<\/span><\/h3>\nGroup 14 in the periodic table consists of elements like Carbon, Silicon, Germanium, Tin, and Lead.<\/p>\n
Carbon is an abundant element that exists in different forms, including coal, diamond, and graphite.<\/p>\n
All the elements that form Group 14 are solids, and carbon is no exception to this rule. It has a small atomic size and high electronegativity.<\/p>\n
Carbon exists in two forms, namely, amorphous and crystalline. Diamond and graphite are crystalline forms of carbon and find various uses, such as for the sharpening of tools used in the manufacture of dyes. It also finds applications in making tungsten filaments.<\/p>\n
Graphite is a soft and slippery element that finds uses in lubricating machines that run at high temperatures.<\/p>\n
Other forms of carbon include wood charcoal and animal charcoal.<\/p>\n
Carbon has hundreds of other uses. For instance, activated charcoal can absorb poisonous gases. So, it finds applications in gas masks. It also finds uses in the decolorization of sugar.<\/p>\n
Likewise, using carbon, you can produce black ink. Carbon also finds applications in the manufacture of automobile tyres.<\/p>\n
Coke, a form of carbon, finds uses in metallurgical processes as a reducing agent.<\/p>\n
Carbon can form hundreds of different compounds when mixed with other substances.<\/p>\n
For instance, when mixed with Oxygen, Carbon forms two prominent compounds that you can find in everyday life, namely, carbon monoxide and carbon dioxide. Both these compounds differ drastically in terms of properties. While carbon monoxide is highly poisonous, carbon dioxide is less toxic. Similarly, carbon monoxide is highly porous and insoluble in water, while carbon dioxide is slightly soluble in water and forms carbonic acid when it dissolves.<\/p>\n
<\/span>Benefits of NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements<\/strong><\/span><\/h2>\nThere are hundreds of benefits of studying NCERT Solutions for Class 11 Chemistry Chapter 11, which is why those who aspire to come out on top prefer them. Listed below are a few benefits of preparing for NCERT Solutions for Class 11 Chemistry Chapter 11.<\/p>\n
<\/span>1. <\/strong>Prepared by Experts<\/strong><\/span><\/h3>\nTeachers with extensive knowledge prepare NCERT solutions. They have only one goal in mind \u2013 To provide students of class 11 with quality educational content. So, NCERT solutions for Chemistry are concise as well as detailed and offer students with enough information to crack the exams.<\/p>\n
<\/span>2. <\/strong>NCERT Solutions Enable Students to Test their Knowledge<\/strong><\/span><\/h3>\nNCERT Solutions for Class 11 Chemistry Chapter 11 come with questions at the end of each chapter. So, it allows students to evaluate their knowledge of the different concepts that appear in the chapter.<\/p>\n
<\/span>3. <\/strong>NCERT Solutions Function as Revision Materials<\/strong><\/span><\/h3>\nNCERT solutions also help students to revise each concept in the subject before the exams begin. So, by going over the solutions, you can ensure that you answer all the questions that appear in the exams.<\/p>\n
So, teachers should recommend to students to go through NCERT solutions for Chemistry so that they are thorough with all the concepts. The solutions are also a means by which you can answer questions that appear at the end of each chapter and assess your knowledge. Additionally, they also serve as reference materials and help clear any doubts that a student may have. So, instead of going to their teachers, students can refer to NCERT solutions and get all their doubts cleared.<\/p>\n
<\/span>Access NCERT Solutions For Class 11 Chemistry Chapter 11<\/span><\/h2>\n<\/span>Question 1. Discuss the pattern of variation in the oxidation states of<\/strong>
(i) B to Tl (ii) C to Pb.<\/strong><\/span><\/h3>\nAnswer: (i) B to Tl<\/strong>
Common oxidation states are +1 and +3. The stability of +3 oxidation state decreases from B to Tl. +1 oxidation state increases from B to Tl.<\/p>\n(ii) C to Pb<\/strong>
The common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.
