(a)
The ionic charge for the given element chlorine
( Cl )
(a)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Chlorine belongs to the group 7A. The elements of group 7A have one electron less than the noble configuration thus tends to gain one electron and complete their octet.
The elements of group 7A gain one electron and form an anion with charge
Conclusion:
Therefore, the ionic charge for chlorine
(b)
The ionic charge for the given element lithium
( Li )
(b)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Lithium belongs to the group 1A. The elements of group 1A have one electron more than the noble configuration thus tends to lose one electron and complete their octet.
The elements of group 1A lose one electron and form a cation with charge
Conclusion:
Therefore, the ionic charge for lithium
(c)
The ionic charge for the given element selenium
( Se )
(c)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Selenium belongs to the group 6A. The elements of group 6A have two electrons less than the noble configuration thus tends to gain two electrons and complete their octet.
The elements of group 6A gain two electrons and form an anion with charge
Conclusion:
Therefore, the ionic charge for selenium
(d)
The ionic charge for the given element strontium
( Sr )
(d)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Strontium belongs to the group 2A. The elements of group 2A have two electrons more than the noble configuration thus tends to lose two electrons and complete their octet.
The elements of group 2A lose two electrons and form a cation with charge
Conclusion:
Therefore, the ionic charge for strontium
(e)
The ionic charge for the given element gallium
( Ga )
(e)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Gallium belongs to the group 3A. The elements of group 3A have three electrons more than the noble configuration thus tends to lose three electrons and complete their octet.
The elements of group 3A lose three electrons and form a cation with charge
Conclusion:
Therefore, the ionic charge for gallium
(f)
The ionic charge for the given element phosphorous
( P )
(f)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Phosphorous belongs to the group 5A. The elements of group 5A have three electrons less than the noble configuration thus tends to gain 3 electrons and complete their octet.
The elements of group 5A gain 3 electrons and form an anion with charge
Conclusion:
Therefore, the ionic charge for phosphorous
(g)
The ionic charge for the given element radon
( Rn )
(g)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Radon belongs to the group 8A. The elements of group 8A are called noble gas elements. They have completely filled octet and are very stable. They do not gain or lose electrons because they are already very stable.
Thus, the elements of group 8A have an ionic charge of
Conclusion:
Therefore, the ionic charge for radon
(h)
The ionic charge for the given element silicon
( Si )
(h)
Answer to Problem 14E
Explanation of Solution
Given Info: Refer to the periodic table figure 11.24.
Explanation:
Silicon belongs to the group 4A. Elements of group 4A have 4 valence electrons and 4 electrons less than noble configuration. Elements of group 4A do not form ions generally because it is very difficult to gain or lose 4 electrons to complete the octet. Thus, the elements of group 4A have an ionic charge of
Conclusion:
Therefore, the ionic charge for silicon
Want to see more full solutions like this?
Chapter 12 Solutions
INTRO TO PHYSICAL SCIENCE W/MINDTAP
- Find the total capacitance in micro farads of the combination of capacitors shown in the figure below. 2.01 0.30 µF 2.5 µF 10 μF × HFarrow_forwardI do not understand the process to answer the second part of question b. Please help me understand how to get there!arrow_forwardRank the six combinations of electric charges on the basis of the electric force acting on 91. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them. ▸ View Available Hint(s) [most negative 91 = +1nC 92 = +1nC 91 = -1nC 93 = +1nC 92- +1nC 93 = +1nC -1nC 92- -1nC 93- -1nC 91= +1nC 92 = +1nC 93=-1nC 91 +1nC 92=-1nC 93=-1nC 91 = +1nC 2 = −1nC 93 = +1nC The correct ranking cannot be determined. Reset Help most positivearrow_forward
- Part A Find the x-component of the electric field at the origin, point O. Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz = Η ΑΣΦ ? N/C Submit Part B Now, assume that charge q2 is negative; q2 = -6 nC, as shown in (Figure 2). What is the x-component of the net electric field at the origin, point O? Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz= Η ΑΣΦ ? N/Carrow_forward1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm. Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from the center of the sphere. (a) = = (b) E = (c)Ẻ = = NC NC NCarrow_forward1. A long silver rod of radius 3.5 cm has a charge of -3.9 ис on its surface. Here ŕ is a unit vector ст directed perpendicularly away from the axis of the rod as shown in the figure. (a) Find the electric field at a point 5 cm from the center of the rod (an outside point). E = N C (b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point) E=0 Think & Prepare N C 1. Is there a symmetry in the charge distribution? What kind of symmetry? 2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ from a?arrow_forward
- 1. Determine the electric flux through each surface whose cross-section is shown below. 55 S₂ -29 S5 SA S3 + 9 Enter your answer in terms of q and ε Φ (a) s₁ (b) s₂ = -29 (C) Φ զ Ερ (d) SA = (e) $5 (f) Sa $6 = II ✓ -29 S6 +39arrow_forwardNo chatgpt pls will upvotearrow_forwardthe cable may break and cause severe injury. cable is more likely to break as compared to the [1] ds, inclined at angles of 30° and 50° to the vertical rings by way of a scaled diagram. [4] I 30° T₁ 3cm 3.8T2 cm 200 N 50° at it is headed due North and its airspeed indicat 240 km/h. If there is a wind of 100 km/h from We e relative to the Earth? [3]arrow_forward
- Can you explain this using nodal analysis With the nodes I have present And then show me how many KCL equations I need to write, I’m thinking 2 since we have 2 dependent sourcesarrow_forwardstate the difference between vector and scalar quarrow_forwardPlease don't use Chatgpt will upvote and give handwritten solutionarrow_forward
An Introduction to Physical SciencePhysicsISBN:9781305079137Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar TorresPublisher:Cengage Learning
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill