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4vir4ik [10]
3 years ago
15

.If the atoms that share electrons have an unequal attraction for the electrons, the bond is calleda. nonpolar. c. ionic.b. pola

r. d. dipolar
Physics
1 answer:
Fittoniya [83]3 years ago
3 0

Answer:

The right option is the 'polar' option.

Explanation:

- The right answer cannot be ionic as sharing of electrons is mentioned. Ionic bonds don't involve the sharing of electrons

- Polarity in terms of bonding is used to describe the situation when one of the atoms involved in the sharing of electrons is more electronegative than the other and pulls the shared electrons more towards itself, thereby inducing a slight positive charge on the other atom and gaining a slight negative charge itself.

- In the absence of this kind of scenario in a covalent bonding that sharing occurs, then it is described as non polar. The two atoms that participate in the sharing of electrons here have similar electronegativities and thereby leave the shared electrons at the middle.

- Dipolar compound is used to refer to organic molecules that are essentially neutral but carry a positive & negative charge in one of their various existence forms.

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What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
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Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

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