Details can be seen from the text part.<\/p>\n<\/span>Question 2. How can you explain higher stability of BCl3 as compared to TlCl3<\/sub>?<\/strong><\/span><\/h3>\nAnswer:<\/strong>\u00a0BCl3<\/sub>\u00a0is quite stable. Because there is absence of d- and f-electrons in boron three valence electrons (2s2<\/sup>\u00a02px1<\/sub>) are there for bonding with chlorine atom. In Tl the valence s-electron (6s2<\/sup>) are experiencing maximum inert pair effect. Thus, only 6p1<\/sup>\u00a0electron is available for bonding. Therefore, BCl3<\/sub>\u00a0is stable but TlCl3<\/sub>\u00a0is comparatively unstable.<\/p>\n<\/span>Question 3. Why does borontrifluori.de behave as a Lewis acid?<\/strong><\/span><\/h3>\nAnswer:<\/strong>\u00a0In BF3<\/sub>, central atom has only six electrons after sharing with the electrons of the F atoms. It is an electron-deficient compound and thus behaves as a Lewis acid.<\/p>\n<\/span>Question 4. Consider the compounds, BCl3,<\/sub>\u00a0and CCl4<\/sub>. How will they\u00a0 \u00a0behave with water justify?<\/strong><\/span><\/h3>\nAnswer:\u00a0<\/strong>In BCl3<\/sub>, there is only six electrons in the valence shell of B atom. Thus, the octet is incomplete and it can accept a pair of electrons from water and hence BCl3<\/sub>\u00a0undergoes hydrolysis. Whereas, in CCl4<\/sub>, C atom has 8 electrons and its octet is complete. That\u2019s why it has no tendency to react with water.
CCl4<\/sub>\u00a0+ H2<\/sub>0 \u2014\u2014\u2014\u2014\u2014> No reaction<\/p>\n<\/span>Question 5. Is boric acid a protonic acid? Explain.<\/strong><\/span><\/h3>\nAnswer:\u00a0<\/strong>Boric acid is a Lewis acid, it is not a protonic acid.
Boric acid accepts electrons from hydroxyl ion of H2<\/sub>O molecule.
B (OH)3<\/sub>\u00a0+ 2HOH \u2014\u2014\u2014\u2014> [B (OH)4<\/sub>]\u2013<\/sup>\u00a0+ H3<\/sub>O+<\/sup><\/p>\n<\/span>Question 6. Explain what happens when boric acid is heated.<\/strong><\/span><\/h3>\nAnswer:\u00a0<\/strong>On heating boric acid above 370 K, it forms metabolic acid, HB02 which on further heating yields boric oxide B2<\/sub>O3<\/sub>.
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-11-The-p-Block-Elements-Q6.png\")
<\/p>\n<\/span>Question 7. Describe the shapes of BF3<\/sub>\u00a0and BH4<\/sub>\u2013<\/sup>. Assign the hybridisation of boron in these species.<\/strong><\/span><\/h3>\nAnswer:\u00a0<\/strong>In BF3<\/sub>, boron is SP2<\/sup>\u00a0hybridized.
\u2234 shape of BF3<\/sub>\u00a0= planar.
In [BH4<\/sub>]\u2013<\/sup>, boron is sp3 hybridized, thus the shape is tetrahedral.
![\"NCERT](\"https:\/\/www.learncbse.in\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-for-Class-11-Chemistry-Chapter-11-The-p-Block-Elements-Q7.png\")
<\/p>\n<\/span>Question 8. Write reactions to justify amphoteric nature of aluminium.<\/strong><\/span><\/h3>\nAnswer:\u00a0<\/strong>Aluminium reacts with acid as well as base. This shows amphoteric nature of aluminium.
2Al(s) + 6HCl(dil.) \u2014\u2014> 2AlCl3<\/sub>(aq) + 3H2<\/sub>(g)
2Al(s) + 2NaOH(aq) + 6H2<\/sub>O(l) \u2014\u2014\u2014-> 2Na+ [Al(OH)4<\/sub>]\u2013<\/sup>(aq) + 3H
![\"NCERT](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2020\/09\/NCERT-Solutions-For-Class-11-Chemistry-Chapter-11-The-p-Block-Elements-1.jpg\")
<\/p>\n