Classification of Elements and Periodicity in Prop
Get insights from 123 questions on Classification of Elements and Periodicity in Prop, answered by students, alumni, and experts. You may also ask and answer any question you like about Classification of Elements and Periodicity in Prop
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New question posted
5 months agoNew answer posted
5 months agoContributor-Level 10
3.42. (a) Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have an incrementally increasing number of electrons from left to right, and, therefore, have different valences.
New question posted
5 months agoNew question posted
5 months agoNew answer posted
5 months agoContributor-Level 10
3.40. Within a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is: F > O > Cl > N, i.e., option (b) is correct.
New answer posted
5 months agoContributor-Level 10
3.39. In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C > B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F > N > C > B > Si, i.e., option (c) is correct.
New answer posted
5 months agoContributor-Level 10
3.38. In a period, metallic character decreases as we move from left to right. Therefore, metallic character of K, Mg and Al decreases in the order: K > Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is: K > Mg > Al > B, i.e., option (d) is correct
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers