SSLC:SCERT Chemistry
Unit 1: Nomenclature of Organic Compounds and Isomerism
1. A chain having 6 carbon atoms is given below. C C C C C C a) Complete the structure by adding hydrogen atoms to each carbon atom. b) Write the molecular formula of this compound. c) How many carbon atoms are there in the main chain of this compound? d) Write its IUPAC name.
2. Solution: a) The given carbon chain appears to be a branched alkane. Assuming the image intends for a continuous 4-carbon chain with two single-carbon branches (methyl groups) attached to the second carbon from either end (2,3-dimethylbutane configuration based on the diagram), the completed structure with hydrogen atoms would be:
3. b) To find the molecular formula, count the total carbon and hydrogen atoms from the structure above. Total Carbon atoms = 6 Total Hydrogen atoms = 3 (from first CH3) + 1 (from first CH) + 1 (from second CH) + 3 (from second CH3) + 3 (from top CH3) + 3 (from bottom CH3) = 14. Molecular Formula: C6H14 c) To find the main chain, select the longest continuous chain of carbon atoms. In the given structure (assuming 2,3-dimethylbutane), the longest chain has 4 carbon atoms. d) To write the IUPAC name: * Main chain: Butane (4 carbon atoms). * Branches: Two methyl groups (CH3). * Numbering: Number the main chain to give the branches the lowest possible numbers. If numbering from left, branches are at positions 2 and 3. If numbering from right, they are also at 2 and 3. * Name: Since there are two identical methyl branches, use 'di'. The positions are 2,3. IUPAC Name: 2,3-Dimethylbutane.
4. Write down the IUPAC names of the given compounds. a) CH3 CH2 CH2 CH CH2 CH3 CH3 b) CH3 CH2 CH2 CH = CH CH3 c) CH3 CH2 CH2 C ≡ C CH3 d) CH3 CH2 CH2 CH2 COOH e) CH3 CH2 CH2 CHO f) CH3 CH2 CH2 CO CH3 g) Cl Cl CH3 CH2 C CH3 h) CH3 CH2 O CH2 CH3 i) CH3 CH3 C = CH CH3 j) CH3 CH ≡ C CH CH3
5. Solution: a) CH3 CH2 CH2 CH CH2 CH3 CH3 * Longest chain: 6 carbons (Hexane). * Branch: Methyl group (CH3). * Position of branch: Number from right to give the branch the lowest number (position 3). IUPAC Name: 3-Methylhexane. b) CH3 CH2 CH2 CH = CH CH3 * Longest chain containing the double bond: 6 carbons (Hexene). * Position of double bond: Number from the right to give the double bond the lowest position number (between 2 and 3, so position 2). IUPAC Name: Hex-2-ene. c) CH3 CH2 CH2 C ≡ C CH3 * Longest chain containing the triple bond: 6 carbons (Hexyne). * Position of triple bond: Number from the right to give the triple bond the lowest position number (between 2 and 3, so position 2). IUPAC Name: Hex-2-yne. d) CH3 CH2 CH2 CH2 COOH * Functional group: Carboxyl group (–COOH). * Type: Carboxylic acid. * Main chain: 5 carbons including the carboxyl carbon (Pentanoic acid). IUPAC Name: Pentanoic acid. e) CH3 CH2 CH2 CHO * Functional group: Aldehyde group (–CHO). * Type: Aldehyde. * Main chain: 4 carbons including the aldehyde carbon (Butanal). IUPAC Name: Butanal. f) CH3 CH2 CH2 CO CH3 * Functional group: Keto group (C=O). * Type: Ketone. * Main chain: 5 carbons (Pentanone). * Position of keto group: Number from the right to give the keto group the lowest position number (position 2). IUPAC Name: Pentan-2-one. g) Cl Cl CH3 CH2 C CH3 * Functional group: Halo group (–Cl). * Type: Halo compound. * Main chain: 4 carbons (Butane). * Branches: Two chloro groups. * Position of branches: Number from the right to give the chloro groups the lowest position number (both at position 2). * If a carbon atom has two identical branches, their position numbers should be repeated. IUPAC Name: 2,2-Dichlorobutane. h) CH3 CH2 O CH2 CH3 * Functional group: Alkoxy group (–O–R). * Type: Ether. * Naming ethers: Alkoxyalkane, with the longer alkyl group as alkane and the shorter as alkoxy group. Here, both sides are ethyl groups. IUPAC Name: Ethoxyethane. i) CH3 CH3 C = CH CH3 * Longest chain containing the double bond: 4 carbons (Butene). * Branch: Methyl group (CH3). * Position of double bond: Number from the right to give the double bond the lowest position (position 2). * Position of branch: Numbering from the right places the methyl group at position 2. IUPAC Name: 2-Methylbut-2-ene. j) CH3 CH ≡ C CH CH3 * Longest chain containing the triple bond: 4 carbons (Butyne). * Branch: Methyl group (CH3). * Position of triple bond: Number from the left to give the triple bond the lowest position (position 1). * Position of branch: Numbering from the left places the methyl group at position 3. IUPAC Name: 3-Methylbut-1-yne.
6. Write the structural formulae of the compounds given below. a) 2,3,3–Trimethylhexane b) Ethoxybutane c) Butan-2-one d) Pent–1–yne e) Hexan–2–ol f) 3–Bromoheptane g) Pentanal
7. Solution: a) 2,3,3–Trimethylhexane. * Hexane indicates a main chain of 6 carbon atoms. * Trimethyl indicates three methyl branches. * Positions 2, 3, 3 indicate the locations of the methyl branches.
CH3 | CH3 - CH - C - CH2 - CH2 - CH3 | CH3 b) Ethoxybutane. * Butane indicates the longer alkyl group as the alkane part (4 carbons). * Ethoxy indicates the shorter alkyl group (ethyl) with an oxygen linkage (2 carbons with -O-). CH3 - CH2 - O - CH2 - CH2 - CH2 - CH3 c) Butan-2-one. * Butane indicates a 4-carbon chain. * -one indicates a keto group (C=O). * Position 2 indicates the location of the keto group. CH3 - CO - CH2 - CH3 d) Pent–1–yne. * Pentane indicates a 5-carbon chain. * -yne indicates a triple bond. * Position 1 indicates the location of the triple bond. CH ≡ C - CH2 - CH2 - CH3 e) Hexan–2–ol. * Hexane indicates a 6-carbon chain. * -ol indicates a hydroxyl group (–OH). * Position 2 indicates the location of the hydroxyl group. CH3 - CH - CH2 - CH2 - CH2 - CH3 | OH f) 3–Bromoheptane. * Heptane indicates a 7-carbon chain. * Bromo indicates a bromine atom. * Position 3 indicates the location of the bromine atom. CH3 - CH2 - CH - CH2 - CH2 - CH2 - CH3 | Br g) Pentanal. * Pentane indicates a 5-carbon chain. * -al indicates an aldehyde group (–CHO), which is always at the end of the chain and takes position 1. CH3 - CH2 - CH2 - CH2 - CHO8. The structural formulae and IUPAC names of certain compounds are given. Identify the wrong ones and correct them. i) CH3 CH2 CH2 CH CH3 CH2 CH3 2–Ethylpentane ii) CH3 CH2 CH2 CO CH3 Pentan–2–one iii) CH3 CH2 CH2 C ≡ C CH3 Hex–3–yne iv) 2,3–Dichloropentane Cl Cl Cl CH3 CH2 CH C CH3
9. Solution: i) Wrong. CH3 CH2 CH2 CH CH3 CH2 CH3 * The longest continuous carbon chain needs to be identified first. The given name "2-Ethylpentane" suggests a 5-carbon main chain with an ethyl group at position 2. However, if you count the carbons in a straight line from the end and include the ethyl branch, you find a longer chain (e.g., CH3-CH2-CH2-CH-CH2-CH3, where CH2-CH3 is part of the main chain, making it 6 carbons long). * The longest chain is 6 carbons long, which makes it a hexane derivative. * Numbering from the right gives the methyl branch at position 3. Correct Name: 3-Methylhexane.
10. ii) Correct. CH3 CH2 CH2 CO CH3 * Main chain: 5 carbons (pentane). * Functional group: Ketone (C=O). * Position of ketone: Numbering from the right gives the ketone at position 2. IUPAC Name: Pentan-2-one.
11. iii) Wrong. CH3 CH2 CH2 C ≡ C CH3 * Main chain: 6 carbons (hexyne). * Triple bond: Numbering from the left gives the triple bond at position 3. Numbering from the right also gives it at position 3. * The name Hex-3-yne is correct according to the numbering. However, the rule states that numbering should give the functional group the lowest possible number. In this case, 3 is the lowest. So the numbering is correct. Let's re-evaluate. * The original name provided Hex-3-yne is correct. The instruction was to identify wrong ones and correct them. It seems this one is correctly named. IUPAC Name: Hex-3-yne.
12. iv) Wrong. Cl Cl Cl CH3 CH2 CH C CH3 * Main chain: 5 carbons (pentane). * Halo groups: Three chloro groups. * Position of branches: Number from the right to give the branches the lowest position numbers. * Right to left: Chloro at 2, 2, 3. * Left to right: Chloro at 3, 4, 4. * Correct positions: 2,2,3. Correct Name: 2,2,3-Trichloropentane.
13. i) CH3 – CH2 – O CH2 CH3 ii) CH3 – CH2 – CH2 – CH2 OH a) What type of isomerism do these compounds exhibit? b) Write the structural formula of the metamer of compound (i).
14. Solution: a) Compound (i) is an ether (Ethoxyethane). Compound (ii) is an alcohol (Butan-1-ol). * Their molecular formulas need to be checked. * (i) Ethoxyethane: C4H10O (C2H5-O-C2H5) * (ii) Butan-1-ol: C4H10O (CH3-CH2-CH2-CH2-OH) * They have the same molecular formula (C4H10O) but different functional groups (ether vs. alcohol). Type of Isomerism: Functional Isomerism. b) A metamer has the same molecular formula but different alkyl groups attached to either side of the bivalent functional group. For compound (i) CH3 – CH2 – O – CH2 – CH3 (Ethoxyethane), the ether linkage (-O-) has two ethyl groups on either side. A metamer would have different alkyl groups around the ether linkage while maintaining the same molecular formula (C4H10O). Structural formula of a metamer of compound (i): CH3 – O – CH2 – CH2 – CH3 (Methoxypropane).
15. The structural formulae of two compounds are given. i) CH3 – CH2 – CH2 – CH2 CHO ii) CH3 – CH2 – CH2 – CO CH3 a) What is the IUPAC name of the first compound? b) These two compounds are isomers. Why? c) What type of isomerism do these compounds exhibit? d) Write the structural formula of the position isomer of the second compound.
16. Solution: a) IUPAC name of the first compound: * CH3 – CH2 – CH2 – CH2 CHO * Functional group: Aldehyde (–CHO). * Main chain: 5 carbons including the aldehyde carbon (Pentanal). IUPAC Name: Pentanal. b) These two compounds are isomers because they have the same molecular formula but different structural formulae, leading to differences in chemical and physical properties. * (i) Pentanal: C5H10O * (ii) CH3 – CH2 – CH2 – CO CH3 (Pentan-2-one): C5H10O Both compounds have the molecular formula C5H10O. c) Type of isomerism: * Compound (i) has an aldehyde functional group (–CHO). * Compound (ii) has a keto functional group (C=O). * They have the same molecular formula but different functional groups. Type of Isomerism: Functional Isomerism. d) Structural formula of the position isomer of the second compound: * The second compound is Pentan-2-one (CH3 – CH2 – CH2 – CO – CH3). Its functional group (keto group) is at position 2. * A position isomer has the same molecular formula and functional group but differs in the position of the functional group. * For Pentan-2-one, the keto group could be moved to position 3 on the 5-carbon chain to form a position isomer (Pentan-3-one). Structural Formula: CH3 – CH2 – CO – CH2 – CH3 (Pentan-3-one).
17. Examine the compounds given below and identify the isomeric pairs. What type of isomerism is shown by each pair? a) Methoxypropane b) 2,3–Dimethylbutane c) Propan–1–ol d) Ethoxyethane e) Propan–2–ol f) Hexane
18. Solution: Let's determine the molecular formula for each compound first: a) Methoxypropane: CH3–O–CH2–CH2–CH3. Molecular Formula: C4H10O. b) 2,3–Dimethylbutane: CH3–CH(CH3)–CH(CH3)–CH3. Molecular Formula: C6H14. c) Propan–1–ol: CH3–CH2–CH2–OH. Molecular Formula: C3H8O. d) Ethoxyethane: CH3–CH2–O–CH2–CH3. Molecular Formula: C4H10O. e) Propan–2–ol: CH3–CH(OH)–CH3. Molecular Formula: C3H8O. f) Hexane: CH3–CH2–CH2–CH2–CH2–CH3. Molecular Formula: C6H14.
19. Now, let's identify isomeric pairs and their types:
◦ Pair 1: a) Methoxypropane (C4H10O) and d) Ethoxyethane (C4H10O)
▪ Both are ethers with the same molecular formula.
