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Maharashtra Board Class 11 Hindi परिशिष मुद्रित शोधन चिह्नदर्शक तालिका

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष मुद्रित शोधन चिह्नदर्शक तालिका Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष मुद्रित शोधन चिह्नदर्शक तालिका

मुद्रण सही ढंग से न हो तो अशुद्धियाँ रह जाती हैं। इससे मुद्रित सामग्री की रोचकता तथा सहजता कम हो जाती है। कभी–कभी किसी शब्द के अशुद्ध रहने से अर्थ बदल जाता है या किसी शब्द के रह जाने से अर्थ का अनर्थ हो जाता है। इस दृष्टि से मुद्रण प्रक्रिया में मुद्रित शोधन का अत्यधिक महत्त्व है।

जिस प्रकार मन की सुंदरता न हो तो तन की सुंदरता अर्थहीन हो जाती है। उसी प्रकार पुस्तक बाहर से भले ही कितनी ही आकर्षक हो; भाषा की अशुद्धता के कारण वह प्रभावहीन हो जाती है।

मुद्रित शोधन के लिए आवश्यक योग्यताएँ :

मुद्रित शोधन का कार्य अत्यंत दायित्वपूर्ण ढंग से निभाया जाने वाला कार्य है। अत: इस कार्य के लिए मुद्रित शोधक में कतिपय योग्यताओं का होना आवश्यक है। जैसे –

  • मुद्रित शोधक को संबंधित भाषा एवं व्याकरण की समग्र और भली–भाँति जानकारी होनी चाहिए।
  • उसे प्रिंटिंग मशीन पर होने वाले कार्य का परिचय होना चाहिए।
  • उसे टाइप के प्रकारों, संकेत चिह्नों और अक्षर विन्यास की पूर्ण जानकारी होनी चाहिए।
  • मुद्रित शोधक को पांडुलिपि में स्वयं कोई परिवर्तन नहीं करना चाहिए। यदि कहीं उसे अशुद्धियाँ लगें या वाक्य सही/शुद्ध न लगे तो इसकी ओर लेखक का ध्यान आकृष्ट करना चाहिए।

मुद्रित शोधन चिह्नदर्शक तालिका : –
Maharashtra Board Class 11 Hindi परिशिष मुद्रित शोधन चिह्नदर्शक तालिका 1

Maharashtra Board Class 11 Hindi परिशिष हिंदी साहित्यकारों के मूल नाम और उनके विशेष नाम

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष हिंदी साहित्यकारों के मूल नाम और उनके विशेष नाम Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष हिंदी साहित्यकारों के मूल नाम और उनके विशेष नाम

  • अब्दुल हसन – अमीर खुसरो
  • मलिक मुहम्मद – जायसी
  • अब्दुर्रहीम खानखाना – रहीम
  • सय्यद इब्राहिम – रसखान
  • चंद्रधर शर्मा – ‘गुलेरी’
  • पांडेय बेचन शर्मा – ‘उग्र’
  • राजेंद्रबाला घोष – बंग महिला
  • बदरीनारायण चौधरी – प्रेमधन
  • गयाप्रसाद शुक्ल – ‘स्नेही’
  • अयोध्यासिंह उपाध्याय – ‘हरिऔध’
  • मोहनलाल महतो – वियोगी
  • धनपतराय – ‘प्रेमचंद’
  • रामधारी सिंह – ‘दिनकर’
  • शिवमंगल सिंह – ‘सुमन’
  • रामेश्वर शुक्ल – ‘अंचल’
  • बालकृष्ण शर्मा – ‘नवीन’
  • कन्हैयालाल मिश्र – ‘प्रभाकर’
  • फणीश्वरनाथ – ‘रेणु’
  • वैद्यनाथ मिश्र – नागार्जुन
  • सूर्यकांत त्रिपाठी – ‘निराला’
  • सच्चिदानंद हीरानंद वात्सायन – अज्ञेय
  • वासुदेव सिंह – त्रिलोचन
  • गोपाल दास सक्सेना – ‘नीरज’
  • महेंद्रकुमारी – मन्नू भंडारी
  • श्रीराम वर्मा – अमरकांत
  • उपेंद्रनाथ – ‘अश्क’
  • सुदामा पांडेय – धूमिल

Maharashtra Board Class 11 Hindi परिशिष ज्ञानपीठ पुरस्कार प्राप्त हिंदी साहित्यकार

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष ज्ञानपीठ पुरस्कार प्राप्त हिंदी साहित्यकार Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष ज्ञानपीठ पुरस्कार प्राप्त हिंदी साहित्यकार

ज्ञानपीठ पुरस्कार प्राप्त हिंदी साहित्यकार

साहित्यकार साहित्यिक कृति वर्ष
सुमित्रानंदन पंत चिदंबरा १९६८
रामधारी सिंह ‘दिनकर’ उर्वशी १९७२
‘अज्ञेय’ कितनी नावों में कितनी बार १९७८
महादेवी वर्मा यामा १९८२
नरेश मेहता समग्र साहित्य १९९२
निर्मल वर्मा समग्र साहित्य १९९९
कुँवर नारायण समग्र साहित्य २००५
अमरकांत समग्र साहित्य २००९
श्रीलाल शुक्ल राग दरबारी २००९
केदारनाथ सिंह अकाल में सारस २०१३
कृष्णा सोबती जिंदगीनामा २०१७

Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष पारिभाषिक शब्दावली Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष पारिभाषिक शब्दावली

1. बैंक तथा वाणिज्य से संबंधित शब्द

  • Account = लेखा
  • Accountant = लेखापाल
  • Act = अधिनियम Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली
  • Affidavit = शपथपत्र
  • Agreement = अनुबंध/करार
  • Annexure = परिशिष्ट
  • Audit = लेखा परीक्षण
  • Average = औसत
  • Session = सत्र
  • Advocate General = महाधिवक्ता
  • Foreign Exchange = विदेशी विनिमय
  • Fund Sinking = निक्षेप निधि
  • Finance Commissioner = वित्त आयुक्त
  • Deduction = कटौती
  • Dividend = लाभांश
  • Domicile Certificate = अधिवास प्रमाणपत्र
  • Draft = मसौदा/प्रारूप
  • Gazette = राजपत्र
  • Investment = निवेश
  • Management = प्रबंधन
  • Revenue = राजस्व
  • Clearing = समाशोधन
  • Attestation = साक्ष्यांकन
  • Cheque = धनादेश (चैक)
  • Advance = अग्रिम
  • Capital = पूँजी Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली
  • Cashier = रोकड़िया/कोषाध्यक्ष
  • Amount = धनराशि, रकम
  • Custom Duty = सीमा शुल्क
  • Credit Amount = जमा रक्कम
  • Finance Bill = वित्त विधेयक
  • Finance Statement = वित्तीय विवरण
  • Pension = निवृत्ति वेतन
  • Service Charges = सेवा भार
  • Corporation-Tax = नगर निगम कर
  • Trade Mark = व्यापार चिह्न
  • Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली

2. विधि से संबंधित शब्द

  • Bailable Offence = जमानती अपराध
  • Defendent = ufdact
  • Accused = अभियुक्त
  • Bench = न्यायपीठ
  • Show Cause = कारण बताओ
  • Custody (Police) = पुलिस हिरासत
  • Formal Investigation = औपचारिक जाँच
  • Validity = वैधता
  • Advocate General = Halfeta chall
  • Judicial Power = न्यायालयीन अधिकार
  • Ordinance = अध्यादेश

3. प्रशासनिक

  • Chancellor = कुलाधिपति
  • Deputation = प्रतिनियुक्ति
  • Director = निदेशक
  • Surveyor = सर्वेक्षक
  • Supervisor = पर्यवेक्षक
  • Governor = राज्यपाल
  • Secretary = सचिव
  • Eligibility = अर्हता
  • Memorandum = ज्ञापन
  • Notification = अधिसूचना
  • Registrar = कुलसचिव
  • Administration = प्रशासन
  • Commission = आयोग
  • Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली

4. वैज्ञानिक एवं तकनीकी शब्दावली

  • Mechanics = यांत्रिक
  • Gravitation = गुरुत्वाकर्षण
  • Orbit = कक्षा
  • Satellite = उपग्रह
  • Nerve = तंत्रिका
  • Nutrition = पोषण
  • Radiation = विकिरण
  • Tissue = ऊतक
  • Fertility = उर्वरता
  • Genetics = अनुवांशिकी

5. कंप्यूटर (संगणक) विषयक

  • Internet = अंतरजाल
  • Control Section = नियंत्रण अनुभाग
  • Hard Copy = मुद्रित प्रति
  • Storage = भंडार
  • Data = आँकड़ा
  • Software = प्रक्रिया सामग्री
  • Output = निर्गम
  • Screen = प्रपट्ट Maharashtra Board Class 11 Hindi परिशिष पारिभाषिक शब्दावली
  • Network = संजाल
  • Command = समादेश

Maharashtra Board Class 11 Hindi परिशिष रेडियो जॉकी और रेडियो संहिता

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष रेडियो जॉकी और रेडियो संहिता Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष रेडियो जॉकी और रेडियो संहिता

रेडियो जॉकी

Maharashtra Board Class 12 Hindi परिशिष रेडियो जॉकी और रेडियो संहिता 1

Maharashtra Board Class 11 Hindi परिशिष रेडियो जॉकी और रेडियो संहिता

रेडियो संहिता

रेडियो श्राव्य माध्यम है। इसलिए श्राव्य माध्यम के अनुकूल संहिता होती है। इसमें शब्दों के साथ ध्वनि संकेत, ठहराव, मौन, अंतराल आदि के संकेत भी होने चाहिए। गीत– संगीत के बीच में चलनेवाली आर.जे. की बातचीत कम शब्दों में रोचक, चटपटी और मिठास भरी होनी चाहिए।

भाषा प्रवाहमयी हो। शब्द सरल हों। संहिता लयात्मकता के साथ कार्यक्रम को आगे बढ़ाने में सहायक होनी चाहिए। रेडियो संहिता के तीन हिस्से होते हैं। आरंभ, मध्य और अंत। आरंभ जितना आकर्षक, उतना ही अंत भी आकर्षक होना चाहिए। मध्य में विषयवस्तु कार्यक्रम की लंबाई पर निर्भर है।

हिंदी में रेडियो चैनल के लिए जो संहिता होती है, वह बहुत ही सधी हुई होती है। रेडियो की लोकप्रियता का सबसे बड़ा कारण है – कार्यक्रमों की प्रस्तुति, संयोजन और भाषा का नयापन। संहिता की भाषा गतिशील और अनौपचारिक होनी चाहिए।

Maharashtra Board Class 11 Hindi परिशिष रेडियो जॉकी और रेडियो संहिता

कुछ चैनलों पर जिस हिंदी भाषा का प्रयोग किया जाता है वह ‘प्रोमो’ हिंदी है। ‘प्रोमो’। अर्थात ‘पोस्ट मॉडर्न’ – उत्तर आधुनिक हिंदी। इस हिंदी भाषा में चुलबुलापन, मसखरापन, मस्ती और लय होती है। इसकी अपनी एक अलग पहचान है।

Maharashtra Board Class 11 Hindi परिशिष भावार्थ : भक्ति महिमा और बाल लीला

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष भावार्थ : भक्ति महिमा और बाल लीला Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष भावार्थ : भक्ति महिमा और बाल लीला

भावार्थ : पाठ्यपुस्तक पृष्ठ क्रमांक २० : पाठ – भक्ति महिमा – संत दादू दयाल

जो माया–मोह का रस पीते रहे, उनका मक्खन–सा हृदय सूखकर पत्थर हो गया किंतु जिन्होंने भक्ति रस का पान किया, उनका पत्थर हृदय गलकर मक्खन हो गया। उनका हृदय प्रेम से भर गया।

Maharashtra Board Class 12 Hindi परिशिष भावार्थ : भक्ति महिमा और बाल लीला

अहंकारी व्यक्ति से प्रभु दूर रहता है। जो व्यक्ति प्रभुमय हो जाता है, फिर उसमें अहंकार नहीं होता। मनुष्य का हृदय एक ऐसा सँकरा महल है, जिसमें प्रभु और अहंकार दोनों साथ–साथ नहीं रह सकते। अहंकार का त्याग करना अनिवार्य है।

दादू मगन होकर प्रभु का कीर्तन कर रहे हैं। उनकी वाणी ऐसे मुखरित हो रही है जैसे ताल बज रहा हो। यह मन प्रेमोन्माद में नाच रहा है। दादू के सम्मुख दीन–दुखियों पर विशेष कृपा करने वाला प्रभु खड़ा है।

जिन लोगों ने भक्ति के सहारे भवसागर पार कर लिया, उन सभी की एक ही बात है कि भक्ति का संबल लेकर ही सागर को पार किया जा सकता है। सभी संतजन भी यही बात कहते हैं। अन्य मार्गदर्शक, जीवन के उद्धार के लिए जो दूसरे अनेक मार्ग बताते हैं, वे भ्रम में डालने वाले हैं। प्रभु स्मरण के सिवा अन्य सभी मार्ग दुर्गम हैं।

प्रेम की पाती (पत्री) कोई विरला ही पढ़ पाता है। वही पढ़ पाता है, जिसका हृदय प्रेम से भरा हुआ है। यदि हृदय में जीवन और जगत के लिए प्रेम भाव नहीं है तो वेद–पुराण की पुस्तकें पढ़ने से क्या लाभ ?

कितने ही लोगों ने वेद–पुराणों का गहन अध्ययन किया और उसकी व्याख्या करने में लिख–लिखकर कागज काले कर दिए लेकिन उन्हें जीवन का सच्चा मार्ग नहीं मिला। वे भटकते ही रहे, जिसने प्रिय प्रभु का एक अक्षर पढ़ लिया, वह सुजान–पंडित हो गया।

Maharashtra Board Class 12 Hindi परिशिष भावार्थ : भक्ति महिमा और बाल लीला

मेरा अहंकार – “मैं’ ही मेरा शत्रु निकला, जिसने मुझे मार डाला, जिसने मुझे पराजित कर दिया। मेरा अहंकार ही मुझे मारने वाला निकला, दूसरा कोई और नहीं।

अब मैं स्वयं इस ‘मैं’ (अहंकार) को मारने जा रहा हूँ। इसके मरते ही मैं मरजीवा हो जाऊँगा। मरा हुआ था फिर से जी उठूगा। एक विजेता बन जाऊँगा।

हे सृष्टिकर्ता ! जिनकी रक्षा तू करता है, वे संसार सागर से पार हो जाते हैं।

और जिनका तू हाथ छोड़ देता है, वे भवसागर में डूब जाते हैं। तेरी कृपा सज्जनों पर ही होती है।

रे नासमझ ! तू क्यों किसी को दुख देता है। प्रभु तो सभी के भीतर हैं। क्यों तू अपने स्वामी का अपमान करता है? सब की आत्मा एक है। आत्मा ही परमात्मा है। परमात्मा के अलावा वहाँ दूसरा कोई नहीं।

इस संसार में केवल ऐसे दो रत्न हैं, जो अनमोल हैं। एक है सबका स्वामी–प्रभु। दूसरा स्वामी का संकीर्तन करने वाला संतजन, जो जीवन और जगत को सुंदर बनाता है।

इन दो रत्नों का न कोई मोल है, न कोई तोल ! न इनका मूल्यांकन हो सकता है, न इन्हें खरीदा जा सकता है, न तौला जा सकता है।

भावार्थ : पाठ्यपुस्तक पृष्ठ क्रमांक २४ : पाठ – बाल लीला – संत सूरदास

1. यशोदा अपने पुत्र को चुप करने के लिए बार–बार समझाती है। वह कहती है – “चंदा आओ ! तुम्हें मेरा लाल बुला रहा है। यह मधु मेवा, पकवान, मिठाई स्वयं खाएगा और तुम्हें भी खिलाएगा। (मेरा लाल) तुम्हें हाथ में रखकर खेलेगा; तुम्हें जरा भी भूमि पर नहीं बिठाएगा।”

Maharashtra Board Class 12 Hindi परिशिष भावार्थ : भक्ति महिमा और बाल लीला

यशोदा हाथ में पानी का बर्तन उठाकर कहती है – “चंद्रमा ! तुम शरीर धारण कर आ जाओ।’ फिर उन्होंने जल का पात्र भूमि पर रख दिया और उसे दिखाने लगी – ‘बेटा देखो ! मैं वह चंद्रमा पकड़ लाई हूँ।’ अब सूरदास के प्रभु श्रीकृष्ण हँस पड़े और मुस्कुराते हुए उस पात्र में बार–बार दोनों हाथ डालने लगे।

2. हे श्याम ! उठो, कलेवा (नाश्ता) कर लो। मैं मनमोहन के मुख को देख–देखकर जीती हूँ। हे लाल ! मैं तुम्हारे लिए छुहारा, दाख, खोपरा, खीरा, केला, आम, ईख का रस, शीरा, मधुर श्रीफल और चिरौंजी लाई हूँ। अमरूद, चिउरा, लाल खुबानी, घेवर–फेनी और सादी पूड़ी खोवा के साथ खाओ।

मैं बलिहारी जाऊँ। गुझिया, लड्डू बनाकर और दही लाई हूँ। तुम्हें पूड़ी और अचार बहुत प्रिय हैं। इसके बाद पान बनाकर खिलाऊँगी। सूरदास कहते हैं कि मुझे पानखिलाई मिले।

Maharashtra Board Class 11 Hindi परिशिष मुहावरे

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest परिशिष मुहावरे Notes, Questions and Answers.

