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andrew11 [14]
3 years ago
10

Describe the electron distribution in a polar-covalent bond and it's effect on the partial charges of the compound?

Chemistry
1 answer:
Artemon [7]3 years ago
5 0

Answer:

In polar Covalent bonds, the electrons which are in bonded shifts towards an atom which has more valance electrons.

<u>Explanation:</u>

We know if an atom takes the electron it acquires a negative charge whereas if it gives an electron it acquires a positive charge in the ionic bond. But here we are talking about covalent bonds. Covalent bonds are those in which atoms share the electron instead of completely giving off the electron. If the atoms are identical in case of covalent bond that is 2 hydrogen atoms then this type of bonding is called pure covalent bonds but if the atoms linked in covalent bonds are different then it is called polar covalent bonds.

In this, the bonding electrons will shift towards an atom which has more valence electron thereby acquiring the partial negative charges and the other atom will acquire a partial positive charge. For example, HCl. In this the Chlorine atom is having more valence electron than hydrogen atom, and hence Chlorine atom has a partial negative charge and Hydrogen atom has a partial positive charge.

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Using a 300 MHz NMR instrument:
Softa [21]

Answer:

a. 750Hz, b. 4.0ppm, c. 600Hz

Explanation:

The Downfield Shift (Hz) is given by the formula

Downfield Shift (Hz) = Chemical Shift (ppm) x Spectrometer Frequency (Hz)

Using the above formula we can solve all three parts easily

a. fspec = 300 MHz, Chem. Shift = 2.5ppm, 1MHz = 10⁶ Hz, 1ppm (parts per million) = 10⁻⁶

Downfield Shift (Hz) = 2.5ppm x 300MHz x (1Hz/10⁶MHz) x (10⁻⁶/1ppm)

Downfield Shift = 750 Hz

The signal is at 750Hz Downfield from TMS

b. Downfield Shift = 1200 Hz, Chemical Shift = ?

Chemical Shift = Downfield shift/Spectrometer Frequency

Chemical Shift = (1200Hz/300MHz) x (1ppm/10⁻⁶) = 4.0 ppm

The signal comes at 4.0 ppm

c. Separation of 2ppm, Downfield Shift = ?

Downfield Shift (Hz) = 2(ppm) x 300 (MHz) x (1Hz/10⁶MHz) x (10⁻⁶/1ppm) = 600 Hz

The two peaks are separated by 600Hz

6 0
2 years ago
The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
Rus_ich [418]

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}

\log\frac{0.25}{a-x}=0.15

\frac{0.25}{a-x}=1.41

(a-x)=0.17M

b) for concentration to decrease from 0.25M to 0.15M

t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

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What are some physical properties of metal
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Calculate the energy of a quantum of radiant energy with a frequency of 5.00x1011/s
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