Modern periodic table class 11 chemistry chapter 7 exercise solutions

Modern periodic table class 11 chemistry chapter 7 exercise solutions
modern periodic table class 11
modern periodic table class 11 exercise
class 11 chemistry chapter 7 modern periodic table exercise solutions
modern periodic table exercise solutions
class 11 chemistry exercise chapter 7 modern periodic table
maharashtra board class 11 chemistry chapter 7
class 11 chemistry chapter 7 exercise solutions maharashtra board
Class 11
Chemistry
7. Modern Periodic Table
Exercise solutions
By: solutions made easy
11th standard Chemistry ch. 7 Modern periodic Table Exercise By Study With Vishal
Modern periodic table 11 Class 7.Chapter Chemistry Exercise (Science) Question Answer HSC board
Exercise solving of Modern Periodic Table Question No. 1 to 4

1. Explain the following
A. The elements Li, B, C, Be and N have the electronegativities 1.0, 2.0, 1.5 and 3.0, respectively on the Pauling scale.
Answer:
• Li, B, Be and N belong to the same period.
• As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily and therefore, electronegativity increases.
Hence, the elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.
Please share this video video your friends
B. The atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.
Answer:
• Cl, I and Br belong to group 17 (halogen group) in the periodic table.
• As we move down the group from top to bottom in the periodic table, a new shell gets added in the atom of the elements.
• As a result, the effective nuclear charge decreases due to increase in the atomic size as well as increased shielding effect.
• Therefore, the valence electrons experience less attractive force from the nucleus and are held less tightly resulting in the increased atomic radius.
• Thus, their atomic radii increases in the following order down the group.
Solutions made easy
C. The ionic radii of F and Na⊕ are 133 and 98 pm, respectively.
Answer:
• F– and Na+ are isoelectronic ions as they both have 10 electrons.
• However, the nuclear charge on F– is +9 while that of Na+ is +11.
• In isoelectronic species, larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.
Thus, F– has larger ionic radii (133 pm) than Na+ (98 pm).
Please share this video video your friends
D. 13Al is a metal, 14Si is a metalloid and
15P is a nonmetal.

#solutions_made_easy #maharashtrastateboard
#ashish_madhwai #answers #solutions
#practical # maharashtraboard # class11 #class12

Check out my Instagram profile and follow for direct messages.🤗

join telegram group