▪ In Methoxypropane, the ether linkage has a methyl group on one side and a propyl group on the other.
▪ In Ethoxyethane, the ether linkage has two ethyl groups on either side.
▪ They differ in the alkyl groups attached to the ether linkage. Type of Isomerism: Metamerism.
◦ Pair 2: b) 2,3–Dimethylbutane (C6H14) and f) Hexane (C6H14)
▪ Both are alkanes with the same molecular formula.
▪ Hexane is a straight chain alkane.
▪ 2,3-Dimethylbutane is a branched chain alkane.
▪ They differ in the structure of their carbon chain. Type of Isomerism: Chain Isomerism.
◦ Pair 3: c) Propan–1–ol (C3H8O) and e) Propan–2–ol (C3H8O)
▪ Both are alcohols with the same molecular formula and functional group (hydroxyl group).
▪ Propan-1-ol has the –OH group at position 1.
▪ Propan-2-ol has the –OH group at position 2.
▪ They differ in the position of the functional group. Type of Isomerism: Position Isomerism.
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Unit 2: Chemical Reactions of Organic Compounds
1. a) In which of the following situation is methane converted to chloromethane? (i) Chlorine + sunlight (ii) Hydrochloric acid + sunlight (iii) Oxygen + temperature (iv) Heating in the absence of oxygen b) Write the name of such types of reactions.
2. Solution: a) Methane is converted to chloromethane when it reacts with chlorine in the presence of sunlight. Answer: (i) Chlorine + sunlight. b) In this reaction, a hydrogen atom in methane is replaced by a chlorine atom. Name of reaction: Substitution reaction.
3. a) How many hydrogen molecules are required to convert CH ≡ CH (ethyne) into C2H6 (ethane)? b) Write the chemical equation of the reaction. c) To which category does this chemical reaction belong?
4. Solution: a) Ethyne (CH≡CH) has a triple bond, and ethane (C2H6) is a saturated compound. To convert a triple bond to a single bond, two molecules of hydrogen (H2) are required to break both pi bonds. Answer: 2 hydrogen molecules. b) Chemical Equation: H – C ≡ C – H + 2H2 → CH3 – CH3. c) This is a reaction where an unsaturated compound combines with another molecule to form a saturated compound. Category: Addition reaction.
5. a) Complete the chemical equation. i) CH ≡ CH + HCl → A ii) n A → B b) Write the IUPAC names of the molecules A and B. c) To which category does each of these chemical reactions belong?
6. Solution: a) i) CH ≡ CH + HCl → CH2 = CHCl (Ethene reacts with HCl to form chloroethene or vinyl chloride). ii) n CH2 = CHCl → [CH2 – CH Cl]n (Polymerization of vinyl chloride). b)
◦ A: Chloroethene (or Vinyl chloride).
◦ B: Polychloroethene (or Polyvinyl chloride - PVC). c)
◦ Reaction i): Addition reaction (unsaturated compound combines with HCl).
◦ Reaction ii): Polymerisation (specifically, Addition polymerisation) (simple molecules join to form a large complex molecule).
7. a) Which of the given polymers is used to coat the inner surface of cookware? (Polythene, polyvinyl chloride, teflon) b) What is the monomer of this polymer?
8. Solution: a) Teflon is used to coat the inner surface of non-stick cookware. b) The monomer of Teflon (Polytetrafluoroethene) is tetrafluoroethene (CF2=CF2).
9. a) Which among the following is a condensation polymer? (Polyvinyl chloride, nylon 66, teflon) b) What are the monomers of nylon 66?
10. Solution: a) Among the given options, nylon 66 is a condensation polymer. Polyvinyl chloride and teflon are addition polymers. b) The monomers of nylon 66 are adipic acid and hexamethylenediamine.
11. C12H22O11 + H2O A C6H12O6 B + .......... C6H12O6 C 2C2H5OH D + .......... a) Identify A, B, C and D in the given chemical reactions. b) What is wash? c) How is rectified spirit obtained from wash? d) What is the purpose of denaturing rectified spirit?
12. Solution: a) This reaction sequence describes the industrial preparation of ethanol from molasses.
◦ C12H22O11 (Sugar solution, e.g., from molasses) + H2O --(A)--> C6H12O6 (Glucose) + C6H12O6 (Fructose)
▪ A: Invertase (enzyme from yeast that converts sugar to glucose and fructose).