Maharashtra State Board 11th Hindi परिशिष मुहावरे

मुहावरा वह वाक्यांश जो सामान्य अर्थ को छोड़कर किसी विशेष अर्थ में प्रयुक्त होता है; मुहावरे में उसके लाक्षणिक और व्यंजनात्मक अर्थ को ही स्वीकार किया जाता है। वाक्य में प्रयुक्त किए जाने पर ही मुहावरा सार्थक प्रतीत होता है।

Maharashtra Board Class 11 Hindi परिशिष मुहावरे

  • अंकुर जमाना – प्रारंभ करना।
  • अपने पैरों पर खड़ा होना – आत्मनिर्भर होना।
  • आँच न आने देना – संकट न आने देना।
  • आँखों में सैलाब उमड़ना – फूट–फूटकर रोना।
  • आँखें फटी रहना – आश्चर्यचकित रह जाना।
  • आईने में मुँह देखना – अपनी योग्यता जाँचना।
  • आसमान के तारे तोड़ना – असंभव कार्य करना।
  • ईंट का जवाब पत्थर से देना – कड़ा जवाब देना
  • उधेड़ बुन में लगना – सोच–विचार करना।
  • एक आँख से देखना – समान रूप से देखना।
  • एक और एक ग्यारह होना – एकता में बल होना।
  • कदम बढ़ाना – प्रगति करना।
  • कमर कसना – प्रगति करना।
  • कमर सीधी करना – आराम करना, सुस्ताना।
  • कलई खुलना – भेद प्रकट होना।
  • कान देना – ध्यान से सुनना।
  • किस्मत खुलना – भाग्य चमकना।
  • गले का हार होना – अत्यंत प्रिय होना।
  • गागर में सागर भरना – थोड़े में बहुत कहना।
  • घी के दीये जलाना – खुशी मनाना।
  • चिकना घड़ा होना – निर्लज्ज होना।
  • चुटकी लेना – व्यंग्य करना।
  • जबान देना – वचन देना। Maharashtra Board Class 11 Hindi परिशिष मुहावरे
  • झंडे गाड़ना – पूर्ण रूप से प्रभाव जमाना।
  • डंका पीटना – प्रचार करना।
  • तितर–बितर होना – बिखर जाना।
  • हजारों दीप जल उठना – आनंदित हो उठना।
  • रुपये दाँत से पकड़ना – कंजूसी करना।
  • दूध का दूध, पानी का पानी करना – इनसाफ करना, न्याय करना।
  • नाम कमाना – यश प्राप्त करना।
  • पाँचों उंगलियाँ घी में होना – हर तरफ से लाभ होना।
  • फूला न समाना – अत्यधिक प्रसन्न होना।
  • बीड़ा उठाना – किसी काम को करने की ठान लेना।
  • बाँछे खिलना – अत्यधिक प्रसन्न होना।
  • मरजीवा होना – कठोर साधना से लक्ष्य तक पहुँचने वाला होना।
  • मल्हार गाना – आनंद मनाना।
  • राई का पहाड़ बनाना – बात को बढ़ा–चढ़ाकर कहना।
  • लोहा मानना – श्रेष्ठता स्वीकार करना।
  • सफेद झूठ बोलना – पूरी तरह से झूठ बोलना।
  • सिर खपाना – ऐसे काम में समय लगाना जिसमें कोई लाभ नहीं।
  • सिर पर सेहरा बाँधना – अधिक यश प्राप्त करना।
  • सोना उगलना – बहुत अधिक लाभ होना।
  • सौ बात की एक बात – असली बात, निचोड़।
  • हाथ–पैर मारना – बहुत प्रयत्न करना।
  • हौसला बुलंद होना – उत्साह बना रहना।
  • श्रीगणेश करना – कार्य आरंभ करना।
  • दाँतों तले उँगली दबाना – आश्चर्यचकित होना।
  • अंधे की लाठी होना – निराधार का सहारा बनना।
  • आग से खेलना – मुसीबत मोल लेना।
  • मुट्ठी गर्म करना – रिश्वत देना।
  • इतिश्री होना – समाप्त होना। Maharashtra Board Class 11 Hindi परिशिष मुहावरे
  • उड़ती चिड़िया पहचानना – तीक्ष्ण बुद्धि वाला होना।
  • हथेली पर सरसों जमाना – कठिन कार्य करना।
  • कंचन बरसना – धन–दौलत से परिपूर्ण होना।
  • कानों कान खबर न होना – बिल्कुल पता न चलना।
  • गाल बजाना – अपनी प्रशंसा आप करना।
  • घड़ों पानी पड़ना – बहुत लज्जित होना।
  • चिकनी–चुपड़ी बातें करना – चापलूसी करना, मीठी-मीठी बातें बोलना।
  • छाती पर साँप लोटना – ईर्ष्या होना।
  • तूती बोलना – प्रभाव होना।
  • दो टूक जवाब देना – स्पष्ट बोलना।
  • नुक्ताचीनी करना – आलोचना करना।

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 7 Modern Periodic Table Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 1.
Mention features of Mendeleev’s periodic table.
Answer:
Features of Mendeleev’s periodic table:

  • In Mendeleev’s periodic table, all 63 elements were arranged in an increasing order of their atomic masses. The serial or ordinal number of an element in the increasing order of atomic mass was referred to as its atomic number.
  • Mendeleev’s periodic table consisted of vertical groups and horizontal series (now called periods).
  • Elements belonging to the same group showed similar properties.
  • Properties of elements in a series/period showed gradual variation from left to right.

Question 2.
Why was Mendeleev’s periodic table readily accepted by scientific community?
Answer:
Mendeleev’s periodic table was readily accepted by scientific community due to the following advantages:
i. Mendeleev had left some gaps corresponding to certain atomic numbers in the periodic table so as to maintain periodicity of the properties. When the elements corresponding to these atomic numbers were discovered, they fitted well into the gaps with their properties as predicted by Mendeleev’s periodic law.

ii. Mendeleev did not predicted presence of inert gases, however, they were discovered in later years. It was possible to accommodate inert gases in Mendeleev’s periodic table by creating an additional group without disturbing the position of other elements in his periodic table.

Question 3.
Give reason: Mendeleev’s periodic law was modified into modern periodic law.
Answer:

  • Henry Moseley in 1913, studied X-ray spectra of large number of elements.
  • He observed that the frequency of X-ray emitted from an element is related to atomic number (Z) of an element and not its atomic mass.
  • Therefore, the atomic number, Z, was considered as more fundamental property of the atom than the atomic mass.
  • As a result, Mendeleev’s periodic law was modified.

Question 4.
Define atomic number.
Answer:
Atomic number (Z) is the total number protons present in the atom of an element.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 5.
State the modern periodic law.
Answer:
Modern periodic law: “The physical and chemical properties of elements are a periodic function of their atomic numbers”.

Question 6.
Periods and groups present in the modern periodic table are numbered based on whose recommendation ?
Answer:
Numbering of the periods and groups in the modem periodic table is based on the recommendation provided by the International Union of Pure and Applied Chemistry (IUPAC).

Question 7.
Write a note on: Structure of the modern periodic table.
Answer:
Structure of the modern periodic table:
i. The modem periodic table also known as the Tong form of periodic table’ has number of boxes formed by the intersection of horizontal rows and vertical columns.
ii. The horizontal rows are called periods and the vertical columns are called groups.
iii. There are seven periods numbered from 1 to 7 and eighteen groups numbered from 1 to 18.
iv. There are total 118 boxes in the modem periodic table which are filled with 118 elements discovered till now including manmade elements.
v. The modem periodic table is divided into four blocks i.e., s-block, p-block, d-block and f-block.

  • Two groups on the extreme left of the modem periodic table form the s-block.
  • Six groups on the extreme right constitute the p-block.
  • Ten groups in the centre form the d-block
  • Two series at the bottom of the modem periodic table constitute the f-block. It contains fourteen elements in each series.

Question 8.
State the relationship between the modern periodic table and electronic configuration in periods.
Answer:

  • The modem periodic table is based on the atomic numbers of the elements. When elements are arranged in an increasing order of atomic number (Z), periodicity is observed in their electronic configurations which reflects in the characteristic structure of the modem periodic table.
  • The location of elements in the modem periodic table is correlated to quantum numbers of the last filled orbital.
  • Along a period, the atomic number increases by one and one electron is added to the outermost shell which forms neutral atom of the next element.
  • The period number is same as the principal quantum number ‘n’ of the valence shell of the elements.
  • A period begins with filling of a particular shell and ends when the valence shell attains complete octet configuration (or duplet, in case of the first period).
  • The next period begins with addition of electron to the next shell of higher energy compared to the previous period. e. g. First shell of the elements gets filled along the first period while second shell starts filling in the second period and addition of electrons continues till second shell (valence shell) attains stable electronic configuration.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 9.
Give reason: First period in the modern periodic table contains only two elements.
Answer:

  • Elements present in the first period i.e., H and He contain only one shell which is also their valence shell and can accommodate maximum two electrons.
  • As first shell can accommodate only two electrons, first period ends at He which has a complete duplet. Hence, first period in the modem periodic table contains only two elements.

Question 10.
Write names and electronic configurations of elements of first period in the modern periodic table. Identify which of them has the stable complete electronic configuration.
Answer:

  • Hydrogen (H) : 1s2, Helium (He) : 2s2
  • Since helium has a complete duplet i.e., two electrons in its valence shell, it has the stable complete electronic configuration.

Question 11.
Explain how does the filling of electrons takes place in the modern periodic table across:
i. Second period
ii. Third period
Answer:
i. Filling of electrons in the second period:

  • In the second period, electrons are filled in the second shell i.e., n = 2.
  • This shell can accommodate a maximum of eight electrons and gets filled as the atomic number increases along the second period.
  • The second period begins with Li (Z = 3): 1s2 2s1 and ends up with Ne (Z = 10): 1s2 2s2 2p6.
  • Neon has complete octet with 8 electrons in its valence shell. Therefore, the second period contains eight elements.

ii. Filling of electrons in the third period:

  • The third period corresponds to the filling of the third shell i.e. n = 3.
  • The third period also contains eight elements.
  • It begins with the filling of electrons in the first element Na (Z = 11) : [Ne] 3s1 and ends with the last element Ar (Z = 18) = [Ne] 3s2 3p6.
  • The condensed electronic configurations for the elements of third period is [Ne] 3s1-2 3p1-6.

Question 12.
There are 18 elements in the fourth period of the modern periodic table. Explain.
Answer:

  • The fourth period corresponds to the filling of fourth shell, n = 4.
  • Therefore, it begins with filling of 4s subshell. The first two elements of the fourth period are K (Z = 19) : [Ar] 4s1 and Ca (Z = 20) : [Ar] 4s2.
  • According to the aufbau principle, the next higher energy subshell is 3d, which can accommodate up to 10 electrons. Thus, filling of the 3d subshell results in the next 10 elements of the fourth period i.e., from Sc (Z = 21) : [Ar] 4s23d1 to Zn (Z = 30): [Ar] 4s23d10.
  • After filling of 3d subshell, the electrons enter the 4p subshell which can accommodate maximum 6 electrons. Hence, filling of 4p subshell results in the next 6 elements i.e., from Ga (Z = 31): [Ar] 4s23d10 4p1 to Kr (Z = 36): [Ar] 4s2 3d10 4p6.
  • Thus, the elements in fourth period are: 2 elements (with 4s subshell), 10 elements (with 3d subshell) and 6 elements (with 4p subshell).
  • Hence, there are 18 elements in the fourth period of the modem periodic table.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 13.
Why does the fifth period of the modern periodic table contain 18 electrons?
Answer:
The fifth period accommodates 18 elements as a result of successive filling of electrons in the 5s, 4d and 5p subshells.

Question 14.
What is the general trend followed while filling of electrons across a period in the modern periodic table.
Answer:

  • A period begins by filling of one electron to the ‘s’ subshell of a new shell and ends when an element corresponding to the same shell attains complete octet (or duplet).
  • Between these two ‘s’ and ‘p’ subshell of the valence shell, the inner subshells ‘d’ and ‘f’ are filled successively following the aufbau principle.

Question 15.
What is the subshell in which the last electron of the first element in the 6th period enters?
Answer:
The 6th period begins by filling the last electron in the shell with n = 6. The lowest energy subshell of any shell is ‘s’. Therefore, the last electron of the first element in the 6th period enters the subshell ‘6s’.

Question 16.
How many elements are present in the 6th period? Explain.
Answer:

  • The 6th period begins by filling the last electron in the subshell ‘6s’ and ends by completing the subshell ‘6p’. Therefore, the sixth period has the subshells filled in increasing order of energy as 6s < 4f < 5d < 6p.
  • The electron capacities of these subshells are 2, 14, 10 and 6, respectively. Therefore, the total number of elements in the 6th period is 2 + 14 + 10 + 6 = 32.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 17.
How does electronic configuration vary down a group in the modern periodic table?
Answer:

  • As we move from top to bottom in a group, a new shell gets added successively in the atom of an element. Therefore, the last electron enters in a new shell down the group.
  • Hence, the general outer electronic configuration of the elements in a group remains the same. This holds true for groups 1, 2 and 3 elements.
  • In the groups 13 to 18 the appropriate inner ‘d’ and ‘f’ subshells are completely filled and the general outer electronic configuration is the same down the groups 13 to 18.
  • However, in the groups 4 to 12, the ‘d’ and ‘f subshells are introduced at a later stage (4th period for ‘d’ and 6th period for ‘f’) down the group. As a result, variation in the general outer configuration is introduced only at the

Note: General outer electronic configuration in groups 1 to 3 and 13 to 18.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 1

Question 18.
On what basis is the modern periodic table divided into four blocks?
Answer:
The modem periodic table is divided into four blocks based on the subshell in which the last electron enters.

Question 19.
Why elements of group 1 and group 2 are known as s-block elements?
Answer:

  • The subshell in which the last electron enters decides the block to which an element belongs.
  • In group 1 and group 2 elements, the last electron is filled in the s subshell.

Therefore, the elements of group 1 and group 2 are known as s-block elements.

Question 20.
Elements belonging to which groups constitute the p-block and why?
Answer:

  • Elements belonging to groups 13, 14, 15, 16, 17 and 18 constitute the p-block.
  • The last electron in the p-block elements is filled in p subshell.
  • As p subshell contains three degenerate p orbitals, it can accommodate up to 6 electrons.
  • Therefore, the p-block elements belonging to six groups i.e., groups 13, 14, 15, 16, 17 and 18 in which last electron enters in p subshell constitute the p-block.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 21.
Give reason: Helium which is the first element of group 18 is placed in the p-block even though its last electron enters in s subshell.
Answer:

  • Electronic configuration of helium is 1s2 which indicates that it has a stable electronic configuration i.e., a complete duplet.
  • The p-block ends with group 18 which is a family of inert gases having stable electronic configuration (complete octet except helium).
  • Therefore, helium is placed with group 18 elements in p-block due to its stable electronic configuration even though its last electron enters in s subshell.

