elements are given below. (The symbols ore not real) A – 8 B – 10 C – 12D – 18 a. Write the sub-shell electronic configuration of the elements, b. Which of them are inert gases? c. Write the chemical formula of the compound formed by two elements other than inert gases. |
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Answer» a. A – 1s2 2s2 2p4 B – 1s2 2s2 2p6 C – 1s2 2s2 2p6 3s2 D – 1s2 2s2 2p6 3s2 3p6 b. B, D c. CA, (C2 A2 is simplified and written as CA) |
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| 7720. |
Complete the tableCompoundOxidation state of Fe Symbol of Fe ionsFeCl2FeCl3How does Fe change to Fe2+ ? |
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By losing 2 electrons from 4s valence subshell |
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| 7721. |
Complete the TableShell number123 4Maximum number of electrons that can be accommodated in each shell281832Sub shell1s2s2p3s3p3d4s4p4d4fmaximum number of electrons that can be accommodated in each sub shell22....................1. What is the maximum number of electronics that can be accommodated in the ‘s’?2. What may be the maximum number of electrons to be filled in the ‘p’ subshell? |
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1. 2 2. 6 |
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| 7722. |
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl2 , YCl2 , YCl3 are formed. a. What is the specialty of the oxidation number of Y, compared to that of X?b. Explain the reason for this, on the basis of the subshell based electronic configuration. |
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Answer» a. Element X has constant oxidation state. Y shows variable oxidation states. b. X20 = 1s2 , 2s2 , 2p6 , 3s2 , 3p6 , 4s2 X26 = 1s2, 2s2 , 2p6 , 3s2 , 3p6 , 4s2 , 3d6 Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part |
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| 7723. |
Select the suitable one from the following columns A, B, C. A B CFeF[Xe]4f1,5d1,6s2Cshigh atomic weightActinoidsp-blockf-blocktransition elementsSemi metals-blockLanthanoidscolourNon-metal |
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| 7724. |
Atomic number of iron is 26. It exhibits Fe2+ , Fe3+ oxidation state. Write the sub shell electronic configuration. Sub shell electronic configuration Fe Fe2+ Fe3+ |
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| 7725. |
24Cr – [Ar] 3d5 4s1 29Cu – [Ar] 3d10 4s1Why chromium and copper exhibits such electronic configuration ? |
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Answer» Half filled and completely filled subshells are most stable. Change in the electronic configuration of 24Cr &, 29Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability |
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| 7726. |
Manganese, a d-block element exhibits I different oxidation state. Why? a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.CompoundsOxidation state Subshell electronic configurationMnCl2MnO2KMnO4b. Write the oxidation number and subshell electronic configuration K, Cl and O. |
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Answer» Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions. a.
b.
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| 7727. |
The acid that is believed to be mainly responsible for the damage of Taj Mahal is(A) Sulfuric acid (B) Hydrofluoric acid (C) Phosphoric acid (D) Hydrochloric acid |
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Answer» Correct option is (A) Sulfuric acid CaCO3 + H2SO4 → CaSO4 + H2O + CO2 |
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| 7728. |
Polar cloud in stratosphere helps in the formation of which molecule (1) ClO (2) ClONO2 (3) CH4 (4) HOCl |
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Answer» Correct option is (1) ClO Polar stratospheric clouds (PSCs) surfaces act as catalysts that convert more benign forms of human made chlorine into active free radicals (for example CIO, chlorine monoxide). During the return of spring sunlight these radicals destroy many ozone molecules in a series of chain reactions. |
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| 7729. |
Which of the following is responsible for the secretion of pepsin?(a) Histamine (b) Cimetidine (c) Zantac(d) Serotonin |
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Answer» (a) Histamine Histamine, stimulates the secretion of pepsin and hydrochloric acid in the stomach. |
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| 7730. |
Why a drug should not be taken without consulting a doctor ? Give two reasons. |
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Answer» [Hint : (i) To avoid side effects caused by drug. |
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| 7731. |
The acid that is believed to be mainly responsible for the damage of Taj Mahal is(1) Phosphoric acid(2) Hydrochloric acid(3) Hydrofluoric acid(4) Sulphuric acid |
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Answer» Correct option is (4) Sulphuric acid The acid rain reacts with marble, CaCO3 of Taj Mahal (CaCO3 + H2SO4 → CaSO4 + H2O + CO2) causing damage to this wonderful monument that has attracted people from around the world. As a result, the monument is being slowly disfigured and the marble is getting discoloured and lustreless. |
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| 7732. |
An ideal gas (density = 0.121 gram/ml) at 257°C temperature have pressure 100 mm of Hg then find molar mass of gas. [Given R = 0.082 Latam / mole.k.] [Report your answer to nearest integer] |
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Answer» For ideal gas PM = dRT \((\frac{100}{760})\)M = 0.121 x 0.082 x 530 M = d x 330.296 M = 39.97 ≈ 40 |
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| 7733. |
Which of the following leads to the secretion of pepsin in stomach. (1) Histamine (2) Antihistamine (3) Cimetidine(4) Zintac |
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Answer» Correct option is 1) Histamine Excess of acidity (pepsin in stomach) is due to release of excess of histamine. Therefore modern synthetic drugs are antihistamines for the treatment of gastic ulcers by blocking the acid release action of histamine |
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| 7734. |
Why synthetic detergents are preferred over soaps for use in washing machines ? |
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Answer» [Hint : They work well even with hard water and not form any scum.] |
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| 7735. |
From the given examples - ciprofl oxacin, phenelzine, morphine, ranitidine -choose the drug used for :(i) treating allergic conditions (ii) to get relief from pain |
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Answer» [Hint : (i) Ranitidine (ii) Morphine] |
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| 7736. |
How is acidity cured with cimetidine ? |
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Answer» [Hint : Cimetidine prevents the interaction of histamines with the receptors present in stomach wall.] |
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| 7737. |
Identify the class of drug :(i) Phenelzine (Nardin)(ii) Aspirin(iii) Cimetidine |
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Answer» [Hint : (i) Antidepressant drug (ii) Analgesics and antipyretic (iii) Antihistamine] |
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| 7738. |
Classify the following drugs into analgesics and antibiotics.1. Ofloxacin 2. Morphine 3. Ampicillin 4. Chloramphenicol |
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Answer» 1. Ofloxacin – Antibiotic 2. Morphine – Analgesic 3. Ampicillin – Antibiotic 4. Chloramphenicol – Antibiotic |
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| 7739. |
Explain the bonding in coordination compounds in terms of Werner’s postulates. |
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Answer» Werner’s postulates explain the bonding in coordination compounds as follows: |
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| 7740. |
Write any three postulates of Werner’s theory of complexes. |
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Answer» 1. Co-ordination compounds metals show two types of Valencies primary and secondary 2. Primary valencies are ionisable 3. Secondary valencies are non-ionisable |
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| 7741. |
Define ligand. Write four postulates of Werner’s theory. |
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Answer» Ligands: The molecules or ions which are coordinated to the central atom or ion in the coordination compound are called ligands or donor groups. Postulates of Werner’s theory: a. Two types of valencies are shown by most metallic elements: 1. Primary valence or principal valence 2. Secondary valence. b. The tendency of every metal is to satisfy both primary and secondary valences. c. The number of secondary valence shown by each metal is fixed. d. The secondary valence is always directed towards fixed positions in space. |
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| 7742. |
What is metamerism? Write the structure and IUPAC name of methyl-n-propyl ether. What is the action of hot HI on it? |
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Answer» a. Metamerism (positional isomerism): Ethers having same molecular formula but different alkyl groups attached on either side of the oxygen atom are called metamers of each other. This phenomenon is called metamerism (positional isomerism). b. Methyl n-propyl ether: Structure: CH3 – O – CH2 – CH2 – CH3 IUPAC Name: 1-Methoxypropane c. Action of hot HI on methyl n-propyl ether: CH3 – O – CH2 – CH2 – CH3 (Methyl n-propyl ether) + 2HI (Conc.) \(\overset{Hot/373 K}\longrightarrow\) CH3I (Methyl iodide)+ CH3 – CH2 – CH2 – I (Propyl iodide)+ H2O |
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| 7743. |
Write reactions involved in preparation of potassium dichromate from chrome iron ore. |
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Answer» Reactions involved in the preparation of potassium dichromate from chrome iron ore (chromite ore): a. Conversion of chrome iron ore into sodium chromate (Roasting): 4FeO.Cr2O3 (Chrome iron ore) + 8Na2CO3 (Sodium carbonate) + 7O2 (Oxygen) → 2Fe2O3 (Ferric oxide) + 8Na2CrO4 (Sodium chromate) + 8CO2 (Carbon dioxide) b. Conversion of sodium chromate into sodium dichromate: 2Na2CrO4 (Sodium chromate) + H2SO4 (conc.) → Na2Cr2O7 (Sodium dichromate)+ Na2SO4 (Sodium sulphate) + H2O c. Conversion of sodium dichromate into potassium dichromate: Na2Cr2O7 (Sodium dichromate) + 2KCl (Potassium chloride) → K2Cr2O7 (Potassium dichromate) + 2NaCl (Sodium chloride) |
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| 7744. |
Which of the following is observed when Cl2 reacts with hot and concentrated NaOH?(a) NaCl , NaOCl (b) NaCl , NaClO2(c) NaCl , NaClO3 (d) NaOCl , NaClO3 |
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Answer» (c) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O (hot and conc.) |
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| 7745. |
I2 is more soluble in KI than in water. Why ? |
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Answer» [Hint : KI + I2 -> KI3] |
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| 7746. |
Electron gain enthalpy of fluorine (F) is less negative than that of chlorine (Cl). Why ? |
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Answer» [Hint : Due to small size of F atom and compact 2p orbitals there are strong interelectronic repulsions in the relatively smaller 2p orbitals of fluorine. So the incoming electron does experience more repulsion in F than in Cl.] |
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| 7747. |
What is cause of bleaching action of chlorine water ? Explain it with chemical equation. |
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Answer» [Hint : Formation of nascent oxygen.] |
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| 7748. |
Bleaching powder on standing forms mixture of :(a) CaO+ Cl2 (b) CaO+ CaCl2(c) HOCl + Cl2 (d) CaCl2 + Ca(ClO3)2 |
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Answer» (d) 6CaOCl2→ Ca(ClO3)2 + 5CaCl2 . It is autooxidation. |
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| 7749. |
Which is correct order of priority of functional groups as per IUPAC(1) RSO3H> RCOOR > RCOCI > RCONH2(2) RCOOR > RCONH2 > RSO3H > RCOCI(3) RCOOH > RCOCI > NH2 > C = O(4) RCOOR > RSO3H > RCOCI > RCONH2 |
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Answer» Correct option is (1) RSO3H> RCOOR > RCOCI > RCONH2 The correct order of priority of functional groups as per IUPAC is RSO3H> RCOOR > RCOCI > RCONH2 |
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| 7750. |
Which of the following oxide is amphoteric ?(a) SnO2 (b) CaO (c) SiO2 (d) CO2 |
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Answer» (a) SnO2 is an amphoteric oxide because it reacts with acids as well as with bases to form corresponding salts. SnO2 + 2H2 SO4 (conc) → Sn(SO4 ) 2 + 2H2O , SnO2 + 2NaOH → Na2 SnO3 + H2O |
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