▪ B: Fructose (the other product along with Glucose from the hydrolysis of sugar).
◦ C6H12O6 (Glucose/Fructose) --(C)--> 2C2H5OH (Ethanol) + (D)
▪ C: Zymase (enzyme from yeast that converts glucose/fructose to ethanol).
▪ D: 2CO2 (carbon dioxide, the other product of fermentation). b) Wash is the 8-10% ethanol solution obtained after the fermentation of molasses. c) Rectified spirit (95.6% ethanol) is obtained by subjecting the 'wash' to fractional distillation. d) Denaturing rectified spirit (ethanol) is done by adding toxic substances like methanol, pyridine, or rubber distillate to prevent its misuse as a beverage.
13. a) How is ethanoic acid prepared industrially? b) 5 - 8% ethanoic acid is called ...................
14. Solution: a) Ethanoic acid can be prepared industrially by treating methanol with carbon monoxide in the presence of a catalyst. b) 5 - 8% ethanoic acid is called vinegar.
15. CH3 – COOH + CH3 – OH conc. H2SO4 CH3 – COO – CH3 + H2O Analyse the equation and answer the following questions. a) What is the name of this chemical reaction? b) What is the name of the ester formed? c) Write any two uses of esters.
16. Solution: a) This reaction, where an alcohol (methanol) reacts with a carboxylic acid (ethanoic acid) to form an ester, is called esterification. b) The ester formed is derived from ethanoic acid and methanol. The alcohol part (methyl) comes first, followed by the acid part (ethanoate). Name of the ester formed: Methyl ethanoate. c) Two uses of esters are: * Used to make artificial perfumes due to their fragrance of flowers and fruits. * Used to make artificial juices due to their fragrance of flowers and fruits.
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Unit 3: Periodic Table and Electron Configuration
1. The element X having 3 shells belongs to group 17. a) Write the subshell electron configuration of this element. b) To which block does this element belong? c) What is its period number? d) Write the molecular formula of the compound formed when X reacts with an atom of element Y which belongs to the third period and has one electron in its p subshell.
2. Solution: a) Element X has 3 shells, meaning its principal quantum number (n) for the outermost shell is 3. It belongs to Group 17, which is a p-block group. For p-block elements, the group number is 10 + (number of electrons in outermost s and p subshells). Group 17 means 10 + (s + p) = 17, so s + p = 7. Since it's in the 3rd period, the outermost shell is n=3, so the configuration is 3s² 3p⁵. The full subshell electron configuration would be 1s² 2s² 2p⁶ 3s² 3p⁵. Subshell electron configuration of X: 1s² 2s² 2p⁶ 3s² 3p⁵. b) The last electron is added to the p subshell. Block: p-block. c) The highest shell number in its subshell electron configuration is 3. Period number: 3. d) Element X is Chlorine (Cl) from its electron configuration (1s² 2s² 2p⁶ 3s² 3p⁵). Element Y belongs to the third period (n=3) and has one electron in its p subshell (3p¹). This means its configuration is 3s² 3p¹. The full configuration is 1s² 2s² 2p⁶ 3s² 3p¹. This element has 3 electrons in its outermost shell (2 from 3s and 1 from 3p), which means it belongs to Group 13 (10 + 3 = 13). This element is Aluminium (Al). Chlorine (X) typically forms a -1 ion (gains 1 electron to complete its octet). Aluminium (Y) typically forms a +3 ion (loses 3 electrons). To balance the charges (Al³⁺ and Cl⁻), 1 Aluminium atom will combine with 3 Chlorine atoms. Molecular formula of the compound: AlCl3 (Aluminium Chloride).
3. A few subshells are given. 3p, 4d, 3f, 2d, 2p a) Among these, which subshells are not possible? b) Explain the reason.
4. Solution: a) The values for the azimuthal quantum number (l) range from 0 to (n-1).
◦ For 3p: n=3, l=1. Possible.
◦ For 4d: n=4, l=2. Possible.
◦ For 3f: n=3, l=3. Not possible, because for n=3, l can only be 0, 1, 2 (i.e., s, p, d).
◦ For 2d: n=2, l=2. Not possible, because for n=2, l can only be 0, 1 (i.e., s, p).
◦ For 2p: n=2, l=1. Possible. Subshells not possible: 3f and 2d. b) Reason: The azimuthal quantum number (l) for a given principal quantum number (n) can only take values from 0 up to (n-1).