Question 22.
State the general outer electronic configuration of s-block and p-block elements.
Answer:
The general outer electronic configuration of s-block elements is ns1-2.
The general electronic configuration for the p-block elements is ns2np1-6.

Question 23.
There are total 10 groups in the d-block of the modern periodic table. Explain.
Answer:

  • The d-block in the modem periodic table is formed as a result of filling the last electron in d orbital.
  • As there are five orbitals in a d subshell, 10 electrons can successively be accommodated.

Hence, there are total 10 groups in the d-block of the modem periodic table i.e., group 3 to 12.

Question 24.
The last electron enters a (n-1)d orbital only after the ns subshell is completely filled. Explain.
Answer:
A d subshell is present in the shells with n ≥ 3 and according to the (n+1) rule, the energy of ns orbital is less than that of the (n-1)d orbital. As a result, the last electron enters a (n-1)d orbital only after the ns subshell is completely filled.

Question 25.
Chromium exhibit 4s1 3d5 electronic configuration instead of 4s2 3d4. Explain.
Answer:

  • Completely filled or half-filled subshells are highly stable.
  • In 4s1 3d5 configuration, both s and d subshells are half-filled.
  • Thus, due to the extra stability associated with half-filled subshells, chromium exhibits 4s1 3d5 electronic configuration instead of 4s2 3d4.

Question 26.
What is the general outer electronic configuration of d-block and f-block elements?
Answer:
The general outer electronic configuration of the d-block elements is ns0-2 (n-1)d1-10 while the general outer electronic configuration of the f-block elements is ns2 (n-1)d0-1 (n-2)f1-14.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 27.
Expected outer electronic configuration of europium (Eu) is 6s2 4f6 5d1. However, it exhibits different than expected outer electronic configuration.
i. Write the observed outer electronic configuration of Eu.
ii. What is the reason for this variation in electronic configuration?
Answer:
i. Observed outer electronic configuration of europium (Eu) is 6s2 4f7 5d0.
ii. In the observed electronic configuration of Eu, 4f subshell is half-filled which is a highly stable configuration. Therefore, observed electronic configuration of Eu varies than expected.

Question 28.
Name the two series that constitute f-block.
Answer:
The f-block constitutes two series of 14 elements called the lanthanide and the actinide series which are placed one below the other.

Question 29.
State whether the following statements are true or false. Correct if false.
i. Position of the elements in the modern periodic table is related to the quantum number of their last filled orbital.
ii. Group number is same as the principal quantum number ‘n’ of the valence shell of the elements.
Answer:
i. True
ii. False
Period number is same as the principal quantum number ‘n’ of the valence shell of the elements.

Question 30.
Name the following.
i. Shortest period in the modern periodic table.
ii. Block which is placed separately at the bottom of the modern periodic table.
Answer:
i. First period
ii. f-Block

Question 31.
How can a period, group and block of the element be determined?
Answer:
The group, period and the block of the element can be determined on the basis of its electronic configuration.
i. Period: The principal quantum number of the valence shell corresponds to the period of the element.
e. g. The principal quantum number (n) of the valence shell (3s1) of Na (1s2 2s2 2p6 3s1) is 3. This corresponds to third period.

ii. Block: The subshell in which the last electron enters, corresponds to the block of the elements (with exception being He).
e. g. The subshell 3d (in which the last electron enters) for Sc (1s2 2s2 2p6 3s2 3p6 3d1 4s2) corresponds to d block.

iii. Group: The group of the element is determined on the basis of number of electrons present in the outermost or penultimate [next to outermost, i.e. (n-1)] shell:

  • For s-block elements, group number = number of valence electrons.
  • For p-block elements, group number = 18 – number of electrons required to complete octet.
  • For d-block elements, group number = 2 + number of (n-1)d electrons.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 32.
Outer electronic configurations of a few elements are given below. Explain them and identify the period, group and block in the periodic table to which they belong.
2He: 1s2, 54Xe: 5s25p6, 16S: 3s23p4, 79Au: 6s15d10
Answer:
i. 2He: 1s2
Here, n = 1. Therefore, 2He belongs to the 1st period.
The shell n = 1 has only one subshell, namely 1s. The outer electronic configuration 1s2 of ‘He’ corresponds to the maximum capacity of 1s, the complete duplet. Therefore, He is placed at the end of the 1st period in the group 18 of inert gases. So, ‘He’ belongs to p-block.

ii. 54Xe: 5s25p6
Here, n = 5. Therefore, 54Xe belongs to the 5th period.
The outer electronic configuration. 5s25p6 corresponds to complete octet. Therefore, 54Xe is placed in group 18 and belongs to p-block.

iii. 16S: 3s23p4
Here, n = 3. Therefore, 16S belongs to the 3rd period. The 3p subshell in ‘S’ is partially filled and short of completion of octet by two electrons. Therefore, ‘S’ belongs to (18 – 2) = 16th group and p-block.

iv. 79AU: 6s15d10
Here n = 6. Therefore, ‘Au’ belongs to the 6th period.
The sixth period begins with filling of electron into 6s and then into 5d orbital.
The outer configuration of ‘Au’: 6s1 5d10 implies that (1 + 10) = 11 electrons are filled in the outer orbitals to give ‘Au’. Therefore ‘Au’ belongs to the group 11.
As the last electron has entered ‘d’ orbital ‘Au’ belongs to the d-block.

Question 33.
Predict the block, periods and groups to which the following elements belong.
i. Mg (Z = 12)
ii. V (Z = 23)
iii. Sb (Z = 51)
iv. Rn (Z = 86)
v. Na (Z = 11)
vi. Cl (Z = 17)
Answer:
i. Mg (Z = 12): Atomic number of Mg is 12. Electronic configuration is 1s2 2s2 2p6 3s2.
Block: Since the last electron enters s subshell (3 s), Mg belongs to s-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For s-block element, group number = number of valence electrons = 2. Hence, it belongs to group 2.

ii. V (Z = 23): Atomic number of V is 23. Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2.
Block: Since the last electron enters d subshell (3d), V belongs to d-block.
Period: n = 4. Therefore, it is present in the fourth period.
Group: For d-block elements, group number = 2 + number of (n – 1)
d electrons = 2 + 3 = 5. Hence, it belongs to group 5.

iii. Sb (Z = 51): Atomic number of Sb is 51.
Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p3.
Block: Since the last electron enters p subshell (5p), Sb belongs to p-block.
Period: n = 5. Therefore, it is present in the fifth period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 3 = 15. Hence it belongs to group 15.

iv. Rn (Z = 86): Atomic number of Rn is 86.
Electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6.
Block: Since the last electron enters p subshell (6p), Rn belongs to p-block.
Period: n = 6. Therefore, it is present in the sixth period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 0 = 18. Hence, it belongs to group 18.

v. Na (Z = 11): Atomic number of Na is 11. Electronic configuration is 1s2 2s2 2p6 3s1.
Block: Since the last electron enters s subshell (3s), Na belongs to s-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For s-block element, number of the group = number of valence electrons = 1. Hence, it belongs to group 1.

vi. Cl (Z = 17): Atomic number of Cl is 17. Electronic configuration is 1s2 2s2 2p6 3s2 3p5.
Block: Since the last electron enters p subshell (3p), Cl belongs to p-block.
Period: n = 3. Therefore, it is present in the third period.
Group: For p block elements, group number = 18 – number of electrons required to complete octet
= 18 – 1 = 17. Hence, it belongs to group 17.

Question 34.
State the characteristics of s-block elements.
Answer:

  • The s-block contains the elements of group 1 (alkali metals) and group 2 (alkaline earth metals).
  • They occur in nature only in combined state as they are reactive elements.
  • Except Li and Be, compounds formed by all other s-block elements are predominantly ionic in nature.
  • This is because they have only one or two valence electrons which they can lose readily forming M+ or M2+ ions.
  • Since they can lose electrons easily, they have low ionization enthalpies, which decreases down the group resulting in increased reactivity.

Question 35.
What are main group elements?
Answer:
The p-block elements together with s-block elements are called main group elements or representative elements.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 36.
Give reason: Group 18 elements do not participate in chemical reactions readily.
Answer:

  • Group 18 is the last group of p-block and include noble or inert gases.
  • They have closed valence shells (complete duplet in the case of ‘He’ and complete octet in the case of the other noble gases).
  • Therefore, they show very low chemical reactivity and thus, do not participate in chemical reactions readily.

Question 37.
Why nonmetals present in group 17 and 16 in the modern periodic table are highly reactive?
Answer:

  • Nonmetals present in group 17 (halogen family) and group 16 (chalcogens) have highly negative electron gain enthalpies.
  • As a result, they readily accept one or two electrons and form anions (X or X2-) that have complete octet.
  • Therefore, nonmetals present in group 17 and 16 are highly reactive.

Question 38.
Explain the composition of the p-block in the modern periodic table.
Answer:

  • The p-block contains elements of groups 13 to 18.
  • It contains all the three types of elements i.e., metals, nonmetals and metalloids.
  • In the p-block, metals and nonmetals are separated from each other by a zig-zag line. The metals are present on the left and the nonmetals are present on the right side while the metalloids are present along the zig-zag line.

Question 39.
State whether the following statements are true or false. Correct if false.
i. Nonmetallic character increases from left to right across a period.
ii. Nonmetallic character increases down a group.
Answer:
i. True
ii. False
Nonmetallic character decreases down a group.

Question 40.
Differentiate between s-block and p-block elements.
Answer:
s-Block elements:

  • s-Block contains group 1 and group 2 elements.
  • It contains only metals.
  • The last electron in the s-block elements enters in s orbital.
  • General outer electronic configuration of s-block elements is ns1-2.
  • e.g. Na, K, Ca, Mg, etc.

p-Block elements:

  • p-Block contains elements from groups 13 to 18.
  • It contains metals, nonmetals as well as metalloids.
  • The last electron in the p-block elements enters in p orbital.
  • General outer electronic configuration of p-block elements is ns2 np1-6.
  • e.g. C, N, O, F, etc.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 41.
Write a note on the characteristics of the d-block elements.
Answer:

  • The d-block contains elements of the groups 3 to 12 which are all metals. They are also known as transition elements or transition metals.
  • They form a bridge between chemically reactive s-block elements and less reactive elements of groups 13 and 14.
  • Most of the d-block elements possess partially filled inner d orbitals. As a result, the d-block elements have properties such as variable oxidation state, paramagnetism, ability to form coloured ions. They can be used as catalysts.
  • Zn, Cd, and Hg with configuration ns2 (n-1)d10, (completely filled s and d subshells) do not show characteristic properties of transition metals as they are stable.

Question 42.
Explain in brief about the f-block elements.
Answer:

  • The elements present in f-block are all metals and are placed in the two rows called lanthanide series (58Ce to 71Lu) and actinide series (90Th to 103Lr).
  • The lanthanides are also known as rare earth elements while the actinide elements beyond 92U are called transuranium elements.
  • All the transuranium elements are manmade and radioactive.
  • The last electron of the elements of these series is filled in the (n-2)f subshell, and therefore, these are called inner transition elements.
  • These elements have very similar properties within each series.

Question 43.
What is lanthanide and actinide series?
Answer:
i. Lanthanide series: The fourteen elements after lanthanum (Z = 57) i.e., from cerium (Z = 58) to lutetium (Z = 71) are named after their preceding member (57La) present in the third group and 6th period and are called lanthanides. They are kept in separate series called lanthanide series at the bottom of the modem periodic table.

ii. Actinide series: The fourteen elements after actinium i.e., from thorium (Z = 90) to lawrencium (Z = 103) are named after 89Ac present in third period and 7th group. They are kept in separate series called actinide series at the bottom of the modem periodic table.

Question 44.
Differentiate between d-block and f-block elements.
Answer:
d-Block elements:

  • d-Block contains elements from group 3 to group 12.
  • It is present in the middle of the modern periodic table.
  • They are also known as transition elements.
  • The last electron in the d-block elements enters in d orbital.
  • General outer electronic configuration of d-block elements is ns0-2 (n-1)d1-10 .
  • e.g. Cu, Zn, Cr, Ti, V, etc.

f-Block elements:

  • f-Block contains elements of lanthanide and actinide series.
  • f-block elements are present below the modern periodic table as two separate rows.
  • They are also known as inner transition elements.
  • The last electron in the f-block elements enters in f orbital.
  • General outer electronic configuration of f-block elements is ns2 (n-1) d0-1 (n-2) f114.
  • e.g. Ce, Pr, Nd Th, U, Np, etc.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 45.
Chlorides of two metals are common laboratory chemicals and both are colourless. One of the metals reacts vigorously with water while the other reacts slowly. Place the two metals in the appropriate block in the periodic table. Justify your answer.
Answer:
i. Metals are present in all the four blocks of the periodic table.
ii. Salts of metals in the f-block and p-block (except AlCl3) are not common laboratory chemicals. Therefore, the choice is between s- and d-block.
iii. From the given properties their placement is done as shown below:

  • s-block: Metal that reacts vigorously with water.
  • d-block: Metal that reacts slowly with water.

iv. The colourless nature of the less reactive metal in the d-block implies that the inner d subshell is completely filled.

Question 46.
What are periodic properties?
Answer:

  • The elements in the modem periodic table (long form of periodic table) are arranged in such a way that on moving across a period or down the group, several properties of elements vary in regular fashion. These properties are called periodic properties.
  • Atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valency and oxidation states are several properties of elements that show periodic variations.

Question 47.
What leads to the phenomena called effective nuclear charge and screening effect in an atom?
Answer:

  • The periodic trends are explained in the terms of two fundamental factors, namely, attraction of extranuclear electrons towards the nucleus and repulsion between electrons belonging to the same atom.
  • These attractive and repulsive forces operate simultaneously in an atom.
  • This results in two interrelated phenomena called effective nuclear charge and screening effect in an atom.

Question 48.
Explain the concept of effective nuclear charge in detail.
Answer:
i. In a multi-electron atom, the positively charged nucleus attracts the negatively charged electrons around it, and there is mutual repulsion between the negatively charged extranuclear electrons.
ii. The repulsion by inner shell electrons results in pushing the outer shell electrons further away from the nucleus. Thus, the outer shell electrons are held less tightly by the nucleus.
iii. As a result, the attraction of the nucleus for an outer electron is partially cancelled and hence, an outer shell electron does not experience the actual positive charge present on the nucleus. This effect of the inner electrons on the outer electrons is called screening effect or shielding effect.
iv. The net nuclear charge actually experienced by an electron is called the effective nuclear charge (Zeff).
The effective nuclear charge is lower than the actual nuclear charge (Z).
v. Effective nuclear charge (Zeff) = Z – electron shielding
= Z – σ
Here σ (sigma) is called shielding constant or screening constant and the value of σ depends upon type of the orbital that the electron occupies.

Question 49.
Define:
i. Effective nuclear charge (Zeff)
ii. Screening effect (or shielding effect)
Answer:
i. Effective nuclear charge (Zeff): In multi-electron atom, the net nuclear charge actually experienced by an electron is called the effective nuclear charge (Zeff).
ii. Screening effect (or shielding effect): In multi-electron atom, the effect of the inner electrons on the outer electrons is called screening effect or shielding effect of the inner/core electrons.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 50.
Explain the variations in effective nuclear charge
i. Across a period
ii. Down a group
Answer:
i. Across a period:

  • As we move across a period, atomic number increases by one and thus, actual nuclear charge (Z) increases by +1 at a time.
  • However, the valence shell remains the same and the newly added electron gets accommodated in the same shell. There is no addition of electrons to the core i.e., inner shells. Thus, shielding due to core electrons remains the same even though the actual nuclear charge increases.
  • As a result, the effective nuclear charge (Zeff) goes on increasing across a period.

ii. Down a group:

  • As we move down a group, a new larger valence shell is added. As a result, there is an additional shell in the core.
  • The shielding effect of the increased number of core electrons outweighs the effect of the increased nuclear charge. Thus, the effective nuclear charge experienced by the outer electrons decreases largely down a group.
  • Hence, the effective nuclear charge (Zeff) decreases down a group.

