Question

Notify Me Bookmark

Sulfur trioxide is known to be planar with all the oxygen atoms equidistant from the central sulfur atom. On the basis of these facts, which of the following statements are CORRECT? 1. It can be represented by three equivalent resonance structures. 2. The sulfur atom is sp2 hybridized. The molecule is nonpolar. 1 only 1, 2, and 3 2 only 3 only 1 and 2

Sulfur trioxide is known to be planar with all the oxygen atoms equidistant from the central sulfur atom. On the basis of these facts, which of the following statements are CORRECT? 1. It can be represented by three equivalent resonance structures. 2. The sulfur atom is sp2 hybridized. The molecule is nonpolar. 1 only 1, 2, and 3 2 only 3 only 1 and 2

Image text
Sulfur trioxide is known to be planar with all the oxygen atoms equidistant from the central sulfur atom. On the basis of these facts, which of the following statements are CORRECT?
  1. It can be represented by three equivalent resonance structures.
  2. The sulfur atom is s p 2 hybridized.
  3. The molecule is nonpolar.
    1 only
    1,2 , and 3
    2 only
    3 only
    1 and 2

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

Which of the following statements are CORRECT? Ionic bonds form when one or more valence electrons are transferred from one atom to another. Covalent bonds involve sharing of electrons between atoms. Ionic bond formation is always exothermic, covalent bond formation is always endothermic. 1 and 2 1,2 , and 3 3 only 1 only 2 only

Solution
0 0

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn
Solution
3 0

Design Case: The I-V characteristics of a NMOS device is to be studied. Especially, the relationships between IDS, VGS, and VDS are to be found. Find the I-V characteristics of a NMOS with size W/L = 2u/0.4u by performing the following simulations: 1. Run a DC simulation and demonstrate IDS curve when VGS is swept from 0 V to 2 V. > Set VDS=1 V. > Plot the VGS VS. IDS curve. > Q1.1: What is the threshold voltage (VTH) of the given transistor? Q1.2: Explain how gm can be calculated from the VGS Vs. IDS curve. 2. Run a DC simulation and show IDS curve when VDS is swept from 0V to 2 V. > Set VGS = 700mV. > Plot VDS VS. IDS Curve. > Q2.1: At what VDS does the transistor transit from linear region to saturation region? > Q2.2: Does the given device suffer from channel length modulation effect? Q Q2.3: If it suffers from channel length modulation, how can it be fixed? Q2.4: Explain how ro can be calculated from the VDS vs. IDS curve. * For all design sessions, use nfet33 for NMOS and pfet33 for PMOS from "cmrf8sf" library.

Design Case: The I-V characteristics of a NMOS device is to be studied. Especially, the relationships between IDS, VGS, and VDS are to be found. Find the I-V characteristics of a NMOS with size W/L = 2u/0.4u by performing the following simulations: 1. Run a DC simulation and demonstrate IDS curve when VGS is swept from 0 V to 2 V. > Set VDS=1 V. > Plot the VGS VS. IDS curve. > Q1.1: What is the threshold voltage (VTH) of the given transistor? Q1.2: Explain how gm can be calculated from the VGS Vs. IDS curve. 2. Run a DC simulation and show IDS curve when VDS is swept from 0V to 2 V. > Set VGS = 700mV. > Plot VDS VS. IDS Curve. > Q2.1: At what VDS does the transistor transit from linear region to saturation region? > Q2.2: Does the given device suffer from channel length modulation effect? Q Q2.3: If it suffers from channel length modulation, how can it be fixed? Q2.4: Explain how ro can be calculated from the VDS vs. IDS curve. * For all design sessions, use nfet33 for NMOS and pfet33 for PMOS from "cmrf8sf" library.
Solution
0 0