◦ For 3f, n=3. The maximum possible value for l is (3-1) = 2. But 'f' subshell corresponds to l=3. Since 3 > 2, 3f is not possible.
◦ For 2d, n=2. The maximum possible value for l is (2-1) = 1. But 'd' subshell corresponds to l=2. Since 2 > 1, 2d is not possible.
5. The position of two elements A and B in the periodic table are given. (Symbols are not real) A - 4th period and 2nd group B - 2nd period and 16th group a) Write the subshell electron configuration of A and B. b) Write the values of n and l of electrons in the outermost subshell of A. c) How many orbitals are there in the subshell of B having the outermost electron? Find it on the basis of magnetic quantum number ‘m’. d) Write the molecular formula of the compound formed by the combination of A and B. e) What type of chemical bond is present in this compound?
6. Solution: a) Element A: 4th period, 2nd group. * 4th period means highest n value is 4. * 2nd group means it's an s-block element with 2 electrons in its outermost s subshell (ns²). So, 4s². * Subshell electron configuration of A: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². (This is Calcium, Ca). Element B: 2nd period, 16th group. * 2nd period means highest n value is 2. * 16th group means it's a p-block element. Group number = 10 + (s + p outermost electrons). So, 16 = 10 + (2s + 2p), which means 2s² 2p⁴. * Subshell electron configuration of B: 1s² 2s² 2p⁴. (This is Oxygen, O). b) For element A (1s² 2s² 2p⁶ 3s² 3p⁶ 4s²), the outermost subshell is 4s. * n = 4 (principal quantum number). * l = 0 (azimuthal quantum number for s subshell). c) For element B (1s² 2s² 2p⁴), the outermost electrons are in the 2p subshell. * For a p subshell, l = 1. * The number of orbitals for a particular value of l is (2l + 1). * So, for l = 1, number of orbitals = (2 × 1 + 1) = 3. Number of orbitals: 3. d) Element A (Calcium, Ca) is a Group 2 metal and forms a +2 ion (Ca²⁺). Element B (Oxygen, O) is a Group 16 non-metal and forms a -2 ion (O²⁻). To balance charges, one Ca²⁺ combines with one O²⁻. Molecular formula: AO (e.g., CaO). e) A metal (A) reacting with a non-metal (B) usually forms an ionic bond through the transfer of electrons [Information not directly in sources, but implied by discussion of oxidation states and ion formation in 140, 144]. Type of chemical bond: Ionic bond.
7. The subshell electron configuration of a few elements are written on the basis of noble gas. (i) [Ne] 3s2 3p6 (ii) [He] 2s1 (iii) [Ar] 3d2 4s2 (iv) [Kr] 5s2 a) Write the complete subshell electron configuration. b) Find the symbols of these elements with the help of periodic table.
8. Solution: a) Complete subshell electron configurations:
◦ (i) [Ne] 3s² 3p⁶: Neon ([Ne]) has the configuration 1s² 2s² 2p⁶. So, the complete configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
◦ (ii) [He] 2s¹: Helium ([He]) has the configuration 1s². So, the complete configuration is 1s² 2s¹.
◦ (iii) [Ar] 3d² 4s²: Argon ([Ar]) has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶. So, the complete configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s².
◦ (iv) [Kr] 5s²: Krypton ([Kr]) has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ [Not explicitly in source, but implied by the periodic table context for higher atomic numbers. Standard knowledge beyond source.]. So, the complete configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s².
9. b) Symbols of these elements:
◦ (i) 1s² 2s² 2p⁶ 3s² 3p⁶: Total electrons = 18. This is Argon (Ar).
◦ (ii) 1s² 2s¹: Total electrons = 3. This is Lithium (Li).
◦ (iii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²: Total electrons = 22. This is Titanium (Ti).
◦ (iv) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s²: Total electrons = 38. This is Strontium (Sr).
10. The last electron of an atom is added to the 3d subshell. There are 7 electrons in this subshell. Answer the following questions regarding this atom. a) What is its atomic number? b) Write its complete electron configuration. c) To which block does it belong? d) Find its period number. e) What is its group number?
11. Solution: a) If the last electron is in 3d subshell with 7 electrons, it means 4s subshell is filled before 3d (as per energy order, 4s < 3d) and all subshells before 4s are also filled. Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷. Total number of electrons = 2 + 2 + 6 + 2 + 6 + 2 + 7 = 27. Atomic Number: 27. (This element is Cobalt, Co). b) Complete electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ 4s². c) The last electron is added to the d subshell. Block: d-block. d) The highest shell number (n) in the configuration is 4 (from 4s²). Period Number: 4. e) For d-block elements, the group number is the sum of electrons in the outermost s subshell and the preceding d subshell. In this case, 3d⁷ 4s², so 7 (from 3d) + 2 (from 4s) = 9. Group Number: 9.