Question 51.
Define atomic radius.
Answer:
Atomic radius is one half of the internuclear distance between two adjacent atoms of a metal or two single bonded atoms of a nonmetal.

Question 52.
What is meant by covalent radius of the atom? Explain with suitable examples.
Answer:

  • In the case of nonmetals (except noble gases), the atoms of an element are bonded to each other by covalent bonds.
  • Bond length of a single bond is taken as sum of radii of the two single bonded atoms. This is called covalent radius of the atom.
  • For example: Bond length of C-C bond in diamond is 154 pm. Therefore, atomic radius of carbon is estimated to be 77 pm which is half of bond length (\frac {1}{2} × 154 = 77).

Question 53.
How is atomic radius of a nonmetallic element estimated?
Answer:

  • The atomic size of a nonmetallic element is estimated from the distance between the two atoms bound together by a single covalent bond. From this, the covalent radius of the element is estimated.
  • The internuclear distance in a diatomic molecule of an element is its covalent bond length. Half the covalent bond length gives the covalent radius.
  • Bond length of Cl-Cl bond in Cl2 is measured as 198 pm. Therefore, the atomic radius of Cl is estimated to be 99 pm.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 2

Question 54.
Define metallic radius.
Answer:
One half of the distance between the centres of nucleus of the two adjacent atoms of a metallic crystal is called as a metallic radius.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 55.
How is metallic radius determined in the case of metals? Give suitable example.
Answer:

  • In the case of metals, distance between the adjacent atoms in metallic sample is measured. One half of this distance is taken as the metallic radius.
  • For example: In beryllium, distance between the adjacent Be atoms is measured. One half of this distance is taken as the metallic radius of a Be atom.
  • Distance between two adjacent Be atoms is 224 pm. Therefore, metallic radius of a Be atom is 112 pm.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 3

Note: Atomic radii of some elements are given in the table below.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 4

Question 56.
How is a cation and an anion formed?
Answer:
A cation (or positively charged ion) is formed by the removal of one or more electrons from the atom of an element whereas an anion (or negatively charged ion) is formed when the atom of an element gains one or more electrons.

Question 57.
Give reasons: Radius of a cation is smaller and that of an anion is larger as compared to that of their parent atoms.
Answer:

  • A cation is formed by the loss of one or more electrons, therefore, it contains fewer electrons that the parent atom but has the same nuclear charge.
  • As a result, the shielding effect is less and effective nuclear charge is larger within a cation. Thus, radius of a cation is smaller than the parent atom.
  • However, an anion is formed by the gain of one or more electrons and therefore, it contains a greater number of electrons than the parent atom.
  • Due to these additional electrons, anion experiences increased electronic repulsion and decreased effective nuclear charge. As a result, an anion has larger radius than its parent atom.

Hence, radius of a cation is smaller and that of an anion are larger as compared to that of their parent atoms.

Question 58.
Define: Isoelectronic species
Answer:
The atoms or ions which have the same number of electrons are called isoelectronic species.

Question 59.
Explain with example why the radii of isoelectronic species vary.
Answer:
i. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
ii. For example, F and Na+ both have 10 electrons but the nuclear charge on F is +9 which is smaller than that of Na+ which has the nuclear charge +11.
Hence, F has larger ionic radii (133 pm) than Na+ (98 pm).

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 60.
What is the trend observed in the ionic size of the following isoelectronic species? Explain.
i. Na+, Mg2+, Al3+ and Si4+
ii. O2-, F, Na+ and Mg2+
Answer:
i. Na+, Mg2+, Al3+ and Si4+
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 5
a. Among the given ions, the nuclear charge varies but the number of electrons remains the same and therefore, these are isoelectronic species.
b. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
c. The nuclear charge increases in the order Na+ < Mg2+ < Al3+ < Si4+ and thus, the ionic size decreases in the order Na+ > Mg2+ > Al3+ > Si4+.

ii. O2-, F, Na+ and Mg2+
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 6
a. Among the given ions, the nuclear charge varies but the number of electrons remains the same and therefore, these are isoelectronic species.
b. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
c. The nuclear charge increases in the order O2- < F < Na+ < Mg2+ and thus, the ionic size decreases in the order O2- > F > Na+ > Mg2+.

Question 61.
Identify the species having larger radius from the following pairs:
i. Na and Na+
ii. Na+ and Mg2+
Answer:
i. The nuclear charge is the same in Na and Na+. But Na+ has a smaller number of electrons and a smaller number of occupied shells (two shells in Na+, while three shells in Na). Therefore, radius of Na is larger.
ii. Na+ and Mg2+ are isoelectronic species. Mg2+ has a larger nuclear charge than that of Na+. Therefore, Na+ has larger radius.

Question 62.
Which of the following species will have the largest and the smallest size? Why?
Mg, Mg2+, Al, Al3+
Answer:

  • Atomic radius decreases across the period. Hence, the atomic radius of Mg is larger than that of Al.
  • Parent atoms have larger radius than their corresponding cations. Hence, the radius of Mg is larger than that of Mg2+ and the radius of Al is larger than that of Al3+.
  • Mg2+ and Al3+ are isoelectronic. Among isoelectronic species, the one with larger nuclear charge will have smaller radius. Al3+ (Z = 13) has a larger nuclear charge than that of Mg2+ (Z = 12). Hence, the ionic radius of Al3+ is smaller than Mg2+.
  • Therefore, the decreasing order of radius is Mg > Al > Mg2+ > Al3+.

Hence, species with the largest size is Mg and with the smallest size is Al3+.

Question 63.
Identify the element with more negative value of electron gain enthalpy from the following pairs. Justify.
i. Cl and Br
ii. F and O
Answer:
i. Cl and Br belong to the same group of halogens with Br having higher atomic number than CL As the atomic number increases down the group, the effective nuclear charge decreases. The increased shielding effect of core electrons can be noticed. The electron has to be added to a farther shell, which releases less energy and thus, electron gain enthalpy becomes less negative down the group. Therefore, Cl has more negative electron gain enthalpy than Br.

ii. F and O belong to the same second period with F having higher atomic number than O. As the atomic number increases across a period, atomic radius decreases, effective nuclear charge increases and electron can be added more easily. Therefore, more energy is released with gain of an electron as we move towards right in a period. Therefore, F has more negative electron gain enthalpy than O.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 64.
Explain the importance of electronegativity.
Answer:

  • When two atoms of different elements form a covalent bond, the electron pair is shared unequally.
  • Electronegativity represents attractive force exerted by the nucleus on shared electrons. Electron sharing between covalently bonded atoms takes place using the valence electron.
  • It depends upon the effective nuclear charge experienced by electron involved in formation of the covalent bond.
  • Electronegativity predicts the nature of the bond, or, how strong is the force of attraction that holds two atoms together.

Question 65.
Explain the trend in electronegativity
i. across a period
ii. down a group
Answer:
i. Across a period:
a. As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily.
b. Hence, due to the increase in effective nuclear charge, the tendency to attract shared electron pair in a covalent bond increases i.e., electronegativity increases from left to right across a period.
e. g. Li < Be < B < C < N < O <F.

ii. Down a group:
a. As we move down the group from top to bottom in the periodic table, the size of the valence shell goes on increasing.
b. However, the effective nuclear charge decreases as the shielding effect of the core electrons increases due to the increase in the size of the atoms.
c. Thus, the tendency to attract shared electron pair in a covalent bond decreases, decreasing the electronegativity down the group.
e.g. F > Cl > Br > I > At.

Question 66.
Explain the terms:
i. Valency of an element
ii. Oxidation state (or oxidation number)
iii. Chemical reactivity
Answer:
i. Valency of an element:

  • Valency of an element indicates the number of chemical bonds that the atom can form giving a molecule.
  • The most fundamental chemical property of an element is its combining power. This property is numerically expressed in terms of valency or valence.
  • Valence does not have any sign associated with it.
  • Valency of the main group elements is usually equal to the number of valence electrons (outer electrons) and/or equal to difference between 8 and the number of valence electrons.

ii. Oxidation state (or oxidation number):

  • The oxidation state or oxidation number is a frequently used term related to valence.
  • Oxidation number has a sign, + or – which is decided by the electronegativities of atoms that are bonded.

iii. Chemical reactivity:

  • Chemical reactivity is related to the ease with which an element loses or gains the electrons.
  • Chemical properties of elements depend on their electronic configuration.

Question 67.
What is the trend observed in the valency of main group elements?
Answer:
i. Valency of the main group elements is usually equal to the number of valence electrons (outer electrons) or it is equal to the difference between 8 and the number of valence electrons.
i.e., (8 – number of valence electrons).
ii. The valency remains the same down the group and shows a gradual variation across the period as atomic number increases from left to right.

Note: Periodic trends in valency of main group elements.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 7

Question 68.
Give any two distinguishing points between metals and nonmetals.
Answer:
Metals:

  1. Generally, metals exhibit good electrical conductivity.
  2. They can form compounds by loss of valence electrons.

Nonmetals:

  1. Generally, nonmetals exhibit poor electrical conductivity.
  2. Nonmetals can form compounds by gain of valence electrons in valence shell.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 69.
Explain the variation of the following property of elements down a group and across a period.
i. Metallic character
ii. Nonmetallic character
Answer:
The variation observed in the metallic and nonmetallic character of elements can be explained in the terms of ionization enthalpy and electron gain enthalpy.
i. Metallic character:

  • The ionization enthalpy decreases down the group. Thus, the tendency to lose valence electrons increases down the group and the metallic character increases down a group.
  • However, the ionization enthalpy increases across the period and as a result metallic character decreases across a period.

ii. Nonmetallic character:

  • Electron gain enthalpy becomes less negative as we move down the group and hence, nonmetallic character decreases down the group.
  • However, electron gain enthalpy becomes more and more negative across the period and thus, nonmetallic character increases across the period.

Question 70.
Justify the position of most reactive and least reactive elements in the modern periodic table.
Answer:

  • Chemical reactivity of elements depends on the ease with which it attains electronic configuration of the nearest inert gas by gaining or losing electrons.
  • The elements preceding an inert gas react by gaining electrons in the outermost shell, whereas the elements which follow an inert gas in the periodic table react by loss of valence electrons. Thus, the chemical reactivity is decided by the electron gain enthalpy and ionization enthalpy values, which in turn, are decided by effective nuclear charge and finally by the atomic size.
  • The ionization enthalpy is the smallest for the element on the extreme left in a period, whereas the electron gain enthalpy is the most negative for the second last element on the extreme right, (preceding to the inert gas which is the last element of a period).
  • Thus, the most reactive elements lie on the extreme left and the extreme right (excluding inert gases) of the periodic table.

Question 71.
How can we predict chemical reactivity of elements based on their oxide formation reactions and the nature of oxides formed?
Answer:

  • The chemical reactivity can be illustrated by comparing the reaction of elements with oxygen to form oxides and the nature of the oxides.
  • Alkali metals present on the extreme left of the modem periodic table are highly reactive and thus, they react vigorously with oxygen to form oxides such as Na2O which reacts with water to form strong bases like NaOH.
  • The reactive elements on the right i.e., halogens react with oxygen to form oxides such as Cl2O7 which on reaction with water form strong acids like HClO4.
  • The oxides of the elements in the centre of the main group elements are either amphoteric (Al2O3) neutral (CO, NO) or weakly acidic (CO2).

Question 72.
Write the chemical equations for reaction, if any, of (i) Na2O and (ii) Al2O3 with HCl and NaOH both. Correlate this with the position of Na and Al in the periodic table, and infer whether the oxides are basic, acidic or amphoteric.
Answer:
i. Na2O + 2HCl → 2NaCl + H2O
Na2O + NaOH → No reaction
As Na2O reacts with an acid to form salt and water it is a basic oxide. This is because Na is a reactive metal lying on the extreme left of the periodic table.

ii. Al2O3 + 6HCl → 2AlCl3 + 3H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O
As Al2O3 reacts with an acid as well as base to form a salt and water. It is an amphoteric oxide. Al is a moderately reactive element lying in the centre of main group elements in the periodic table.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

Question 73.
Comment on the chemical reactivity of d-block and f-block elements.
Answer:

  • d-block (transition) elements and f-block (inner transition) elements exhibit very small change in atomic radii.
  • Therefore, the transition and inner transition elements belonging to the individual series have similar chemical properties.
  • Their ionization enthalpies are intermediate between those of s-block and p-block elements. Thus, d-block and f-block elements generally show moderate reactivity.

Question 74.
Ge, S and Br belong to the groups 14, 16 and 17, respectively. Predict the empirical formulae of the compounds those can be formed by (i) Ge and S, (ii) Ge and Br.
Answer:
From the group number we understand that the general outer electronic configuration and number of valence electrons and valencies of the three elements are:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 8
i. S is more electronegative than Ge. Therefore, the empirical formula of the compound formed by these two elements is predicted by the method of cross multiplication of the valencies:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 9

ii. Br is more electronegative than Ge. The empirical formula of the compound formed by these two elements is predicted by the method of cross multiplication of valencies:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table 10
[Note: More electronegative element is written on right hand side in cross multiplication method.]

Question 75.
The first ionization enthalpies of 5 elements of second period are given below:

Element 1st IE values (kJ mol-1)
I 520
II 1681
III 1086
IV 2080
V 899

Based on the above data, answer the following questions:
i. Identify the element having highest atomic number.
ii. If element I is lithium, how will you explain its low value of first ionization enthalpy?
iii. Explain why ionization enthalpies are always positive.
Answer:
i. Element IV. The first ionization enthalpy increases with increase in atomic number along a period. Hence, the element IV having highest IE will have highest atomic number among the given elements.
ii. Alkali metals have only one electron in their valence shell which can be easily lost resulting in the stable noble gas configuration. Therefore, lithium shows low value of first ionization enthalpy.
iii. Energy is always required to remove electrons from an atom. Hence, ionization enthalpies have positive value.

Question 76.
From the elements Mg, Ar, Cl, Sr, P and S, choose one that fits each of the below given descriptions:
i. An element having two valence electrons.
ii. An element having properties similar to that of O.
iii. A noble gas.
iv. An alkaline earth metal,
v. An element having electronic configuration 1s22s22p63s23p3.
Answer:
i. Magnesium (Mg)
ii. Sulphur (S)
iii. Argon (Ar)
iv. Strontium (Sr)
v. Phosphorus (P)

Multiple Choice Questions

1. Mendeleev’s periodic table had …………… elements.
(A) 75
(B) 83
(C) 63
(D) 118
Answer:
(C) 63

2. The serial or ordinal number of an element in Mendeleev’s periodic table was recognized as ………….
(A) neutron number
(B) valency
(C) principal quantum number
(D) proton number
Answer:
(D) proton number

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

3. Mendeleev predicted the existence of …………..
(A) aluminium
(B) silicon
(C) tellurium
(D) germanium
Answer:
(D) germanium

4. According to Mendeleev’s periodic law, the physical and chemical properties of elements are the periodic function of their …………..
(A) atomic weights
(B) atomic numbers
(C) molecular formulas
(D) molecular weights
Answer:
(A) atomic weights

5. Moseley showed that the fundamental property of an element is ……………
(A) atomic number
(B) atomic mass
(C) both A and B
(D) none of these
Answer:
(A) atomic number

6. According to periodic law of elements, the variation in properties of elements is related to their ……………
(A) densities
(B) atomic masses
(C) atomic sizes
(D) atomic numbers
Answer:
(D) atomic numbers

7. At present, how many elements are known?
(A) 118
(B) 110
(C) 114
(D) 120
Answer:
(A) 118

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

8. The long form of the periodic table consists of how many periods?
(A) 5
(B) 8
(C) 10
(D) 7
Answer:
(D) 7

9. According to quantum mechanical model of the atom, the properties of elements can be correlated to their …………….
(A) atomic number
(B) atomic mass
(C) valency
(D) electronic configuration
Answer:
(D) electronic configuration

10. The fourth, fifth and sixth periods are long periods and contain ……………
(A) 18, 18 and 36
(B) 18, 28 and 32
(C) 18, 15 and 31
(D) 18, 18 and 32
Answer:
(D) 18, 18 and 32

11. f-block elements are also known as ……………
(A) transition elements
(B) inert gas elements
(C) normal elements
(D) inner transition elements
Answer:
(D) inner transition elements

12. Which of the following forms a bridge between reactive s-block elements and less reactive group 13 and 14 elements?
(A) Inert gases
(B) Transition metals
(C) Halogens
(D) Inner transition metals
Answer:
(B) Transition metals

13. ………… elements are known as chalcogens.
(A) Group 17
(B) Group 18
(C) Group 16
(D) Group 1
Answer:
(C) Group 16

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

14. The name ‘rare earth elements’ is used for …………..
(A) lanthanides only
(B) actinides only
(C) both lanthanides and actinides
(D) alkaline earth metals
Answer:
(C) both lanthanides and actinides

15. Atomic number of V is 23 and its electronic configuration is …………….
(A) 1s2 2s2 2p6 3p6 3d3 4s2
(B) 1s2 2s2 2d3 3p6 2p6 4s2
(C) 2s2 1s2 2p6 3s2 3d3 4s2
(D) 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Answer:
(D) 1s2 2s2 2p6 3s2 3p6 3d3 4s2

16. Aluminium belongs to …………. elements.
(A) s-block
(B) p-block
(C) d-block
(D) f-block
Answer:
(B) p-block

17. In P3-, S2- and Cl ions, the increasing order of size is ………….
(A) Cl < S2- < P3-
(B) P3- < S2- < Cl
(C) S2- < Cl < P3-
(D) S2- < P3- < Cl
Answer:
(A) Cl- < S2- < P3-

18. The CORRECT order of radii is ……………
(A) N < Be < B
(B) F< O2- <N3-
(C) Na < Li < K
(D) Fe3+ < Fe2+ < Fe4+
Answer:
(B) F < O2- <N3-

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

19. Which of the following species will have the largest size Mg, Mg2+, Fe, Fe3+?
(A) Mg
(B) Mg2+
(C) Fe
(D) Fe3+
Answer:
(C) Fe

20. Which one of the following is CORRECT order of the size?
(A) I > I >I+
(B) I > I+ > I
(C) I+ > I > I
(D) I > I > I+
Answer:
(D) I > I > I+

21. The CORRECT order of increasing radii of the elements Na, Si, Al and P is ……………
(A) Si < Al < P < Na
(B) Al < Si < P < Na
(C) P < Si < Al < Na
(D) Al < P < Si < Na
Answer:
(C) P < Si < Al < Na

22. The metallic and nonmetallic properties of elements can be judged by their ……………
(A) electron gain enthalpy
(B) ionization enthalpy
(C) electronegativity
(D) valence
Answer:
(C) electronegativity

Maharashtra Board Class 11 Chemistry Important Questions Chapter 7 Modern Periodic Table

23. Which element has the most negative electron gain enthalpy?
(A) Sulphur
(B) Fluorine
(C) Chlorine
(D) Hydrogen
Answer:
(C) Chlorine

24. Which of the properties remain unchanged on descending a group in the periodic table?
(A) Atomic size
(B) Density
(C) Valency electrons
(D) Metallic character
Answer:
(C) Valency electrons

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 13 Respiration and Energy Transfer Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 1.
What is Phosphorylation?
Answer:
It is the formation of ATP, by addition inorganic phosphate to ADP.
ADP + Pi → ATP
Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 1

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 2.
What are the different ways of Phosphorylation?
Answer:
Phosphorylation occurs in three different ways as – photophosphorylation, substrate-level phosphorylation and oxidative phosphorylation.

Question 3.
What is Substrate-level phosphorylation? Where does it occur?
Answer:
Substrate-level phosphorylation is a direct phosphorylation of ADP by transfer of a phosphate group from any suitable substrate. It occurs in cytoplasm of the cells and matrix of mitochondria.

Question 4.
What is oxidative phosphorylation? Mention the site of this reaction.
Answer:
In oxidative phosphorylation ATP is synthesize by using the energy released during the oxidation of substrates like NADH + H+ and FADH2. This occurs on the inner mitochondrial membrane only.

Question 5.
When is ATP hydrolysed?
Answer:
ATP is hydrolysed whenever energy is required for any metabolic reactions.

Question 6.
What is respiration?
Answer:
Respiration is a catabolic process wherein complex organic substrate is oxidized to simple components to generate biological energy, i.e. ATP.

Question 7.
Give an example for anabolic and catabolic process.
Answer:
Anabolic process- Photosynthesis (Biosynthetic process).
Caiabolic process- Respiration (Breakdown process).

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 8.
Name the following.

  1. The substrate that undergoes oxidation during oxidative phosphorylation
  2. The phosphorylation that occurs only in inner mitochondria! membrane
  3. Two ways of cellular respiration

Answer:

  1. NADH+H+ and FADH2
  2. Oxidative phosphorylation
  3. Anaerobic and aerobic respiration

Question 9.
What is anaerobic respiration?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.

Question 10.
What is glycolysis? Describe various steps involved in glycolysis.
OR
Describe the process involved in formation of pyruvic acid from glucose molecule.
OR
What is glycolysis? Describe the process of glycolysis with the help of schematic representation.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.
Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 2

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 11.
Write down the overall chemical reaction of a process in which glucose in broken down to two molecules of pyruvic acid.
Answer:
Overall reaction of glycolysis:
Glucose + 2 ATP + 2iP + 4 ADP + 2 NAD+ → 2 Pyruvate + 2 ADP + 4 ATP + 2 NADH + 2 Ha+ + 2 H2O

Question 12.
Write explanatorv notes on: Glycolysis
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 13.
How glycolysis is regulated?
Answer:
Glycolysis is strongly regulated by the complex interplay between ATP consumption, NADH2 regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 14.
Where does glycolysis take place in a cell?
Answer:
Glycolysis takes place in the cytoplasm of a cell.

Question 15.
What are the products of cleavage in glycolysis?
Answer:
Dihydroxyacetone phosphate (DHAP) and 3-phosphoglyceraldehyde (3-PGAL) are the products of cleavage in glycolysis.

Question 16.
Where does dehydration occur in glycolysis?
Answer:
In glycolysis, dehydration occurs when 2-Phosphoglyceric acid loses a water molecule (dehydration) to form phosphoenol pyruvic acid in presence of the enzyme enolase.

Question 17.
Name the process which is common to both aerobic and anaerobic respiration.
Answer:
Glycolysis is common to both aerobic and anaerobic respiration.

Question 18.
Name the enzymes that catalyse the irreversible reactions.
Answer:
Hexokinase, Phosphoffuctokinase, Phosphoglycerate kinase and Pyruvate kinase are the enzymes that catalyse the irreversible reactions.

Question 19.
Where glycolysis is the only source of energy production?
Answer:
Glycolysis is only source of energy production in erythrocytes, renal medulla, brain and sperm.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 20.
Which type of muscle fibres are rich in myoglobin?
Answer:
Red muscles are rich in myoglobin.

Question 21.
Which type of muscle fibre mainly performs anaerobic respiration?
Answer:
White muscle fibres mainly performs anaerobic respiration.

Question 22.
What is lactic acid fermentation?
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H+ to form lactic acid as the product and NAD+ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process. It is represented as follows:
Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 3

Question 23.
What is the fate of pyruvate in yeast?
Answer:
Yeast shows both aerobic and anaerobic respiration depending upon the presence or absence of oxygen.
1. In absence of oxygen, the pyruvate undergoes anaerobic respiration where it is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H to ethanol and carbon dioxide. This type of anaerobic respiration is termed alcoholic fermentation.
2. In the presence of oxygen however, it can respire aerobically to produce C02 and H20.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 24.
What is alcoholic fermentation?
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H+ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation. It is represented as follows:
Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 4

Question 25.
Name the process of respiration which does not involve intake of oxygen (02) and release of carbon dioxide (C02).
Answer:
Lactic acid fermentation is the process of anaerobic respiration which does not involve intake of oxygen (02) and release of carbon dioxide (C02).

Question 26.
Why yeast stops multiplying in the culture after alcoholic fermentation?
Answer:
After alcoholic fermentation the multiplication of yeast stops because the alcohol formed during the process kills the yeast cells.

Question 27.
What is aerobic respiration?
Answer:

  1. Aerobic respiration occurs in the presence of free molecular oxygen during oxidation of glucose.
  2. In this type of respiration, the glucose is completely oxidized to C02 and H20 with release of large amount of energy.
  3. It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Question 28.
Where does aerobic respiration occur in eukaryotic cell?
Answer:
Aerobic respiration occurs in the mitochondria in eukaryotes.

Question 29.
Explain the conversion of pyruvic acid to acetyl CoA.
OR
Describe the connecting link between glycolysis and Krebs cycle.
Answer:

  1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
  2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
  3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 5

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 30.
Name the following.
1. Which enzymes converts pyruvic acid to acetyl CoA?
2. Name the coenzyme required by pyruvate dehydrogenase.
Answer:
1. Pyruvate dehydrogenase
2. Thiamin (vitamin Bi)

Question 31.
Why is acetyl Co-A called connecting link between glycolysis and Krebs cycle?
Answer:

  1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
  2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
  3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 32.
Why vitamin Bi is important for maintaining good health?
Answer:
1. Pyruvate dehydrogenase (PDH) requires thiamin (vitamin Bi) as a co-enzyme. It cannot function in absence of vitamin B1.
2. Thiamin deficiency causes many disorders such as pyruvic acidosis and lactic acidosis, which are life threatening conditions. Hence, it is required to maintain good health.

Question 33.
Describe Citric acid cycle.
OR
With the help of schematic representation explain Krebs cycle.
Answer:

  1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
  2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
  3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
  4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
  5. This reaction is an oxidative decarboxylation process and produces H+ ions and electrons along with carbon dioxide.
  6. During the process NAD+ is reduced to NADH+H+.
  7. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
  8. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
  9. Regeneration of oxaloacetic acid occurs to complete the cycle.
  10. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD+ or FAD+ as the coenzyme.
  11. The coenzymes are consequently reduced to NADH+H+ and FADH2 respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
  12. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Question 34.
What is the site of Krebs cycle in mitochondria?
Answer:
Krebs cycle takes place in the mitochondrial matrix.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 35.
Match the following:

Column I Column II Column III
1. Krebs cycle (a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate (p) Pyruvate dehydrogenase
2. Glycolysis (b) Succinyl CoA → Succinate (q) NADH dehydrogenase
3. ETS (c) Pyruvic acid → Acetyl CoA (r) Two ATP are produced
4. Link reaction (d) Transfer of electrons from Complex to Ubiquinone (s) Succinyl CoA – synthetase

Answer:

Column I Column II Column III
1. Krebs cycle (b) Succinyl CoA → Succinate (s) Succinyl CoA – synthetase
2. Glycolysis (a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate (r) Two ATP are produced
3. ETS (d) Transfer of electrons from Complex I to Ubiquinone (q) NADH dehydrogenase
4. Link reaction (c) Pyruvic acid → Acetyl CoA (p) Pyruvate dehydrogenase

Question 36.
Explain ETS.
OR
Illustrate the mechanism of electron transport system.
OR
Give an account of ATP generation steps during ETS.
OR
Explain the mechanism of electron transportation system (ETS).
OR
Explain terminal oxidation.
Answer:

  1. NADH2 and FADH2 produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
  2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
  3. NADH+H+ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
  4. FADH2 is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
  5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube (Tube B).
  6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
  7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
  8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
  9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
  10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Question 37.
Explain the significance of electron transport system.
Answer:
Significance of ETS:

  1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
  2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
  3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H+ and FADH2) for recycling.
  4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
  5. ETS produces water molecules.

Question 38.
What is oxidative phosphorylation?
Answer:
Oxidative phosphorylation: It is a metabolic pathway that uses energy released by the oxidation of substrates to produce ATP.

  1. Oxidative phosphorylation takes place in the mitochondrial membrane.
  2. Many intermediate products during respiration are oxidised and release 2H+.
  3. The released hydrogen is trapped by NAD+ or FAD+.
  4. Electrons pass through electron transport system to produce ATP and metabolic water.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 39.
Give the balance sheet of ATP formed in aerobic respiration.
Answer:

Step of Respiration Consumption Production Total Net
benefit
Substrate level phosphorylation 1 Oxidative Phosphorylation
NADH+ H+ FADH2 Total
Glycolysis 2 4 2 × 3 = 6 ___________ 6 10 8
Pyruvate → AcetylCoA _______________ ___________ 2 × 3 = 6 ___________ 6 6 6
Krebs cycle _______________ 1 × 2 = 2 6 × 3 = 18 2×2 = 4 22 24 24
Total 2             [6]             30 + 4 = [34]             m             38

Question 40.
With the help of an experiment explain how yeast respires anaerobically.
Answer:
Respiration in yeast can be demonstrated with the help of an experiment.
Anaerobic respiration in yeast:

  1. A pinch of dry baker’s yeast suspended in water containing 10ml of 10% glucose in a test tube (test tube A).
  2. The surface of the liquid is covered with oil to prevent entry of air and the test tube is closed tightly with rubber stopper to prevent leakage.
  3. One end of a short-bent glass tube is inserted through it to reach the air inside the tube.
  4. Other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper.
  5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube
    (Tube B).
  6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
  7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
  8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
  9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
  10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 6

Question 41.
With the help of an experiment explain how germinating seeds respire.
Answer:

  1. Seed coats of a few germinating seeds (peas, beans or gram) are removed and are then put in a test tube filled with mercury.
  2. After closing the test tube with the thumb, it is vertically inverted in a trough of mercury and the thumb is carefully removed.
  3. Being lighter than mercury, the seeds rise to the closed upper end of the test Uibe. No gas is seen at first in the test tube.
  4. As germination proceeds, a gas begins to collect at the top of the mercury in the test tube.
  5. On introducing a pellet of potassium hydroxide into the tube, it rises to the top and absorbs the gas. The mercury again fills the tube.
  6. The potassium hydroxide reacts with carbon dioxide gas to produce potassium carbonate and water.
  7. The gas therefore disappears. Evidently germinating seeds produce carbon dioxide by anaerobic respiration in the absence of oxygen in the mercury column.

[Note: Mercury is toxic chemical. Proper precautions must be taken while performing this experiment ]
[Note: Students can scan the adjacent QR code to get conceptual clarity with the aid of a relevant example.]

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 42.
Give the significance of respiration.
OR
Write a short note on significance of respiration.
Answer:

  1. Respiration provides energy for biosynthesis of biomolecules.
  2. It is also a source of energy for cell division, growth, repairs and replacement of worn out parts, movements,
    locomotion etc.
  3. Various intermediates of Krebs cycle are used as building blocks for synthesis of other complex compounds.
  4. Coupled with photosynthesis, it helps to maintain the balance between C02 and O2 in the atmosphere.
  5. Anaerobic respiration (fermentation) is used in various industries such as dairies, bakeries, distilleries, leather industries, paper industries etc. It is used in the commercial production of alcohol, organic acids, vitamins, antibiotics etc.

Question 43.
Match the following:

Column I Column II Column III
1. Krebs cycle (a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate (p) Pyruvate dehydrogenase
2. Glycolysis (b) Succinyl CoA Succinate (q) NADH dehydrogenase
3. ETS (c) Pyruvic acid → Acetyl CoA (r) Two ATP are produced
4. Link reaction (d) Transfer of electrons from Complex I to Ubiquinone (s) Succinyl CoA – synthetase

Answer:

Column I Column II Column III
1. Krebs cycle (b) Succinyl CoA Succinate (s) Succinyl CoA – synthetase
2. Glycolysis (a) 1,3-Bisphosphoglycerate → 3-Phosphoglycerate (r) Two ATP are produced
3. ETS (d) Transfer of electrons from Complex I to Ubiquinone (q) NADH dehydrogenase
4. Link reaction (c) Pyruvic acid → Acetyl CoA (p) Pyruvate dehydrogenase

Question 44.
Apply Your Knowledge

Question 1.
While teaching respiration professor said that oxygen is important for respiration in humans, Rakesh kept thinking, where exactly oxygen is required during cellular respiration? What would be the correct explanation for his doubt?
Answer:
Oxygen is very important for life, we humans cannot survive without oxygen. Glycolysis, link reaction and Krebs cycle do not involve oxygen. In cellular reaction, oxygen is required only during ETS where it acts as an electron acceptor hence leading to terminal oxidation where it gets converted into water called metabolic water.

The process of respiration is very fast and occurs continuously as cell require continuous supply of energy to carry out metabolic activities, thus, we require oxygen for cellular respiration even when we are sleeping.

Question 2.
Sonal while studying ETS had a doubt, why FADH2 yields only 2 ATP’s whereas NADH2 yields three ATP’s? What would be the correct explanation for her doubt?
Answer:

  1. During ETS, NADH2 and FADH2 undergoes oxidation where they are converted to NAD+ and FAD+ respectively.
  2. During this conversion, hydrogen is removed, which latter forms proton and electron. This electron so formed is passed on to different electron carriers. During this transfer of electron, energy is released which is used in ATP synthesis.
  3. In case of NADH2, energy is released at three places hence results into formation of 3 ATPs whereas, in case of FADH2 energy is released at only two places, hence results into formation of only 2 ATPs.
  4. Here, the ATP formation occurs by Chemiosmotic hypothesis.

Question 45.
Quick Review

Various steps involved in glycolysis:

Step(s) Substrate Enzyme End product(s)
Phosphorylation Glucose + ATP Hexokinase Glucose – 6 – Phosphate + ADP
Isomerisation Glucose – 6-Phosphate Phosphohexose isomerase Fructose – 6 – Phosphate
Phosphorylation Fructose – 6-Phosphate + ATP (Phosphate donor) Phosphofructokinase Fructose 1,6-Diphosphate + ADP
Cleavage Fructose -1, 6-Diphosphate Aldolase 3 – Phosphoglyceraldehyde + Dihydroxyacetone phosphate
Phosphorylation and Dehydrogenation 3-PGAL + H3PO4 + NAD+ Triosephosphate

dehydrogenase

1, 3 Diphosphoglyceric acid + NADH + H+
Dephosphorylation 1, 3-DPGA + ADP Diphosphoglycerate kinase (Mg2) ATP + 3-Phosphoglyceric acid
Rearrangement 3-PGA Phosphoglycerate mutase 2-Phosphoglyceric acid
Dehydration 2-PGA Enolase Phosphoenol Pyruvic acid + H20
Dephosphorylation Phosphoenol Pyruvic acid +ADP Pyruvate kinase Pyruvic acid + ATP

Various steps involved in Krebs cycle:

Step(s) Substrate Enzyme End product(s)
Condensation Acetyl-CoA+Oxalo- acetic acid + H20 Citrate synthase Citric acid (6C) + Coenzyme- A
Dehydration
Hydration
It is again completed in two steps:
Citric acid
Cis-Aconitic acid + H20
Aconitase
Aconitase
Cis-Aconitic acid + H20 Iso-citric acid(6C)
Oxidative decarboxylation a. Isocitric acid + NAD+ Isocitrate dehydrogenase Oxalo succinic acid(6C) + NADH + H+
b. Oxalosuccinic acid Isocitrate dehydrogenase a – ketoglutaric acid(5C) + CO
Oxidative Decarboxylation. (-2H) (-C02) a-Ketoglutaric acid + H20 + NAD + Coenzyme A a-Ketoglutarate dehydrogenase complex Succinyl-CoA(4C) + NADH + H++co2
Substrate level phosphorylation Succinyl – Co A + GDP + iP Succinyl CoA synthetase Succinic acid (4C) + Coenzyme-A + GTP
Dehydrogenation (-2H) Succinic acid + FAD (H-acceptor) Succinate dehydrogenase Fumaric acid (4C) + FADH2
Hydration(+H20) Fumaric acid + H20 Hydration(+H20) Fumaric acid + H20
Dehydrogenation (-2H) Malic acid + NAD+ (H-acceptor) Dehydrogenation (-2H) Malic acid + NAD+ (H-acceptor)

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer 7

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 46.
Exercise

Question 1.
Define phosphorylation.
Answer:
It is the formation of ATP, by addition inorganic phosphate to ADP.
ADP + Pi → ATP

Question 2.
Mention the different ways of phosphorylation.
Answer:
Phosphorylation occurs in three different ways as – photophosphorylation, substrate-level phosphorylation and oxidative phosphorylation.

Question 3.
Define substrate-level phosphorylation.
Answer:
Substrate-level phosphorylation is a direct phosphorylation of ADP by transfer of a phosphate group from any suitable substrate. It occurs in cytoplasm of the cells and matrix of mitochondria.

Question 4.
What is fermentation?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.

Question 5.
Where does glycolysis occurs?
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 6.
What is glycolysis? Explain with the help of schematic representation.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question 7.
Define fermentation. What are the different types of fermentation?
Answer:
1. Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation.
2. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol.
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H+ to form lactic acid as the product and NAD+ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H+ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question 8.
Name the products of lactic acid fermentation.
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H+ to form lactic acid as the product and NAD+ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.

Question 9.
Write the chemical reaction of lactic acid fermentation.
Answer:
It is a process of anaerobic respiration where the pyruvic acid undergoes reduction by addition of one proton and two electrons donated by NADH+H+ to form lactic acid as the product and NAD+ as the byproduct of oxidation. Skeletal muscles usually derive energy by this process.

Question 10.
Explain alcoholic fermentation.
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H+ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 11.
Less energy is produced during anaerobic respiration that in aerobic respiration. Justify.
Answer:
Anaerobic respiration produces less energy because:

  1. Incomplete breakdown of respiratory substrate takes place.
  2. Some of the products of anaerobic respiration can be oxidised further to release energy which shows that anaerobic respiration does not liberate the whole energy contained in the respiratory substrate.
  3. NADH2 does not produce ATP, as electron transport is absent.
  4. Only 2 ATP molecules are generated from one molecule of glucose during anaerobic respiration.

Question 12.
Enlist steps involved in aerobic respiration.
Answer:
It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Question 13.
Write the reaction of connecting link between glycolysis and Krebs cycle.
Answer:
This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 14.
How glycolysis is regulated?
Answer:
Glycolysis is strongly regulated by the complex interplay between ATP consumption, NADH2 regeneration and regulation of various glycolytic enzymes like hexokinase, PFK-1, pyruvate kinase, etc. Besides, it is also controlled by hormones like glucagon, epinephrine and insulin.

Question 15.
Write the significance of ETS.
Answer:
Significance of ETS:

  1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
  2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
  3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H+ and FADH2) for recycling.
  4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
  5. ETS produces water molecules.

Question 16.
Give the schematic representation of glycolysis.
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 17.
Describe the formation of Acetyl Co-A in respiration.
Answer:

  1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
  2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
  3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 18.
Give the schematic representation of the overall view of Krebs cycle.
Answer:

  1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
  2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
  3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
  4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
  5. This reaction is an oxidative decarboxylation process and produces H+ ions and electrons along with carbon dioxide. During the process NAD+ is reduced to NADH+H+.
  6. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
  7. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
  8. Regeneration of oxaloacetic acid occurs to complete the cycle.
  9. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD+ or FAD+ as the coenzyme.
  10. The coenzymes are consequently reduced to NADH+H+ and FADH2 respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
  11. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Question 19.
Explain why the respiratory pathway is an amphibolic pathway?
Answer:

  1. Respiration is considered as a catabolic process; however, it is not entirely correct in case of Krebs cycle.
  2. Many reactions of Krebs cycle involve oxidation of acetyl CoA to release energy and C02.
  3. However, the breakdown of respiratory substrates provides intermediates like a-ketoglutarate, oxaloacetate are used as precursors for synthesis of fatty acids, glutamic acid and aspartic acid respectively.
  4. Thus, as the same respiratory process acts as catabolic as well as anabolic pathway for synthesis of various intermediate metabolic products, it is called amphibolic pathway.

Question 20.
Write a detailed note on the connecting link between glycolysis and Krebs cycle.
Answer:

  1. The conversion of pyruvic acid to acetyl CoA is an oxidative decarboxylation reaction.
  2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.
  3. This reaction is called as ‘connecting link’ reaction between glycolysis and Krebs cycle.

Question 21.
Give the diagrammatic representation of ETS.
Answer:

  1. NADH2 and FADH2 produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
  2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
  3. NADH+H+ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
  4. FADH2 is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
  5. During oxidation of NADH+H+ and FADH2, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
  6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
  7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
  8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a3. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
  9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
  10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F0 part of oxysome (complex V) present in inner mitochondrial membrane.
  11. This proton channelling by F0 is coupled to catalytic site of F1 which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
  12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 22.
Explain the process of terminal oxidation. Give its significance (any two points).
Answer:

  1. NADH2 and FADH2 produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
  2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
  3. NADH+H+ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
  4. FADH2 is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
  5. During oxidation of NADH+H+ and FADH2, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
  6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
  7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
  8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a3. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
  9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
  10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F0 part of oxysome (complex V) present in inner mitochondrial membrane.
  11. This proton channelling by F0 is coupled to catalytic site of F1 which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
  12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question 23.
1. Enlist the main steps involved in aerobic respiration and their place of occurrence.
2. Differentiate between photosynthesis and aerobic respiration.
Answer:
1. Glycolysis takes place in the cytoplasm of a cell.
2. It is catalyzed by a multienzyme complex-pyruvate dehydrogenase complex (PDH). This enzyme is present in mitochondria of eukaryotes and cytosol of prokaryotes.,
3. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria. and These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.

Question 24.
1. Give the schematic representation of ETS.
2. Differentiate between respiration and combustion.
Answer:

  1. NADH2 and FADH2 produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
  2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
  3. NADH+H+ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
  4. FADH2 is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
  5. During oxidation of NADH+H+ and FADH2, electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
  6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C.
  7. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
  8. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a3. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
  9. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
  10. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F0 part of oxysome (complex V) present in inner mitochondrial membrane.
  11. This proton channelling by F0 is coupled to catalytic site of F1 which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
  12. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 25.
Enlist the different respiratory substrate.
Answer:
Respiratory substrates are the molecules that are oxidized during respiration to release energy which can be used for ATP synthesis. Carbohydrates, fats and proteins are the common respiratory substrate. Glucose is the most common respiratory substrate.

Question 26.
Define R.Q. What is its value for fats?
Answer:
1. Respiratory quotient (R.Q.) or respiratory ratio is the ratio of volume of CO2 released to the volume of O2 consumed in respiration.
2. R.Q. = Volume of CO2 released/Volume of O2 consumed

Question 27.
Write the significance of respiration.
Answer:

  1. Respiration provides energy for biosynthesis of biomolecules.
  2. It is also a source of energy for cell division, growth, repairs and replacement of worn out parts, movements,
    locomotion etc.
  3. Various intermediates of Krebs cycle are used as building blocks for synthesis of other complex compounds.
  4. Coupled with photosynthesis, it helps to maintain the balance between C02 and O2 in the atmosphere.
  5. Anaerobic respiration (fermentation) is used in various industries such as dairies, bakeries, distilleries, leather industries, paper industries etc. It is used in the commercial production of alcohol, organic acids, vitamins, antibiotics etc.

Question 47.
Multiple Choice Questions:

Question 1.
Respiration is regarded as a ________ process.
(A) catabolic
(B) anabolic
(C) reduction
(D) synthetic
Answer:
(A) catabolic

Question 2.
Anaerobic respiration is also called as ________ .
(A) Glycolysis
(B) fermentation
(C) phosphoryaltion
(D) decarboxylation
Answer:
(B) fermentation

Question 3.
The first compound formed in glycolysis is
(A) Glucose-6-phosphate
(B) Glucose-1,6-biphosphate
(C) Fructose-6-phosphate
(D) Pyruvic acid
Answer:
(A) Glucose-6-phosphate

Question 4.
Which of the following compounds in last step of glycolysis gives pyruvic acid?
(A) 3 – phosphoglyceraldehyde
(B) dihydroxy acetone phosphate
(C) phosphoenolpyruvate
(D) 2-phosphoglycerate
Answer:
(C) phosphoenolpyruvate

Question 5.
In glycolysis, dehydration occurs during formation of
(A) 3 – phosphoglyceraldehyde
(B) 2 – phosphoglycerate
(C) phosphoenolpyruvate
(D) dihydroxyacetone phosphate
Answer:
(B) 2 – phosphoglycerate

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 6.
In which of the following steps dehydrogenation occurs?
(A) Glucose → Glucose 6-phosphate
(B) 3-phophoglcerate → 2-phophoglccratc
(C) phosphoenolpyruvate → pyruvate
(D) 3 – phosphoglyceraldehyde → 1, 3-bisphosphoglycerate
Answer:
(D) 3 – phosphoglyceraldehyde → 1, 3-bisphosphoglycerate

Question 7.
The compound common to both aerobic and anaerobic respiration is
(A) C02
(B) pyruvic acid
(C) acetyl CoA
(D) free oxygen
Answer:
(B) pyruvic acid

Question 8.
Which compound is found both in respiration and photosynthesis?
(A) Phosphoglycerate
(B) Phosphoglyceraldehyde
(C) Both (A) and (B)
(D) Succinic acid
Answer:
(C) Both (A) and (B)

Question 9.
Which type of respiration does not release C02?
(A) Aerobic respiration
(B) Alcoholic fermentation
(C) Lactic acid fermentation
(D) Krebs cycle
Answer:
(C) Lactic acid fermentation

Question 10.
What is the overall goal of glycolysis, Krebs cycle and electron transport system?
(A) Synthesis of ATP in fermentation reaction
(B) Carbohydrates
(C) Nucleic acids
(D) ATP in small stepwise units
Answer:
(D) ATP in small stepwise units

Question 11.
The intermediate between glycolysis and TCA cycle is:
(A) Lactic acid
(B) Acetaldehyde
(C) Fructose-6-phosphate
(D) Acetyl Co-A
Answer:
(D) Acetyl Co-A

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 12.
During alcoholic fermentation, decarboxylation of pyruvate gives
(A) acetaldehyde
(B) lactic acid
(C) ethyl alcohol
(D) methyl alcohol
Answer:
(A) acetaldehyde

Question 13.
Where the link reaction occurs in prokaryotes?
(A) cytoplasm
(B) mitochondrial matrix
(C) cell membrane
(D) mitochondrial membrane
Answer:
(A) cytoplasm

Question 14.
In Krebs cycle, dehydration of substrate occurs
(A) once
(B) twice
(C) thrice
(D) four times
Answer:
(A) once

Question 15.
Which of the following steps generate ATP without ETS?
(A) Pyruvic acid → Acetyl Co-A
(B) ∝-ketoglutarate → Succinic acid
(C) Iso-citric acid → Oxalosuccinic acid
(D) Succinyl Co-A → Succinic acid
Answer:
(D) Succinyl Co-A → Succinic acid

Question 16.
In Krebs cycle, the acid which undergoes oxidative decarboxylation is
(A) citric acid
(B) succinic acid
(C) malic acid
(D) ∝-ketoglutaric acid
Answer:
(D) ∝-ketoglutaric acid

Question 17.
During Krebs cycle, fumaric acid gets converted into malic acid by
(A) decarboxylation
(B) dehydrogenation
(C) dehydration
(D) hydration
Answer:
(D) hydration

Question 18.
Krebs cycle is also called TCA cycle because
(A) the first compound formed is citric acid.
(B) it was discovered by Sir Hans Krebs.
(C) organic acids formed have 3 carboxylic acid groups.
(D) acetyl Co-A is formed
Answer:
(C) organic acids formed have 3 carboxylic acid groups.

Question 19.
Which of the following compound is the acceptor of Acetyl Co-A in Krebs cycle?
(A) Oxaloacetic acid
(B) Fumaric acid
(C) Malic acid
(D) Oxalo succinic acid
Answer:
(A) Oxaloacetic acid

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 20.
Which of the following compounds is formed in Krebs cycle from fumaric acid?
(A) Oxalo acetic acid
(B) Malic acid
(C) ∝-KGA
(D) Citric acid
Answer:
(B) Malic acid

Question 21.
Which of the following step of aerobic respiration would be omitted when fatty acids are used as respiratory substrate?
(A) Glycolysis
(B) Krebs cycle
(C) Electron transfer chain reaction
(D) Terminal oxidation.
Answer:
(A) Glycolysis

Question 22.
During Krebs cycle, decarboxylation occurs _______ time/s.
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(B) 2

Question 23.
The conversion of malic acid to oxalo acetic acid is catalyzed by
(A) malate reductase
(B) malate thiokinase
(C) fumarase
(D) malate dehydrogenase
Answer:
(D) malate dehydrogenase

Question 24.
Electron carriers of oxidative phosphorylation are present on
(A) outer membrane of mitochondria.
(B) inner membrane of mitochondria.
(C) thylakoid membrane of chloroplast.
(D) matrix of mitochondria.
Answer:
(B) inner membrane of mitochondria.

Question 25.
Which of the following derives maximum energy per molecule of glucose?
(A) Alcoholic fermentation.
(B) Lactic acid fermentation.
(C) Aerobic respiration in unicellular organisms.
(D) Glycolysis in liver cells.
Answer:
(C) Aerobic respiration in unicellular organisms.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 26.
The cytochrome which donates de-energised electron to oxygen is
(A) cytochrome-a
(B) cytochrome-b
(C) cytochrome-a3
(D) cytochrome-c
Answer:
(C) cytochrome-a3

Question 27.
In terminal oxidation FADH2 is oxidized by
(A) complex I
(B) complex II
(C) complex III
(D) complex IV
Answer:
(B) complex II

Question 28.
The net gain of energy from a molecule of glucose in the aerobic respiration is
(A) 38
(B) 35
(C) 70
(D) 76
Answer:
(A) 38

Question 29.
Each molecule of NADH2 through ETS yields
(A) 1 ATP
(B) 2 ATPs
(C) 3 ATPs
(D) 4 ATPs
Answer:
(C) 3 ATPs

Question 30.
One glucose molecule, through ETS yields
(A) 2 ATP molecules
(B) 3 ATP molecules
(C) 34 ATP molecules
(D) 38 ATP molecules
Answer:
(C) 34 ATP molecules

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 31.
The adenosine triphosphate (ATP) gain during glycolysis, connecting link and Krebs cycle respectively are _______ .
(A) 8, 6,24
(B) 8,24, 6
(C) 24, 8, 6
(D) 6, 8, 24
Answer:
(A) 8, 6,24

Question 32.
The respiratory quotient (R.Q.) of carbohydrate is ________ .
(A) 0.7
(B) 1
(C) 0.9
(D) 0.1
Answer:
(B) 1

Question 33.
R. Q. for proteins is about _______ .
(A) 0.7
(B) 0.8
(C) 0.9
(D) 1.0
Answer:
(C) 0.9

Question 48.
Competitive Corner:

Question 1.
Which of the following statements regarding mitochondria is INCORRECT?
(A) Inner membrane is convoluted with infoldings.
(B) Mitochondrial matrix contains single circular DNA molecule and ribosomes.
(C) Outer membrane is permeable to monomers of carbohydrates, fats and proteins.
(D) Enzymes of electron transport are embedded in outer membrane.
Hint: Enzymes of electron transport are embedded in inner membrane.
Answer:
(D) Enzymes of electron transport are embedded in outer membrane.

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 2.
Where is the respiratory electron transport system (ETS) located in plants?
(A) Intermembrane space
(B) Mitochondrial matrix
(C) Outer mitochondrial membrane
(D) Inner mitochondrial membrane
Answer:
(D) Inner mitochondrial membrane

Question 3.
In case of anaerobic respiration, the R.Q is _____ .
(A) always less than one
(B) always more than one
(C) always infinity
(D) Variable on the basis of substrate.
Answer:
(C) always infinity

Question 4.
The net gain of ATP molecules during aerobic breakdown of one glucose molecule is _______ .
(A) 40
(B) 38
(C) 36
(D) 30
Answer:
(B) 38

Question 5.
During glycolysis the compounds PGAL and DHAP are formed from fructose 1,6- diphosphate by _______ .
(A) cleavage
(B) isomerisation
(C) phosphorylation
(D) condensation
Answer:
(A) cleavage

Maharashtra Board Class 11 Biology Important Questions Chapter 13 Respiration and Energy Transfer

Question 6.
Number of oxygen molecules utilized in glycolysis is ________ .
(A) 0
(B) 2
(C) 4
(D) 6
Answer:
(A) 0

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Balbharati Maharashtra State Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law Important Questions and Answers.

Maharashtra State Board 11th Information Technology Important Questions Chapter 4 Cyber Law

1. Fill in the Blanks.

Question 1.
__________ is the area of law that deals with the Internet’s relationship to technological and electronic elements.
Answer:
Cyberlaw

Question 2.
__________ is the part of the overall legal system that deals with the Internet, cyberspace, and their respective legal issues.
Answer:
Cyberlaw

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 3.
__________ covers a fairly broad area, encompassing several subtopics including freedom of expression, access to and usage of the Internet, and online privacy.
Answer:
Cyberlaw

Question 4.
The attempts to change outdated laws and provides ways to deal with cybercrimes.
Answer:
Information Technology Act 2000

Question 5.
The Act offers the much-needed legal framework so that information is not denied.
Answer:
Information Technology

Question 6.
The Act has also proposed a legal framework for the authentication and origin of an electronic records.
Answer:
Information Technology

Question 7.
__________ is also called moral philosophy is the discipline concerned with what is morally good and bad, right or wrong.
Answer:
Ethics

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 8.
__________ is the standards of behaviour; principles of right and wrong behaviour.
Answer:
Morals

Question 9.
Thus __________ are dictated by society, culture or religion.
Answer:
Morals

Question 10.
Computer Crime is alternatively referred to as __________
Answer:
Cybercrime

Question 11.
Computer crime is an act performed by a knowledgeable computer user, sometimes referred to as a __________
Answer:
Hacker

Question 12.
A __________ who illegally browses or steals a company’s or individual’s private information.
Answer:
Hacker

Question 13.
__________ can involve criminal activities that are traditional in nature, such as theft, fraud, forgery, defamation and mischief.
Answer:
Cybercrimes

Question 14.
__________ is nothing but copyright violation of software created originally by an individual or an institution.
Answer:
Software piracy

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 15.
__________ includes stealing of codes/programs and other information illegally and creating the imitated copy by unauthorized means.
Answer:
Software piracy

Question 16.
Gaining access without the users’ permission is known as __________
Answer:
Unauthorized Access

Question 17.
__________ means granting access rights to resources.
Answer:
Authorization

Question 18.
A __________ is a legal right that gives the creator of a literary, artistic, musical, or other creative work.
Answer:
Copyright

Question 19.
__________ owners have the right to control the reproduction of their work, including the right to receive the royalty payment for that reproduction.
Answer:
Copyright

Question 20.
Activity such as decipher codes or passwords and breaking security systems for illegal reasons is called __________
Answer:
Cracking

Question 21.
The cracker will use a program or script known as a __________
Answer:
Crack

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 22.
__________ is a criminal practice where an individual uses the Internet to systematically harass or threaten someone.
Answer:
Cyberstalking

Question 23.
__________ is a technique of extracting confidential information such as credit card numbers and username password combos by pretending as a legal enterprise.
Answer:
Phishing

Question 24.
__________ is typically carried out by email spoofing.
Answer:
Phishing

Question 25.
__________ is presenting someone else’s work or idea as your own without their consent.
Answer:
Plagarism

Question 26.
__________ refers to unauthorised intrusion into a computer or a network.
Answer:
Hacking

Question 27.
__________ is a person intensely interested in the deep and hidden or concealed working of any computer operating system and programming language.
Answer:
Hacker

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 28.
__________ is a method of converting the original message into random text.
Answer:
Encryption

Question 29.
Full form of SSL is __________
Answer:
Secure Socket Layer

Question 30.
Through the __________, transmission of data is encrypted.
Answer:
Secure Socket Layer

Question 31.
__________ refers to a network security (Hardware and Software) system which blocks certain types of information.
Answer:
Firewall

Question 32.
__________ is an Act to provide legal recognition for transactions carried out by means of electronic data interchange and other means of electronic communication.
Answer:
Information Technology Act, 2000

Question 33.
__________ aims to provide the legal infrastructure for E-Commerce in India.
Answer:
Information Technology Act, 2000

Question 34.
The __________ have a major impact on E-Businesses and the new economy in India.
Answer:
Cyber laws

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 35.
The __________ also aims to provide for the legal framework.
Answer:
Information Technology Act, 2000

Question 36.
__________ has been replaced with the electronic signature to make it a more technology-neutral act.
Answer:
Digital Signature

2. True or False.

Question 1.
Criminal Law is the area of law that deals with the Internet’s relationship to technological and electronic elements.
Answer:
False

Question 2.
Cyberlaw is the part of the overall legal system that deals with the Internet, cyberspace, and their respective legal issues.
Answer:
True

Question 3.
Cyberlaw covers a fairly broad area, encompassing several subtopics including freedom of expression, access to and usage of the Internet, and online privacy.
Answer:
True

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 4.
The Information Technology Act 2000 attempts to change outdated laws and provides ways to deal with cybercrimes.
Answer:
True

Question 5.
The Information Technology Act offers the much-needed legal framework so that information is not denied.
Answer:
True

Question 6.
The Computer Act has also proposed a legal framework for the authentication and origin of an electronic records.
Answer:
False

Question 7.
Morals are also called moral philosophy is the discipline concerned with what is morally good and bad, right or wrong.
Answer:
False

Question 8.
Ethics is the standards of behaviour; principles of right and wrong behaviour.
Answer:
False

Question 9.
Thus Morals are dictated by society, culture, or religion.
Answer:
True

Question 10.
Computer Crime is alternatively referred to as Hacker.
Answer:
False

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 11.
Computer crime is an act performed by a knowledgeable computer user, sometimes referred to as a Hacker.
Answer:
True

Question 12.
A Cybercrime who illegally browses or steals a company’s or individual’s private information.
Answer:
False

Question 13.
Cybercrimes can involve criminal activities that are traditional in nature, such as theft, fraud, forgery, defamation, and mischief.
Answer:
True

Question 14.
Software piracy is nothing but copyright violation of software created originally by an individual or an institution.
Answer:
True

Question 15.
Unauthorized access includes stealing of codes/programs and other information illegally and creating the imitated copy by unauthorized means.
Answer:
False

Question 16.
Gaining access without the user’s permission is known as Software piracy.
Answer:
False

Question 17.
Authorization means granting access rights to resources.
Answer:
True

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 18.
Cracking is a legal right that gives the creator of a literary, artistic, musical, or other creative work.
Answer:
False

Question 19.
Copyright owners have the right to control the reproduction of their work, including the right to receive the royalty payment for that reproduction.
Answer:
True

Question 20.
Activity such as decipher codes or passwords and breaking security systems for illegal reasons is called copyright.
Answer:
False

Question 21.
The cracker will use a program or script known as a Crack.
Answer:
True

Question 22.
Cyberstalking is a criminal practice where an individual uses the Internet to systematically harass or threaten someone.
Answer:
True

Question 23.
Phishing is a technique of extracting confidential information such as credit card numbers and username password combos by pretending as a legal enterprise.
Answer:
True

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 24.
Phishing is typically carried out by email spoofing.
Answer:
True

Question 25.
Plagiarism is presenting someone else’s work or idea as your own without their consent.
Answer:
True

Question 26.
Hacking refers to unauthorized intrusion into a computer or a network.
Answer:
True

Question 27.
Hacker is a person intensely interested in the deep and hidden or concealed working of any computer operating system and programming language.
Answer:
True

Question 28.
Encryption is a method of converting the original message into random text.
Answer:
True

Question 29.
The full form of SSL is Secure Socket Layer.
Answer:
True

Question 30.
Through the Encryption transmission of data is encrypted.
Answer:
False

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 31.
Encryption refers to a network security (Hardware and Software) system which blocks certain types of information.
Answer:
False

Question 32.
A digital signature is an Act to provide legal recognition for transactions carried out by means of electronic data interchange and other means of electronic communication.
Answer:
False

Question 33.
Information Technology Act, 2000 aims to provide the legal infrastructure for E-Commerce in India.
Answer:
True

Question 34.
Cyber laws have a major impact on E-Businesses and the new economy in India.
Answer:
True

Question 35.
The Information Technology Act, 2000 also aims to provide for the legal framework.
Answer:
True

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 36.
Cyber laws have been replaced with electronic signatures to make it a more technology-neutral act.
Answer:
False

3. Multiple Choice Question (Single Choice)

Question 1.
__________ is the area of law that deals with the Internet’s relationship to technological and electronic elements.
(A) Cyberlaw
(B) Criminal Law
(C) Business Law
(D) Internet Law
Answer:
(A) Cyberlaw

Question 2.
__________ is the part of the overall legal system that deals with the Internet, cyberspace, and their respective legal issues.
(A) Cyberlaw
(B) Criminal Law
(C) Business Law
(D) Internet Law
Answer:
(A) Cyberlaw

Question 3.
__________ covers a fairly broad area, encompassing several subtopics including freedom of expression, access to and usage of the Internet, and online privacy.
(A) Cyberlaw
(B) Criminal Law
(C) Business Law
(D) Internet Law
Answer:
(A) Cyberlaw

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 4.
The __________ attempts to change outdated laws and provides ways to deal with cybercrimes.
(A) Information Technology Act 2000
(B) Cyber Rule
(C) Computer Act
(D) Legal Act
Answer:
(A) Information Technology Act 2000

Question 5.
The __________ Act offers the much-needed legal framework so that information is not denied.
(A) Cyber Rule
(B) Information Technology
(C) Computer Act
(D) Legal Act
Answer:
(B) Information Technology

Question 6.
The __________ Act has also proposed a legal framework for the authentication and origin of an electronic records.
(A) Cyber Rule
(B) Information Technology
(C) Computer Act
(D) Legal Act
Answer:
(B) Information Technology

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 7.
__________ is also called moral philosophy is the discipline concerned with what is morally good and bad, right or wrong.
(A) Ethics
(B) Morals
(C) Society
(D) Culture
Answer:
(A) Ethics

Question 8.
__________ is the standards of behaviour; principles of right and wrong behaviour.
(A) Ethics
(B) Morals
(C) Society
(D) Culture
Answer:
(B) Morals

Question 9.
Thus __________ are dictated by society, culture or religion.
(A) Ethics
(B) Morals
(C) Society
(D) Culture
Answer:
(B) Morals

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 10.
Computer Crime is alternatively referred to as __________
(A) Cybercrime
(B) Hacker
(C) Computer crime
(D) Illegal Act
Answer:
(A) Cybercrime

Question 11.
Computer crime is an act performed by a knowledgeable computer user, sometimes referred to as a __________
(A) Cybercrime
(B) Hacker
(C) Computer crime
(D) Illegal Act
Answer:
(B) Hacker

Question 12.
A __________ who illegally browses or steals a company’s or individual’s private information.
(A) Cybercrime
(B) Hacker
(C) Computer crime
(D) Illegal Act
Answer:
(B) Hacker

Question 13.
__________ can involve criminal activities that are traditional in nature, such as theft, fraud, forgery, defamation, and mischief.
(A) Cybercrimes
(B) Hacker
(C) Computer crime
(D) Illegal Act
Answer:
(A) Cybercrimes

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 14.
__________ is nothing but copyright violation of software created originally by an individual or an institution.
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(A) Software piracy

Question 15.
__________ includes stealing of codes/programs and other information illegally and creating the imitated copy by unauthorized means.
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(A) Software piracy

Question 16.
Gaining access without the users’ permission is known as __________
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(B) Unauthorized access