12. Find the oxidation state of s block elements in the following compounds. a) Na2O b) KBr c) CaO d) MgCl2
13. Solution: s-block elements typically exhibit +1 (Group 1) or +2 (Group 2) oxidation states. a) Na2O: Oxygen usually has an oxidation state of -2. Since there are two Na atoms, each Na must have an oxidation state of +1 (2 × (+1) + (-2) = 0). Sodium (Na) is a Group 1 element. Oxidation state of Na: +1. b) KBr: Bromine (Br) is a halogen and usually has an oxidation state of -1 in compounds with metals. Potassium (K) is a Group 1 element. Oxidation state of K: +1. c) CaO: Oxygen usually has an oxidation state of -2. Calcium (Ca) is a Group 2 element. Oxidation state of Ca: +2. d) MgCl2: Chlorine (Cl) is a halogen and usually has an oxidation state of -1. Since there are two Cl atoms, they contribute 2 × (-1) = -2. Magnesium (Mg) is a Group 2 element. Oxidation state of Mg: +2.
14. Which subshell is represented by each pair of the quantum number values given below? a) n = 1, l = 0 b) n = 2, l = 1
15. Solution: a) n = 1, l = 0: * n = 1 represents the K shell. * l = 0 represents the s subshell. Subshell: 1s. b) n = 2, l = 1: * n = 2 represents the L shell. * l = 1 represents the p subshell. Subshell: 2p.
16. A few subshells are given. Find the values of n and l. a) 2s b) 4p c) 3d d) 5f
17. Solution: a) 2s: * n = 2 (the number before 's'). * l = 0 (for 's' subshell). b) 4p: * n = 4 (the number before 'p'). * l = 1 (for 'p' subshell). c) 3d: * n = 3 (the number before 'd'). * l = 2 (for 'd' subshell). d) 5f: * n = 5 (the number before 'f'). * l = 3 (for 'f' subshell).
18. The subshell electron configuration of certain elements are given. Write the short form of each using the symbol of corresponding inert gas. a) 1s2 2s2 2p4 b) 1s2 2s2 2p6 3s2 3p5 c) 1s2 2s2 2p6 3s2 3p6 4s1 d) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
19. Solution: To write the short form, identify the noble gas (inert gas) that precedes the element and write its symbol in square brackets, followed by the remaining electron configuration. a) 1s² 2s² 2p⁴: The preceding noble gas is Helium (He), with configuration 1s². Short Form: [He] 2s² 2p⁴. b) 1s² 2s² 2p⁶ 3s² 3p⁵: The preceding noble gas is Neon (Ne), with configuration 1s² 2s² 2p⁶. Short Form: [Ne] 3s² 3p⁵. c) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹: The preceding noble gas is Argon (Ar), with configuration 1s² 2s² 2p⁶ 3s² 3p⁶. Short Form: [Ar] 4s¹. d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²: The preceding noble gas is Argon (Ar), with configuration 1s² 2s² 2p⁶ 3s² 3p⁶. Short Form: [Ar] 3d⁵ 4s².
20. Iron (Fe) takes part in chemical reactions and becomes Fe3+ ion. (Atomic number of Fe = 26). a) Write the electron configuration of this ion. b) Write the chemical formula of the compound formed when this ion combines with sulphate ion (SO4²⁻). c) Which is the other oxidation state of this element? Write the electron configuration of the ion thus formed. d) Iron shows variable oxidation states. Why?
21. Solution: Atomic number of Fe = 26. Original electron configuration of Fe (ground state): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s². a) Electron configuration of Fe³⁺ ion: When forming ions, electrons are removed first from the outermost shell (highest 'n' value), which is 4s, and then from the 3d subshell if more electrons are to be lost. To form Fe³⁺, 3 electrons are removed: * 2 electrons from 4s: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶. * 1 electron from 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵. Electron configuration of Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵. b) Chemical formula with sulphate ion (SO4²⁻): Fe³⁺ has a +3 charge. SO4²⁻ has a -2 charge. To balance the charges, we need 2 Fe³⁺ ions (total charge +6) and 3 SO4²⁻ ions (total charge -6). Chemical Formula: Fe2(SO4)3. c) Other common oxidation state of Iron: Iron can also form Fe²⁺ ions. To form Fe²⁺, 2 electrons are removed from the outermost 4s subshell. Electron configuration of Fe²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶. d) Reason for variable oxidation states: In transition elements (d-block elements like Iron), there is only a slight energy difference between the outermost s subshell (e.g., 4s) and the penultimate d subshell (e.g., 3d). As a result, under favourable conditions, electrons from both the outermost s subshell and the preceding d subshell can participate in chemical reactions (i.e., be lost to form ions). This allows them to exhibit multiple oxidation states.
22. A portion of the periodic table is given. Answer the following questions (Symbols are not real). A B D C E F G H I J K 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 a) Which element has the lowest ionisation enthalpy? b) Identify the alkaline earth metal. c) Which are the d block elements? d) Which element has the completely filled d subshell? e) Identify the element having the electron configuration 3d3 4s2? f) Identify the noble gas. g) Which element has only 3 electrons in the outermost p subshell?
23. Solution: First, let's identify the blocks and general trends from the periodic table sketch.
◦ Groups 1 & 2: s-block (A, B).
◦ Groups 3-12: d-block (C, E, F, G, H, I, J, K).
◦ Groups 13-18: p-block (D).
◦ A, B are in Period 4.
◦ C, E, F, G are in Period 4.
◦ D is in Period 4.
◦ H, I, J, K are in Period 5.
24. a) Lowest ionisation enthalpy: Ionisation enthalpy decreases down a group and increases across a period from left to right. The elements with lowest ionisation enthalpy are generally found at the bottom-left of the periodic table. Among the given options, H (Group 1, Period 5) would have the lowest ionization enthalpy as it is furthest down and left. b) Alkaline earth metal: Alkaline earth metals belong to Group 2. In the given table, element B is in Group 2. c) d block elements: These are elements in Groups 3 to 12. So, C, E, F, G, H, I, J, K are the d-block elements. d) Element with completely filled d subshell: In the 4th period d-block elements, the d subshell is completely filled (3d¹⁰) for Zinc (Zn), which is in Group 12. In the given diagram, G is in Group 12. e) Element with electron configuration 3d³ 4s²: This configuration belongs to Vanadium (V). For d-block elements, the group number is the sum of electrons in 3d and 4s. So, 3+2=5. This element is in Group 5. In the given diagram, E is in Group 5. f) Noble gas: Noble gases belong to Group 18. In the given diagram, D is in Group 18. g) Element with only 3 electrons in the outermost p subshell: Outermost p subshell configuration is p³. For p-block elements, Group number = 10 + (s + p). If p=3 and s=2 (as s subshell is filled), then Group = 10 + 2 + 3 = 15. So, the element is in Group 15. In the given diagram, the element directly below B and to the left of D (i.e. if this were a complete table) would be in group 15. However, there is no element explicitly labelled in Group 15 in Period 4 in this specific diagram. If we assume 'D' is meant to represent the p-block column, we are looking for the 15th group element. Let's re-examine the diagram for a potential element in Group 15, Period 4. The arrangement shows A (Group 1), B (Group 2), then a gap for d-block, then D (Group 18). There is no specific label for Group 15. If the question implies an element shown in the diagram, it's ambiguous. If it refers to any element in the periodic table, the element in Group 15 is Phosphorus (P) for 3rd period, Arsenic (As) for 4th period. The question asks for "an element" not necessarily one labelled. Revisiting the image: The diagram shows the groups 1, 2, 3-12 (d-block), 13, 14, 15, 16, 17, 18. 'D' is under 18. The element in the same period as A, B, C, E, F, G but in Group 15 would be the one that ends with ns² np³. If it refers to any element, it's a general question. If it refers to the diagram, no explicit symbol is given. Since the question asks "Which element has...", it implies one from the table. However, without more labels for the p-block, it's not possible to definitively pick one from the labels provided (A, B, C, D, E, F, G, H, I, J, K). Let's assume it means a conceptual element in that location. For an element with 3 electrons in its outermost p subshell, it would have the configuration ...np³. If the outermost shell is 'n', then it would be ns² np³. This would put it in Group 15 (2+3+10 = 15). There is no labelled element in Group 15 in the provided diagram. If the image intends 'D' to be part of group 18, and there are groups 13-17 to its left, then the element would be in group 15, but it's not labeled. I will state the group it belongs to.