Question 17.
__________ means granting access rights to resources.
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(C) Authorization

Question 18.
A __________ is a legal right that gives the creator of a literary, artistic, musical, or other creative work.
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(D) Copyright

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 19.
__________ owners have the right to control the reproduction of their work, including the right to receive the royalty payment for that reproduction.
(A) Software piracy
(B) Unauthorized access
(C) Authorization
(D) Copyright
Answer:
(D) Copyright

Question 20.
Activity such as decipher codes or passwords and breaking security systems for illegal reasons is called __________
(A) Cracking
(B) Crack
(C) Cyberstalking
(D) Phishing
Answer:
(A) Cracking

Question 21.
The cracker will use a program or script known as a __________
(A) Cracking
(B) Crack
(C) Cyberstalking
(D) Phishing
Answer:
(B) Crack

Question 22.
__________ is a criminal practice where an individual uses the Internet to systematically harass or threaten someone.
(A) Cracking
(B) Crack
(C) Cyberstalking
(D) Phishing
Answer:
(C) Cyberstalking

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 23.
__________ is a technique of extracting confidential information such as credit card numbers and username password combos by pretending as a legal enterprise.
(A) Cracking
(B) Crack
(C) Cyberstalking
(D) Phishing
Answer:
(D) Phishing

Question 24.
__________ is typically carried out by email spoofing.
(A) Cracking
(B) Crack
(C) Cyberstalking
(D) Phishing
Answer:
(D) Phishing

Question 25.
__________ is presenting someone else’s work or idea as your own without their consent.
(A) Plagiarism
(B) Hacking
(C) Hacker
(D) A. Cracking
Answer:
(A) Plagarism

Question 26.
__________ refers to unauthorised intrusion into a computer or a network.
(A) Plagiarism
(B) Hacking
(C) Hacker
(D) Cracking
Answer:
(B) Hacking

Question 27.
__________ is a person intensely interested in the deep and hidden or concealed working of any computer operating system and programming language.
(A) Plagiarism
(B) Hacking
(C) Hacker
(D) Cracking
Answer:
(C) Hacker

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 28.
__________ is a method of converting the original message into random text.
(A) Encryption
(B) Secure Socket Layer
(C) Firewall
(D) Hardware
Answer:
(A) Encryption

Question 29.
Full form of SSL is __________
(A) Secure Socket Layer
(B) Secure Switch Layer
(C) Secure Session Lesson
(D) System Secure Layer
Answer:
(A) Secure Socket Layer

Question 30.
Through the __________, transmission of data is encrypted.
(A) Secure Socket Layer
(B) Secure Switch Layer
(C) Secure Session Lesson
(D) System Secure Layer
Answer:
(A) Secure Socket Layer

Question 31.
__________ refers to a network security (Hardware and Software) system which blocks certain types of information.
(A) Encryption
(B) Secure Socket Layer
(C) Firewall
(D) Hardware
Answer:
(C) Firewall

Question 32.
__________ is an Act to provide legal recognition for transactions carried out by means of electronic data interchange and other means of electronic communication.
(A) Information Technology Act, 2000
(B) Cyber laws
(C) Digital Signature
(D) Digital Certificate
Answer:
(A) Information Technology Act, 2000

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 33.
__________ aims to provide the legal infrastructure for e-commerce in India.
(A) Information Technology Act, 2000
(B) Cyber laws
(C) Digital signature
(D) Digital Certificate
Answer:
(A) Information Technology Act, 2000

Question 34.
The __________ have a major impact on E-Businesses and the new economy in India.
(A) Information Technology Act, 2000
(B) Cyber Laws
(C) Digital signature
(D) Digital Certificate
Answer:
(B) Cyber Laws

Question 35.
The also aims to provide for the legal framework.
(A) Information Technology Act, 2000
(B) Cyber Laws
(C) Digital signature
(D) Digital Certificate
Answer:
(A) Information Technology Act, 2000

Question 36.
__________ has been replaced with the electronic signature to make it a more technology-neutral act.
(A) Information Technology Act, 2000
(B) Cyber laws
(C) Digital Signature
(D) Digital Certificate
Answer:
(C) Digital Signature

4. Multiple Choice Two Correct Answers.

Question 1.
The Information Technology Act 2000 attempts to change outdated __________ and provides ways to deal with __________
(A) Authentication
(B) Laws
(C) Electronic record
(D) Cybercrimes
(E) Morals
Answer:
(B) Laws, (D) Cybercrimes

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 2.
The Information Technology Act has also proposed a legal framework for the __________ and origin of __________
(A) Authentication
(B) Laws
(C) Electronic record
(D) Cybercrimes
(E) Morals
Answer:
(A) Authentication, (C) Electronic record

Question 3.
Thus __________ are dictated by society, culture or religion while __________ are chosen by the person himself which governs his life.
(A) Authentication
(B) Laws
(C) Electronic record
(D) Morals
(E) Ethics
Answer:
(D) Morals, (E) Ethics

Question 4.
Hacker is a person intensely interested in the deep and hidden or concealed working of any computer __________ and __________
(A) Operating System
(B) Programming Language
(C) Individual
(D) Institution
(E) Passwords
Answer:
(A) Operating System, (B) Programming Language

Question 5.
Software piracy is nothing but copyright violation of software created originally by an __________ or an __________
(A) Operating system
(B) Programming language
(C) Individual
(D) Institution
(E) Passwords
Answer:
(C) Individual, (D) Institution

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 6.
Activity such as __________ or __________ and breaking security systems for illegal reasons is called cracking.
(A) Operating system
(B) Programming language
(C) Individual
(D) Decipher Codes
(E) Passwords
Answer:
(D) Decipher Codes, (E) Passwords

Question 7.
Phishing is a technique of extracting confidential information such as __________ and __________ combos by pretending as a legal enterprise.
(A) Computer
(B) Credit Card Numbers
(C) Network
(D) Username Password
(E) Programming Language
Answer:
(B) Credit Card Numbers, (D) Username Password

Question 8.
Hacking refers to unauthorised intrusion into a __________ or a __________
(A) Computer
(B) Credit Card Numbers
(C) Network
(D) Username Password
(E) Programming Language
Answer:
(A) Computer, (C) Network

Question 9.
Hacker is a person intensely interested in the deep and hidden or concealed working of any computer __________ and __________
(A) Operating system
(B) Programming language
(C) Individual
(D) Decipher Codes
(E) Passwords
Answer:
(A) Operating system, (B) Programming language

Question 10.
Encryption is a method of converting the __________ message into __________ text.
(A) Original
(B) Random
(C) Hardware
(D) Software
(E) Electronic communication
Answer:
(A) Original, (B) Random

Question 11.
Firewall refers to network security __________ and __________ system which blocks certain type of information.
(A) Original
(B) Random
(C) Hardware
(D) Software
(E) Electronic communication
Answer:
(C) Hardware, (D) Software

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 12.
Information Technology Act, 2000 is an Act to provide legal recognition for transaction carried out by means of __________ and other means of __________
(A) Original
(B) Random
(C) Hardware
(D) Electronic Data Interchange
(E) Electronic Communication
Answer:
(D) Electronic Data Interchange, (E) Electronic Communication

Question 13.
Information Technology Act, 2000 aims to provide the __________ infrastructure for __________ in India.
(A) Legal
(B) E-Commerce
(C) Original
(D) Random
(E) Hardware
Answer:
(A) Legal, (B) E-Commerce

5. Multiple Choice Three Correct Answers.

Question 1.
Cyberlaw is the area of law that deals with the Internet’s relationship to technological and electronic elements, including computers, __________, __________ and __________
(A) Software
(B) Internet
(C) Hardware
(D) Legal Issues
(E) Information Systems
Answer:
(A) Software, (C) Hardware, (E) Information Systems

Question 2.
Cyberlaw is the part of the overall legal system that deals with the __________, __________, and their respective __________
(A) Software
(B) Internet
(C) Cyberspace
(D) Legal Issues
(E) Information Systems
Answer:
(B) Internet, (C) Cyberspace, (D) Legal Issues

Question 3.
Cyber lawcoversafairlybroadarea,encompassingseveralsubtopicsincluding __________, access to and __________, and __________
(A) Freedom of expression
(B) Usage of the Internet
(C) Online privacy
(D) Electronic Crime
(E) Hi-Tech Crime
Answer:
(A) Freedom of expression, (B) Usage of the Internet, (C) Online privacy

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 4.
Computer Crime is alternatively referred to as __________, __________ or __________
(A) Freedom of expression
(B) Usage of the Internet
(C) Cybercrime
(D) Electronic Crime
(E) Hi-Tech Crime
Answer:
(C) Cybercrime, (D) Electronic Crime, (E) Hi-Tech Crime

Question 5.
Activity such as __________ codes or __________ and breaking __________ systems for illegal reasons is called cracking.
(A) Cyber Cafe
(B) Decipher
(C) Cyber Security
(D) Passwords
(E) Security
Answer:
(B) Decipher, (D) Passwords, (E) Security

Question 6.
The IT Act 2000 was amended in 2008 and 2011 and it includes rules for __________, __________, __________ by service provider, Audit of electronic document etc.
(A) Cyber Cafe
(B) Decipher
(C) Cyber Security
(D) Delivery of Services
(E) Security
Answer:
(A) Cyber Cafe, (C) Cyber Security, (D) Delivery of Services

6. Match the following.

Question 1.

Column ‘A’ Column ‘B’
(1) Ethics (A) Computer Crime
(2) Morals (B) Principles of right and wrong
(3) Cyber Crime (C) Moral philosophy
(4) Hacker (D) Illegally browses or steals a company’s or individual’s private information

Answer:

Column ‘A’ Column ‘B’
(1) Ethics (C) Moral philosophy
(2) Morals (B) Principles of right and wrong
(3) Cyber Crime (A) Computer Crime
(4) Hacker (D) Illegally browses or steals a company’s or individual’s private information

Question 2.

Column ‘A’ Column ‘B’
(1) Software Piracy (A) Legal right
(2) Unauthorized access (B) Gaining access without the users’ permission
(3) Authorization (C) Granting access rights to resources
(4) Copyright (D) Copyright violation of software

Answer:

Column ‘A’ Column ‘B’
(1) Software Piracy (D) Copyright violation of software
(2) Unauthorized access (B) Gaining access without the users’ permission
(3) Authorization (C) Granting access rights to resources
(4) Copyright (A) Legal right

Question 3.

Column ‘A’ Column ‘B’
(1) Copyright owners (A) Have the right to control the reproduction of their work
(2) Cracking (B) Will use a program or script known as a crack
(3) Cracker (C) Breaking security systems for illegal reasons
(4) Cyberstalking (D) Uses the Internet to systematically harass or threaten someone

Answer:

Column ‘A’ Column ‘B’
(1) Copyright owners (A) Have the right to control the reproduction of their work
(2) Cracking (C) Breaking security systems for illegal reasons
(3) Cracker (B) Will use a program or script known as a crack
(4) Cyberstalking (D) Uses the Internet to systematically harass or threaten someone

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 4.

Column ‘A’ Column ‘B’
(1) Phishing (A) Technique of extracting confidential information
(2) Plagiarism (B) Presenting someone else’s work or idea as your own without their consent
(3) Hacking (C) Safe and responsible use of information and communication technology
(4) Cyber Safety (D) Refers to unauthorized intrusion into a computer or a network

Answer:

Column ‘A’ Column ‘B’
(1) Phishing (A) Technique of extracting confidential information
(2) Plagiarism (B) Presenting someone else’s work or idea as your own without their consent
(3) Hacking (D) Refers to unauthorized intrusion into a computer or a network
(4) Cyber Safety (C) Safe and responsible use of information and communication technology

Question 5.

Column ‘A’ Column ‘B’
(1) Netiquette (A) Network Security
(2) Encryption (B) Method of converting the original message into random text
(3) Secure Socket Layer (C) Transmission of data is encrypted
(4) Firewall (D) Internet etiquette

Answer:

Column ‘A’ Column ‘B’
(1) Netiquette (D) Internet etiquette
(2) Encryption (B) Method of converting the original message into random text
(3) Secure Socket Layer (C) Transmission of data is encrypted
(4) Firewall (A) Network Security

Question 6.

Column ‘A’ Column ‘B’
(1) IT Act 2000 (A) Provide legal recognition for transactions carried out by means of electronic data interchange
(2) Digital Signature (B) Electronic Signature

Answer:

Column ‘A’ Column ‘B’
(1) IT Act 2000 (A) Provide legal recognition for transactions carried out by means of electronic data interchange
(2) Digital Signature (B) Electronic Signature

7. Answer Briefly.

Question 1.
What is Cyber Law?
Answer:

  1. Cyberlaw is the area of law that deals with the Internet’s relationship to technological and electronic elements.
  2. Cyberlaw is the part of the overall legal system that deals with the Internet, cyberspace, and their respective legal issues.
  3. Cyberlaw covers a fairly broad area, encompassing several subtopics including freedom of expression, access to and usage of the Internet, and online privacy

Question 2.
State the advantages of Cyber Law.
Answer:

  1. The Information Technology Act 2000 attempts to change outdated laws and provides ways to deal with cybercrimes.
  2. The Information Technology Act offers the much-needed legal framework so that information is not denied.
  3. The Information Technology Act has also proposed a legal framework for the authentication and origin of electronic records.

Question 3.
Define Ethics and Morals.
Answer:

  1. Ethics is also called moral philosophy is the discipline concerned with what is morally good and bad, right or wrong.
  2. Morals are the standards of behaviour; principles of right and wrong behaviour.
  3. Thus Morals are dictated by society, culture, or religion.

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 4.
What is Cyber Crime?
Answer:

  1. Computer Crime is alternatively referred to as Cybercrime.
  2. Computer crime is an act performed by a knowledgeable computer user, sometimes referred to as a Hacker.
  3. A Hacker who illegally browses or steals a company’s or individual’s private information.
  4. Cybercrimes can involve criminal activities that are traditional in nature, such as theft, fraud, forgery, defamation, and mischief.

Question 5.
State the examples of Cyber Crime.
Answer:

  1. Software piracy is nothing but copyright violation of software created originally by an individual or an institution.
  2. Software piracy includes stealing codes/programs and other information illegally and creating the imitated copy by unauthorized means.
  3. Gaining access without the users’ permission is known as unauthorized access.
  4. Authorization means granting access rights to resources.
  5. A copyright is a legal right that gives the creator of a literary, artistic, musical, or other creative work.
  6. Copyright owners have the right to control the reproduction of their work, including the right to receive the royalty payment for that reproduction.
  7. Activity such as decipher codes or passwords and breaking security systems for illegal reasons is called cracking.
  8. The cracker will use a program or script known as a Crack.
  9. Cyberstalking is a criminal practice where an individual uses the Internet to systematically harass or threaten someone.
  10. Phishing is a technique of extracting confidential information such as credit card numbers and username password combos by pretending as a legal enterprise.
  11. Phishing is typically carried out by email spoofing.
  12. Plagiarism is presenting someone else’s work or idea as your own without their consent.
  13. Hacking refers to unauthorized intrusion into a computer or a network.
  14. Hacker is a person intensely interested in the deep and hidden or concealed working of any computer operating system and programming language.

Question 6.
Explain Security Procedures.
Answer:

  1. Encryption is a method of converting the original message into random text.
  2. The full form of SSL is Secure Socket Layer.
  3. Through the Secure Socket Layer, the transmission of data is encrypted.
  4. A firewall refers to a network security (Hardware and Software) system which blocks certain types of information.

Maharashtra Board Class 11 Information Technology Important Questions Chapter 4 Cyber Law

Question 7.
Discuss the IT Act of India 2000 and its features.
Answer:

  1. Information Technology Act, 2000 is an Act to provide legal recognition for transactions carried out by means of electronic data interchange and other means of electronic communication.
  2. Information Technology Act, 2000 aims to provide the legal infrastructure for E-Commerce in India.
  3. Cyber Laws have a major impact on e-businesses and the new economy in India.
  4. The Information Technology Act, 2000 also aims to provide for the legal framework.
  5. Digital signature has been replaced with electronic signature to make it a more technology-neutral act.